A4
====it works with
any graphicsNotebookWrite[SelectedNotebook[],
GridBox[{Cell[GraphicsData[PostScript, DisplayString[#,
MPS]], Graphics] & /@ {leftGraphics, rightGraphics}}]] Jens>
(1) Is there a way in Mathematica 4.2 to put two separate
gifs into a> single, cell side by side (with some intervening
space), without> having to combine them in some graphics
program Ūrst?> In particular, Išd like to do that within a
text cell.> Even in a new Input cell, if I Ūrst create a
GridBox (via Inut>Create> Table/Matrix/Palette) and then try
to insert the Ūrst gif (via> Edit>Insert Object>Create from
File ....), Mathematica promptly> crashes. (Mathematica 4.2
under Windows 2000.)> I just donšt see how to get anything
other than a single gif into a> cell.> (2) Is there a way to
cause a gif imported into a Mathematica 4.2> notebook to
become a hyperlink -- so that when the user clicks on the>
gif the hyperlinkšs target is summoned?> (My aim in all this
is to use Mathematica to create web pages with> high
mathematical content -- saving the notebook as HTML+MathML
--> without having to do any extensive editing of the
resulting .xml and> related Ūles. That way as the source
Mathematica notebook changes, I> would need only to re-export
without further tinkering with the .xml> Ūle, etc.)> -->
Murray Eisenberg Internet: murray@math.umass.edu> Mathematics
& Statistics Dept. Voice: 413-545-2859 (W)> University of
Massachusetts 413-549-1020 (H)Reply-To:
kuska@informatik.uni-leipzig.de====set the PlotRange explicit
PlotRange->{{0,largeX},Automatic} Jens> I am using
MultipleListPlot to plot a range of 2D graphs and have found
that> the last graph, which has a much greater (by a factor
of 4) x range than the> others does not plot completely but
is truncated in the x direction. By> removing the Frame, I
can see the points where these lie outside the Frame> but
these are not joined etc. How do I get over this?> -> Malcolm
WoodruffReply-To: kuska@informatik.uni-leipzig.de====what may
be in the *.m Ūles? Mathematica commands ? thatyou can load
into the kernel ? Makr all your input cells in the notebook
as initialization,save it as an *.m Ūle, add a Quit[] as the
last commandto the *.m Ūle and run it withmath <<
yourJustGeneratedMFile JensRegardds> Probably a rather simple
question: what is the easiest way to open a> notebook, execute
it, then quit, from the command line? I would like to be> able
to do this from a MakeŪle.> SidneyReply-To:
kuska@informatik.uni-leipzig.de====do you realy think
that1.07577/( 1+7.12336*10^-7 (1. + z)^2 )with the z inside
is numerical ? or do you meanF[z_?NumericQ]: =
NIntegrate[f[y,z], {y, 0, InŪnity}] Jens> Išve this problem
with NIntegrate:>
**********************************************>
In[1]=f[y_,z_]: = (y^4/( (1+(8.44*10^-4)^2 * (1+z)^2 y^2)
(Exp[z]+1) )> In[2]=F[z_]: = NIntegrate[f[y,z], {z, 0,
InŪnity}> **********************************************>
Mathematica 4.0 says:>
*************************************************************
***************> *> NIntegrate: : inum : Integrand 1.07577/(
1+7.12336*10^-7 (1. + z)^2 ) is not> numerical at {y}={1.}.>
*************************************************************
***************> **> What is it??> --> Rob_jack====>> do you
realy think that>> 1.07577/( 1+7.12336*10^-7 (1. + z)^2 )>>
with the z inside is numerical ? or do you mean>>
F[z_?NumericQ]: = NIntegrate[f[y,z], {y, 0, InŪnity}]>Išm a
newbie, bat I mean:*************************In[1]=
NumericQ[z]Out[1]=False*************************BTW, z is a
real number.--Rob_jackReply-To:
kuska@informatik.uni-leipzig.de====not with Mathematica 4.2.
But the problem can be the page widththat may be used to
convert the Cell[], if you can you shoulduse the
Graphics[]/Graphics3D[] .. inside the Cell orthe PostScript
string in the GraphicsData[] Jens> The following is a button
that will export a selected graphic as a JPEG,>
ButtonBox[JPEG,> ButtonFunction :>> Module[{sel},>
SelectionMove[InputNotebook[], All, Cell];> sel =
NotebookRead[InputNotebook[]];> If[Head[sel] === Cell,>
If[!ValueQ[dpi], dpi = Automatic];> ImageSize -> Automatic,
ImageResolution -> dpi,> ConversionOptions -> {Quality ->
75}]]],> ButtonEvaluator -> Automatic, Active ->
True]//DisplayForm> It works Ūne as long as the graphic is no
wider than about 330 pixels.> If the graphic is wider than 330
pixels, then the size of exported> graphic is correct, but
everything on the right beyond the Ūrst 330> pixels is blank.
There seems to be no problem with the height though.> Any
ideas on how to Ūx or get around this problem?> ---> Selwyn
Hollis====>-----Original Message----->Sent: Wednesday,
September 11, 2002 9:28 AM>>A commonly used symbol for the
Floor function is a square >bracket with the>upper indents
removed. Is this symbol part of Mathematicašs >built-in
symbols? I>think not, so then, can it be constructed by
hand.>>Jack>>Jack, yes it is already built-in! Bring up the
palette with menu: File > Palettes> CompleteCharachters;
there is a section Operators > General, where youšllŪnd what
you want.Alternatively type Œescš l f Œescš <your expression>
Œescš r f Œescš, orinstead of the esc-sequence use the
corresponding mark-ups. For output tryTraditionalForm. If you
donšt like that for all of your output, you may justadd a
formatting rule for Floor:In[7]:=
Unprotect[Floor]In[8]:=Floor /: MakeBoxes[Floor[expr_],
StandardForm] := RowBox[{[LeftFloor], MakeBoxes[expr,
StandardForm], [RightFloor]}]In[9]:= Protect[Floor]In[11]:=
Floor /@ ([Pi] + [Lambda])Out[11]= 3 +
[LeftFloor][Lambda][RightFloor]Same thing with Ceiling,
BTW.--HartmutReply-To: jmt@dxdydz.net====mathematica
-primaryModiŪerMask etcsee man mathematica for other (very
useful) options> PS: One more quick one: Why does the front
end act funny when I have the> Your X server is set up so
that the NumLock key is mapped to Mod2. I> learned by
accident that Mod2 is actually quite useful. Mod2-click on a>
cell selects all cells of that type in the current notebook.
This is an> easy way to delete all the graphics cells and
output cells in a notebook to> reduce Ūle size. An annoying
aspect is that Mod2-click means press> NumLock, click, then
press NumLock again to turn it off. You should be able> to
map a different key to Mod2 using xmodmap. Remapping modiŪer
keys in X> is awfully annoying, though. I recommend xkeycaps>
(http://www.jwz.org/xkeycaps/).> Reply-To:
kuska@informatik.uni-leipzig.de====no because the PostScript
creation know nothingabout the notebook frontend. Mathematica
does itsgraphics in PostScript since version 1.0 but the
notebook frontend was introduced in version 2.x andthe actual
typsetting/markup format was introducedin version 3.0. Jens>
Is it possible to conŪgure a style sheet so that functions or
data shown in> graphs are coloured (colored :)) in working but
black in printing...without> actually re-executing the
function that created the graphics?> thanks>
Mike====>-----Original Message----->Sent: Wednesday,
September 11, 2002 9:28 AM>Is it possible to conŪgure a style
sheet so that functions or >data shown in>graphs are coloured
(colored :)) in working but black in
>printing...without>actually re-executing the function that
created the graphics?>>thanks>>Mike>>Mike, if for some reason
you canšt conŪgure your color printer when printing, youmay
rerender your graphics in Mathematica with different options,
suchavoiding costly recalculation. You may either use
something special, asin...In[1]:=ContourPlot[Sin[x]^2
Sin[y]^2, {x, 0, Pi}, {y, 0, Pi}, ColorFunction -> Hue,
ContourStyle -> GrayLevel[1], PlotPoints ->
100]Out[1]=[SkeletonIndicator]ContourGraphics[
SkeletonIndicator]In[2]:=Show[%, ColorFunction ->
(GrayLevel[1 - #] &)]...using a B/W color function. Or in
general you may use the
optionColorOutput:In[3]:=Plot3D[Sin[x]^2 Sin[y]^2, {x, 0,
Pi}, {y, 0,
Pi}]Out[3]=[SkeletonIndicator]SurfaceGraphics[
SkeletonIndicator]...In[5]:=Show[%%, ColorOutput ->
GrayLevel]--Hartmut====> This graphs f from 0 to 8 Pi on a
logarithmic horizontal scale:>> f[x_] := 1 + Abs[Sin[x]/x]>
Plot[f@Exp@x, {x, 0, Log[8Pi]}, PlotRange -> All];The graph
itself is indeed what is desired, but the labelling of the
x-axisis then incorrect for a logarithmic scale. For example,
regardless of thescale used, any proper graph must show that
the Ūrst minimum occurs atx = Pi. But the method you suggest
makes it seem that it occurs at approx.1.14 (precisely
Log[Pi]).David> -----Original Message----->Reply-To:
kuska@informatik.uni-leipzig.de====Needs[Graphics`Graphics`]
LogPlot[Exp[-x],{x,0,10}]?? Jens> ====Išm trying to write a
fast empirical cummulative distribution function(CDF).
Empirical CDFs are step functions that can be expressed
interms of a Which statement. For example, given the list
ofobservations {1, 2, 3},f = Which[# < 1, 0, # < 2, 1/3, # <
3, 2/3, True, 1]&is the empirical CDF. Note that f /@ {1, 2,
3} returns {1/3, 2/3, 1}and f is continuous from the
right.When the number of observations is large, the Which
statementevaluates fairly slowly (even if it has been
Compiled). SinceInterpolationFunction evaluates so much
faster in general, Išve triedto use Interpolation with
InterpolationOrder -> 0. The problem is thatthe resulting
InterpolatingFunction doesnšt behave the way (I think)it
ought to. For example, letg = Interpolation[{{1, 1/3}, {2,
2/3}, {3, 1}}, InterpolationOrder ->0]Then, g /@ {1, 2, 3}
returns {2/3, 2/3, 1} instead of {1/3, 2/3, 1}.In addition, g
is continuous from the left rather than from the
right.Obviously I am not aware of the considerations that
went intodetermining the behavior of InterpolationFunction
whenInterpolationOrder -> 0.So I have two questions: (1) Does
anyone have any opinions about how InterpolatingFunctionought
to behave with InterpolationOrder -> 0?(2) Does anyone have a
faster way to evaluate an empirical CDF than acompiled Which
function?By the way, herešs my current
version:CompileEmpiricalCDF[list_?(VectorQ[#, NumericQ] &)]
:= Block[{x}, Compile[{{x, _Real}}, Evaluate[ Which @@
Flatten[ Append[ Transpose[{ Thread[x < Sort[list]], Range[0,
1 - 1/#, 1/#] & @ Length[list] }], {True, 1}]]
]]]--Mark====Daniel Lichtblau made two suggestions that allow
one to useInterpolation the way I wanted. First, to make the
resulting functionright-continuous, change the sign twice.
Second, to make the end pointreturn the correct value, add an
extra (phantom) observation (with anextra (irrelevant) value).
(The phantom observation is made at thehigh end because of the
sign reversals.) Herešs the code I cooked upbased on his
suggestions:MakeEmpiricalCDF::usage = MakeEmpiricalCDF[list]
returns a functionthat evaluates the empirical CDF given the
observations in the list.The function is deŪned on the entire
real line.MakeEmpiricalCDF[list_?(VectorQ[#, NumericQ]&)] :=
Module[{n, s, a, r, idata}, n = Length[list]; s = Sort[list];
a = Append[s, s[[-1]] + 1]; (* phantom obs. *) r = Range[1/n,
1 + 1/n, 1/n]; (* phantom value 1 + 1/n *) idata = Last /@
Split[Transpose[{-a, r}], #1[[1]] == #2[[1]]&]; (* -a is the
Ūrst sign change *) Block[{x}, Function @@ {x, Which @@ { x <
s[[ 1]], 0., x > s[[-1]], 1., True, Interpolation[idata,
InterpolationOrder -> 0][-x] (* -x is the second sign change
*) }}] ]The construction Last /@ Split[ ... ] accounts for
duplicate values.Here are two examples.
Needs[Statistics`ContinuousDistributions`]list1 =
RandomArray[NormalDistribution[0, 1], 100];f1 =
MakeEmpiricalCDF[list1];Plot[f1[x], {x, -4, 4}]list2 =
Table[Random[Integer, {1, 10}], {10}];f2 =
MakeEmpiricalCDF[list2];Plot[f2[x], {x, 0, 11}]--Mark> Išm
trying to write a fast empirical cummulative distribution
function> (CDF). Empirical CDFs are step functions that can
be expressed in> terms of a Which statement. For example,
given the list of> observations {1, 2, 3},> f = Which[# < 1,
0, # < 2, 1/3, # < 3, 2/3, True, 1]&> is the empirical CDF.
Note that f /@ {1, 2, 3} returns {1/3, 2/3, 1}> and f is
continuous from the right.> When the number of observations
is large, the Which statement> evaluates fairly slowly (even
if it has been Compiled). Since> InterpolationFunction
evaluates so much faster in general, Išve tried> to use
Interpolation with InterpolationOrder -> 0. The problem is
that> the resulting InterpolatingFunction doesnšt behave the
way (I think)> it ought to. For example, let> g =
Interpolation[{{1, 1/3}, {2, 2/3}, {3, 1}},
InterpolationOrder ->> 0]> Then, g /@ {1, 2, 3} returns {2/3,
2/3, 1} instead of {1/3, 2/3, 1}.> In addition, g is
continuous from the left rather than from the right.>
Obviously I am not aware of the considerations that went
into> determining the behavior of InterpolationFunction when>
InterpolationOrder -> 0.> So I have two questions: > (1) Does
anyone have any opinions about how InterpolatingFunction>
ought to behave with InterpolationOrder -> 0?> (2) Does
anyone have a faster way to evaluate an empirical CDF than a>
compiled Which function?> By the way, herešs my current
version:> CompileEmpiricalCDF[list_?(VectorQ[#, NumericQ] &)]
:=> Block[{x}, Compile[{{x, _Real}}, Evaluate[> Which @@
Flatten[> Append[> Transpose[{> Thread[x < Sort[list]],>
Range[0, 1 - 1/#, 1/#] & @ Length[list]> }],> {True, 1}]]>
]]]> --Mark==== I am new to numerical computing. I have a
equation dQ/dz = exp(-i*4*a^2*z - i*pi/4)*f(z,t)/sqrt(4*pi*z)
where z is the position range from 0 to 10, t is the time
range from0 to 10 f(z,t) = integration of
exp(i*(t-t1-z)^2/z)*g(t1) from t1=-100 to100 g(t1) = d^3
sech(t1) / dt^3 I would like to solve Q(z,t), however, I am a
new to mathematica.Can anyone help me?====I know that there is
an input form. I want the output to have the sameform rather
than look like Floor[x]. Sorry,Jack>> A commonly used symbol
for the Floor function is a square bracket with the> upper
indents removed. Is this symbol part of Mathematicašs
built-in symbols? I> think not, so then, can it be
constructed by hand.>> Jack>>Reply-To:
kuska@informatik.uni-leipzig.de====doesMakeBoxes[Floor[x_],
fmt_:StandardForm] := RowBox[{[LeftFloor], ToBoxes[x, fmt],
[RightFloor]}]help you ? Jens> I know that there is an input
form. I want the output to have the same> form rather than
look like Floor[x]. Sorry,> Jack>> A commonly used symbol for
the Floor function is a square bracket with the> upper indents
removed. Is this symbol part of Mathematicašs built-in
symbols? I> think not, so then, can it be constructed by
hand.>> Jack>>====Recent threads about word processing and
typesetting have got me dreaming again...Wouldnšt it be nice
if someone developed a package that provided a function like
this: DisplayTeX[ some TeX code ]which would cause the kernel
to generate PostScript for the typeset text, to be displayed
by the front end?I doubt that WRI will ever consider this,
but surely therešs someone out there whošs both a TeXpert and
a Mathematica guru who can do it. No doubt therešs money to be
made.Probably just a pipe dream...---Selwyn HollisReply-To:
kuska@informatik.uni-leipzig.de====you mean the inverse of
TeXForm[] ?For what ? A TeXpert && Mathematica guru &&
MathLink experthas a TeX frontend, that translate a
TeXNotebook(that is a TeX Ūle with some Mathematica
Input/Output environments) and send the input cells to a
kerneland insert the output into the Ūnal TeX Ūle ...The most
of the remaining work like creating hyperlinksto references
and Ūgures is done my LaTeX and pdfLaTeX,with help of CTAN
and some macros one generate almost everylayout *and* I have
more than 20 books about TeX/LaTeX (including Don Knuths
excelent manuals) but I have nota single book about the
Mathematica Frontend.The way is not to teach TeX to the
frontend, the way isto teach TeX a bit Mathematica.And a
TeXpert will never switch from his beloved TeXto the Frontend
and itšs typesetting -- thatšs whyhe is a TeXpert. Jens>
Recent threads about word processing and typesetting have got
me> dreaming again...> Wouldnšt it be nice if someone
developed a package that provided a> function like this:>
DisplayTeX[ some TeX code ]> which would cause the kernel to
generate PostScript for the typeset> text, to be displayed by
the front end?> I doubt that WRI will ever consider this, but
surely therešs someone out> there whošs both a TeXpert and a
Mathematica guru who can do it. No> doubt therešs money to be
made.> Probably just a pipe dream...> ---> Selwyn Hollis====>
you mean the inverse of TeXForm[] ?> For what ? A TeXpert &&
Mathematica guru && MathLink expert> has a TeX frontend, that
translate a TeXNotebook> (that is a TeX Ūle with some
Mathematica Input/Output > environments) and send the input
cells to a kernel> and insert the output into the Ūnal TeX
Ūle ...Congratulations to you if you can do that and if
youšre happy with it. Išd prefer life to be a bit simpler.
The point is that many users of Mathematica already have
sufŪcient knowledge of TeX/LaTeX to typeset very complicated
mathematics and get excellent results with little or no
grief. To achieve even close to the same quality with
Mathematica alone can require ten times the effort and a
hundred times the grief.---Selwyn> The most of the remaining
work like creating hyperlinks> to references and Ūgures is
done my LaTeX and pdfLaTeX,> with help of CTAN and some
macros one generate almost every> layout *and* I have more
than 20 books about TeX/LaTeX > (including Don Knuths
excelent manuals) but I have not> a single book about the
Mathematica Frontend.> The way is not to teach TeX to the
frontend, the way is> to teach TeX a bit Mathematica.> And a
TeXpert will never switch from his beloved TeX> to the
Frontend and itšs typesetting -- thatšs why> he is a
TeXpert.> Jens>>Recent threads about word processing and
typesetting have got me>>dreaming again...>>Wouldnšt it be
nice if someone developed a package that provided a>>function
like this:>> DisplayTeX[ some TeX code ]>>which would cause
the kernel to generate PostScript for the typeset>>text, to
be displayed by the front end?>>I doubt that WRI will ever
consider this, but surely therešs someone out>>there whošs
both a TeXpert and a Mathematica guru who can do it.
No>>doubt therešs money to be made.>>Probably just a pipe
dream...>>--->>Selwyn Hollis> ====Mike,I donšt think so.
Style sheets do not control the styles used within anoutput
graphics cell. I think you will have to set the plot style as
astatement in the notebook, change it and re-evaluate for
printing. Thismight not be all that inconvenient because you
probably wonšt be printingthat often.David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/Sender:
steve@smc.vnet.netApproved: Steven M. Christensen
<steve@smc.vnet.net>, Moderator====Dear MathGroup,To plot
FrameTicks outward with the same style as the Frame, I
useExtendGraphics in the following
way.Needs[ExtendGraphics`Ticks`];tf[min_,max_] :=
TickFunction[min, max, MajorStyle -> {Thickness[0.003]},
MajorLength - > {0, 0.01}, MinorStyle ->
{Thickness[0.003]},MinorLength -> {0, 0.005}]Plot [-2.4 (x+6)
(x-8), {x, -6.2, 9.2}, FrameTicks -> {tf, tf, None,
None},AspectRatio -> 18/25, Frame -> True, DefaultFont ->
{Helvetica-Bold, 12},Axes -> None,FrameStyle ->
Thickness[0.003], FrameLabel -> {x, y, None, None},PlotStyle
-> Thickness[0.006],ImageSize -> 504];(instead of the same
Plot statement with FrameTicks -> {Automatic,Automatic, None,
None})This arrangement of major and minor ticks is clearly
unacceptable. Theminor ticks do not divide intervals of
adjacent labelled major ticks into equal parts.Are there
other solutions? I have noticed that some major
statistical(plotting) programs donšt do minor ticks at
all.Tom AldenbergRIVMBilthovenNetherlands====I need to create
an adjacency matrix from my data, which is currently inthe
form of a .txt Ūle and is basically a two column incidence
list.For example: 1 A 1 B 2 B 3 C . . . . . . m n Where 1 to
m represent actors and A to n represent events. My goal is
tohave an (m x m) matrix where cell i,j equals 1 if two
actors areincident to the same event (in the sample above, 1
and 2 are bothincident to B) and 0 otherwise (w/ zeros on the
diagonal). Išm new to Mathmatica, and so Išm on the steep part
of the learningcurve ... All Išve been able to Ūgure out so
far is how to get myincidence list into the program using
Import[Ūlename.txt]. But thenwhat? How do I convert to the
adjacency matrix? Išve found theToAdjacencyMatrix[] command
in DiscreteMath`Combinatorica`, but I canštseem to get it to
work ...
Tom**********************************************Thomas P.
MoliternoGraduate School of ManagementUniversity of
California,
Irvinetmoliter@uci.edu***************************************
*******Reply-To: kuska@informatik.uni-leipzig.de====with
In[]:=lst = {{1, A},{1, B},{2 , B},{3, C}, {3, D}, {1,
D}};and In[]:=AdjacenceMatrix[lst : {{_, _} ..}] := Module[
{actors,events adj}, {events, actors} = Union /@
Transpose[lst]; adj = Table[0, {Length[events]},
{Length[actors]}]; Scan[(Part[adj, Sequence @@ #] = 1) &, lst
/. MapIndexed[Rule[#1, First[#2]] &, actors]]; adj ]you
getIn[]:=AdjacenceMatrix[lst]Out[]={{1, 1, 0, 1}, {0, 1, 0,
0}, {0, 0, 1, 1}} Jens> I need to create an adjacency matrix
from my data, which is currently in> the form of a .txt Ūle
and is basically a two column incidence list.> For example:>
1 A> 1 B> 2 B> 3 C> . .> . .> . .> m n> Where 1 to m
represent actors and A to n represent events. My goal is to>
have an (m x m) matrix where cell i,j equals 1 if two actors
are> incident to the same event (in the sample above, 1 and 2
are both> incident to B) and 0 otherwise (w/ zeros on the
diagonal).> Išm new to Mathmatica, and so Išm on the steep
part of the learning> curve ... All Išve been able to Ūgure
out so far is how to get my> incidence list into the program
using Import[Ūlename.txt]. But then> what? How do I convert
to the adjacency matrix? Išve found the> ToAdjacencyMatrix[]
command in DiscreteMath`Combinatorica`, but I canšt> seem to
get it to work ...> Tom>
**********************************************> Thomas P.
Moliterno> Graduate School of Management> University of
California, Irvine> tmoliter@uci.edu>
**********************************************====>Probably a
rather simple question: what is the easiest way to open
a>notebook, execute it, then quit, from the command line? I
would like to be>able to do this from a MakeŪle.actually,
this is not as simple as it should be and one needs
theextremely useful JLink to really manipulate Notebooks from
the similar under MacOSX and even Windows. Maybe others can
try
this.=================================================Execute
this in the FrontEnd to create a test notebook, then quit the
FrontEnd
completely:NotebookSave[NotebookPut[Notebook[{Cell[<Plot3D[
Sin[x]*Cos[y], {x, 0, 10}, {y, 0, 10}, PlotLabel ->,
Input],Cell[Integrate[Log[1 - x]^2/(1 + x), {x, 0, 1}],
Input]}]],/tmp/test.nb](* ****************************
*)Next, create a Ūle t.m :[rolf@uranus tmp]$ cat t.mnb =
/tmp/test.nb;(* *****************************
*)<<JLink`;UseFrontEnd[n = NotebookOpen[nb];SelectionMove[n,
All, Notebook];SelectionEvaluate[n];m =
NotebookPut[Notebook[{Cell[<en =
First[Select[Select[Notebooks[],
!FreeQ[NotebookInformation[#1], FileName] & ], (WindowTitle
/. NotebookInformation[#1]) === /tmp/test.nb & ]] >,
Input],Cell[NotebookSave[en, Interactive -> False],
Input]}]];SelectionMove[m, All,
Notebook];SelectionEvaluate[m];];Pause[1];(*
**************************************** *)Now you can run it
like this in the background:[rolf@uranus tmp]$ math < t.m >
t.out &This will open /tmp/test.nb , evaluate it and save it
under thesame name.(* ******************************* *)Maybe
there is an easier and cleaner way, but the problem here is
that the kernel and the FrontEnd are really two different
programs, i.e., the kernel is not waiting until the FrontEnd
Ūnished certain tasks (like NotebookSave), i.e., think of it
as threads. Therefore I used the trick with a second steering
notebook.(One usually has similar problems when writing more
intricate buttons.)If you are not, e.g. you work remotely at
a Unix-box without anvirtual frame buffer (Xvfb), like in
webMathematica. ask Wolfram Tech Support.Rolf MertigMertig
ConsultingEfŪcient Softwarehttp://www.mertig.com====> Išm
curious about the different efŪciencies of Export vs. Put;
Import vs.> Get.> Eg. Saving a list called c with approx.
16,000 elements, each element> being a three element list of
reals.> Export[ac.dat,c] takes forever and consumes heaps of
kernel memory> c>>ac.dat all done in an instant (relatively
speaking) and without the> transient burst of kernel memory
usage.> Ditto Import and Get. Can anyone explain this?The
comparison is that of apples and oranges.Get[] and Put[] deal
with reading and writing Mathematica expressions inInputForm
syntax. They are implemented in C, so they are fairly fast,
andthe target format is human readable. However the output
might not bereadily parsable by other computer programs. The
functions may be usedwith Mathematica expressions of all
kinds. In the event that humanreadability and platform
independence are not important, one can useDumpSave[] in lieu
of Put[].Export[] and Import[] for the cases you describe are
using the Table target format. They are implemented in
top-level Mathematica code, so they are not as fast as Get[]
and Put[]. The target is a regularly formatted array of data,
so not all Mathematica expressions are appropriate for this
kind of conversion. In the case of Export[], the expression
is formatted completely in memory using TableForm[] before
being saved out to Ūle. In the case of Import[], there are a
number of heuristics that are applied to each Ūeld to
determine whether the Ūeld should be converted to a number or
a string. This should explain why they are slower and require
more memory.If you are dealing with a large collection of
machine precision numbers, then neither of these approaches
may be suitable for your purposes. You might be better off
using the free add-on FastBinaryFiles, which is available up
on MathSource at this URL:readable by C programs.-- User
Interface Programmer paulh@wolfram.comWolfram Research,
Inc.====> A commonly used symbol for the Floor function is a
square bracket with> the upper indents removed. Is this
symbol part of Mathematicašs> built-in symbols? I think not,
so then, can it be constructed by> hand.Is this what youšre
describing?In[1]:= [LeftFloor]x[RightFloor]Out[1]=
Floor[x]This notation is documented in Section A.2.6 of _The
Mathematica Book_ (Fourth Edition).-- User Interface
Programmer paulh@wolfram.comWolfram Research,
Inc.====Jack,Check pages 959 and 960 in Section 3.10.4 of The
Mathematica Book. You canuse LeftFloor and RightFloor as
bracketing operators, which in turn areintrepreted as
Floor.[LeftFloor] 3.2 [RightFloor]3These can also be entered
as esc-lf-esc and esc-rf-esc.David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/Sender:
steve@smc.vnet.netApproved: Steven M. Christensen
<steve@smc.vnet.net>, Moderator====It turns out that
Mathematica does have the ūoor symbols! Use esc l f escfor
the left ūoor and esc r f esc for the right ūoor. In fact, if
you typein esc l f esc a esc r f esc Mathematica will
interpret the input asFloor[a].Carl WollPhysics DeptU of
Washington>> A commonly used symbol for the Floor function is
a square bracket with the> upper indents removed. Is this
symbol part of Mathematicašs built-insymbols? I> think not,
so then, can it be constructed by hand.>> Jack>>====> A
commonly used symbol for the Floor function is a square
bracket with> the upper indents removed. Is this symbol part
of Mathematicašs built-in> symbols?Sure. Just look for Floor
in A.10 Listing of Major Built-In MathematicaObjects! For
example, you can use [LeftFloor] and [RightFloor] to getthe
symbols you want.DavidReply-To:
kuska@informatik.uni-leipzig.de====what may [LeftFloor] and
[RightFloor]show on screen ? Itšs easy to Ūnd in thecomplete
characters palette | Operators | General Jens> A commonly
used symbol for the Floor function is a square bracket with
the> upper indents removed. Is this symbol part of
Mathematicašs built-in symbols? I> think not, so then, can it
be constructed by hand.> Jack====> A commonly used symbol for
the Floor function is a square bracket with the> upper
indents removed. Is this symbol part of Mathematicašs
built-in symbols? I> think not, so then, can it be
constructed by hand.> Jack[LeftFloor] and [RightFloor] can be
used for this.Look at Chapter Advanced
Mathematics/Mathematical and Other Notations/Operators.Yours,
Alexander-- / Alexander Dreyer, Dipl.-Math. - Abteilung
Adaptive Systeme / Fraunhofer Institut fuer Techno- und
Wirtschaftsmathematik (ITWM) Gottlieb-Daimler-Strasse, Geb.
7^2=49/313 D-67663 Kaiserslautern /====Hugh Goyder and David
Park gave a most elegant function to Ūnd two vectorsthat are
orthogonal to one vector in 3D. The key to coming up with
theelegant solution is an understanding of Mathematicašs
NullSpace function.We can easily make the version from Hugh
and David much more general withthe version
below.-------------OrthogonalUnitVectors[vect__?VectorQ]:=
#/Sqrt[#.#]&/@NullSpace[{vect}]-------------The version above
will give a set of unit orthogonal vectors if given anynumber
of vectors in any dimension. So besides giving it a 3D vector
we can give it the following:
OrthogonalUnitVectors[{2,1,0,-1,1}] or
OrthogonalUnitVectors[{0,1,0,1/2,1},{1,0,-1,1/2}]------------
But the short version above isnšt very robust.(1)
Clear[x,y,z];NullSpace[{{x,y,z}}] returns two vectors
orthogonal to {x,y,z}, but the two vectorsNullSpace returns
arenšt orthogonal to each other. So (OrthogonalUnitVectors)
should only work with numeric vectors.(2) We should ensure
all the vectors have the same dimension and length >1.I give
a less concise version below that corrects these
problems.------------OrthogonalUnitVectors[vect__?(VectorQ[#,
NumericQ]&)]/;
(SameQ@@Length/@{vect})&&(Length[First[{vect}]]>1):=
#/Sqrt[#.#]&/@NullSpace[{vect}]-------------- Ted Ersek Get
Mathematica tips, tricks from
http://www.verbeia.com/mathematica/tips/Tricks.html====I have
installed 128 MB RAM extra in my computer. I had 64 before, so
now Ihave 3 times as much.I expected some calculations that
took very long before, to be executed inless time. To the
contrary, my computer still goes to the hard drive fromemory
for its calculations, and Mathematica stops with an out of
memorymessage.Windows is recognizing this new memory. I dont
know if Mathematica is. Howcan I Ūnd out about
this?Julio====> It seems there is an error in the BinCounts
function of Mathematicašs> standard
<<Statistics`DataManipulation` package where it sometimes>
returns one more than the expected number of histogram bins.
An example> of this is shown in the notebook code below where
the comment in the> last notebook cell explicitly describes
the problem. If anyone can> explain this behavior it would be
appreciated, otherwise Wolfram> Research should be made aware
of this feature of the function when it> is called as:
BinCounts[{x1, x2, ...}, {xmin, xmax, dx}]A developer of
Mathematicašs Standard Packages has requested that
thisresponse be posted to the forum:[begin developeršs
comments]Note that if you donšt apply N when computing
datBinWidth, you will alwaysget the expected number of bins;
i.e., datBinWidth = (datMax -datMin)/datBins (assuming datMax
and datMin are also exact numbers).Given input {xmin, xmax,
dx}, BinCounts determines the number of bins tobe computed
via Ceiling[(xmax - xmin)/dx]. For numericalized values,
itšspossible for dx to end up such that the Ceiling forces an
additional binto be added; presumably, itšs better to have too
many rather than too fewbins -- for example, if the dx was
deliberately given such that the xmaxis not at an integer bin
boundary... If you use exact values for binbounds and
increment, then there will be no problem.[end developeršs
comments]--User Interface Programmer paulh@wolfram.comWolfram
Research, Inc.====I would like to create a 3D chart from an
ASCII Ūle using Mathematica3.0. Can anyone give me some
help?====You might want to tell us how to Ūnd A.10.Anyway,
itšs easier to simply go to Floor in the help browser
(inversion 4.2, anyway).Bobby Treat-----Original
Message-----David====I doubt therešs any money to be made
here. Lots of people have createdvery valuable additions to
Mathematica, but theyšre all being givenaway, so far as I can
tell. Itšs too bad, too.And here we are, giving away our time.
Sigh...Bobby-----Original Message-----there whošs both a
TeXpert and a Mathematica guru who can do it. No doubt
therešs money to be made.Probably just a pipe
dream...---Selwyn Hollis====OK, I give up; how would I Ūnd
Chapter AdvancedMathematics/Mathematical and Other
Notations/Operators in the HelpBrowser?Bobby
Treat-----Original Message-----> think not, so then, can it
be constructed by hand.> Jack[LeftFloor] and [RightFloor] can
be used for this.Look at Chapter Advanced
Mathematics/Mathematical and Other Notations/Operators.Yours,
Alexander-- / Alexander Dreyer, Dipl.-Math. - Abteilung
Adaptive Systeme / Fraunhofer Institut fuer Techno- und
Wirtschaftsmathematik (ITWM) Gottlieb-Daimler-Strasse, Geb.
7^2=49/313 D-67663 Kaiserslautern /====> You might want to
tell us how to Ūnd A.10.Open your fourth edition of _The
Mathematica Book_ to page 1065. Thatšs where part 10 of the
Appendix starts. > Anyway, itšs easier to simply go to Floor
in the help browser (in> version 4.2, anyway).I suppose
youšre right. But I -- call me old-fashioned if you wish --
prefer to use the book.David ====Išll assume you have the
information in a matrix like x:TableForm[x = {{1, A }, {1, B
}, {2, B}, {3, D}, {4, D}, {5, C}}]First Ūnd the highest
actor number (or use what youšve inputelsewhere):m =
Last[Union[x[[All,1]]]]5DeŪne the incidence function:f[x_,
a_, b_] /; a == b := 0f[x_, a_, b_] := If[{} $B!b(B
Intersection[Cases[x, {a, y_} -> y], Cases[x, {b, y_} ->
y]],1, 0]Herešs the incidence matrix:Array[f[x, #1, #2] &,
{m, m}]{{0, 1, 0, 0, 0}, {1, 0, 0, 0, 0}, {0, 0, 0, 1, 0},
{0, 0, 1, 0, 0}, {0, 0, 0, 0, 0}}Once you have the incidence
function, the only reason to store theincidence matrix is to
avoid computing over again the same answers, andthat can be
accomplished in other ways. If you wonšt have other
xmatrices, itšs convenient to deŪne f this way:f[a_, b_] /; a
== b := f[a, b] = 0f[a_, b_] /; a < b := f[a, b] = If[{}
$B!b(B Intersection[Cases[x, {a, y_} -> y], Cases[x, {b,
y_} ->y]], 1, 0]f[a_, b_] := f[b, a]Array[f, {m, m}]{{0, 1,
0, 0, 0}, {1, 0, 0, 0, 0}, {0, 0, 0, 1, 0}, {0, 0, 1, 0, 0},
{0,0, 0, 0, 0}}Whenever you would want the {j,k} element of
the adjacency matrix, justuse f[j,k] instead. Each pair is
computed only once. In this version,Išve taken advantage of
the symmetry of the problem, too, to cut thework in
half.Bobby Treat-----Original Message-----. .. .m nWhere 1 to
m represent actors and A to n represent events. My goal is
tohave an (m x m) matrix where cell i,j equals 1 if two
actors areincident to the same event (in the sample above, 1
and 2 are bothincident to B) and 0 otherwise (w/ zeros on the
diagonal).Išm new to Mathmatica, and so Išm on the steep part
of the learningcurve ... All Išve been able to Ūgure out so
far is how to get myincidence list into the program using
Import[Ūlename.txt]. But thenwhat? How do I convert to the
adjacency matrix? Išve found theToAdjacencyMatrix[] command
in DiscreteMath`Combinatorica`, but I canštseem to get it to
work
...Tom**********************************************Thomas P.
MoliternoGraduate School of ManagementUniversity of
California,
Irvinetmoliter@uci.edu***************************************
*******====Well, I donšt know how fast, but it is fairly
simple, anyway. Suppose youhave a series of values s for
which you wish to obtain the edf.In[1]:=s =
Table[Random[Integer, {0, 3}],
{10}]Out[1]={2,2,1,2,0,0,1,1,1,3}If no speciŪcation is made
about their position on the x-axis, we assumethat they
correspond to the integers from 1 to 10. What we have then is
thecollection of pairsIn[2]:=porig = Transpose[{Range[10],
s}]Out[2]={{1, 2}, {2, 2}, {3, 1}, {4, 2}, {5, 0}, {6, 0},
{7, 1}, {8, 1}, {9, 1}, {10, 3}}The edf gives, for each x,
the proportion of points in s that are less thanor equal to
x, for all x. We obtain these proportions through the
cumulativesumsIn[3]:=N[CumulativeSums[s]/Plus @@
s]Out[3]={
0.153846,0.307692,0.384615,0.538462,0.538462,0.538462,0.615385
,0.692308,0.769231,1.}so that for each of the pairs (x, y)
below, y gives the proportion of pointsin s that are less
than or equal to x:In[4]:=cumporig = Transpose[{Range[10],
N[CumulativeSums[s]/Plus @@ s]}]Out[4]={{1,
0.15384615384615385}, {2, 0.3076923076923077}, {3,
0.38461538461538464}, {4, 0.5384615384615384}, {5,
0.5384615384615384}, {6, 0.5384615384615384}, {7,
0.6153846153846154}, {8, 0.6923076923076923}, {9,
0.7692307692307693}, {10, 1.}}Now shift the x values one unit
to the left, by dropping the last value andprepending 0 to
them:In[5]:=ps = Transpose[{Prepend[Drop[Range[1, 10], -1],
0], CumulativeSums[s]/Plus @@ s}]Out[5]={{0, 2/13}, {1,
4/13}, {2, 5/13}, {3, 7/13}, {4, 7/13}, {5, 7/13}, {6, 8/13},
{7, 9/13}, {8, 10/13}, {9, 1}}Then use Interpolation on this
shifted set of
points:In[6]:=ips=Interpolation[ps,InterpolationOrder[Rule]0]
Out[6]=InterpolatingFunction[{{0,9}},<>]ips[x-1] is the edf
you are looking for, as you may check by plotting it
anddisplaying in the same graph together with the ListPlot of
cumporig above.Tomas GarzaMexico City----- Original Message
-----> and f is continuous from the right.>> When the number
of observations is large, the Which statement> evaluates
fairly slowly (even if it has been Compiled). Since>
InterpolationFunction evaluates so much faster in general,
Išve tried> to use Interpolation with InterpolationOrder ->
0. The problem is that> the resulting InterpolatingFunction
doesnšt behave the way (I think)> it ought to. For example,
let>> g = Interpolation[{{1, 1/3}, {2, 2/3}, {3, 1}},
InterpolationOrder ->> 0]>> Then, g /@ {1, 2, 3} returns
{2/3, 2/3, 1} instead of {1/3, 2/3, 1}.> In addition, g is
continuous from the left rather than from the right.>>
Obviously I am not aware of the considerations that went
into> determining the behavior of InterpolationFunction when>
InterpolationOrder -> 0.>> So I have two questions:>> (1) Does
anyone have any opinions about how InterpolatingFunction>
ought to behave with InterpolationOrder -> 0?>> (2) Does
anyone have a faster way to evaluate an empirical CDF than a>
compiled Which function?>> By the way, herešs my current
version:>> CompileEmpiricalCDF[list_?(VectorQ[#, NumericQ]
&)] :=> Block[{x}, Compile[{{x, _Real}}, Evaluate[> Which @@
Flatten[> Append[> Transpose[{> Thread[x < Sort[list]],>
Range[0, 1 - 1/#, 1/#] & @ Length[list]> }],> {True, 1}]]>
]]]>> --Mark>>====I have no opinion on the behavior
associated with InterpolationOrder->0,except that it should
be DOCUMENTED, but isnšt.Meanwhile, try this:lst = {{1, 1/3},
{2, 2/3}, {3, 1}}; ClearAll[empiricalCDF]empiricalCDF[{x_List,
y_List}] := Compile[ {{z, _Real}}, Evaluate[ Which @@ Flatten[
Transpose[ {(z < #1 & ) /@ x, y}]]]]empiricalCDF[v:{{_, _}..}]
:= empiricalCDF[v] = empiricalCDF[ ({Join[#1[[1]], {InŪnity}],
Join[{0}, #1[[ 2]]]} & )[Transpose[
lst]]]empiricalCDF[lst]Plot[empiricalCDF[lst][x], {x, 1,
3}];I split the work into two deŪnitions for readability. If
you evaluate:?empiricalCDFafter doing the plot above, youšll
see that empiricalCDF[{{1, 1/3}, {2,2/3}, {3, 1}}] has been
compiled and saved for later use, and that ittakes precedence
over the SetDelayed rules listed after it. Hence,
thecompilation only occurs once for each list.As written,
your CompileEmpiricalCDF would be compiled all over
againevery time itšs used -- for each and every point you
plot -- so ofcourse itšs slow. There are other problems, too.
For instance, thepatternlist_?(VectorQ[#, NumericQ] &)makes no
sense at all. Maybe you
meantlist_?And[VectorQ[#],NumericQ[#]]&Bobby
Treat-----Original Message-----evaluates fairly slowly (even
if it has been Compiled). SinceInterpolationFunction
evaluates so much faster in general, Išve triedto use
Interpolation with InterpolationOrder -> 0. The problem is
thatthe resulting InterpolatingFunction doesnšt behave the
way (I think)it ought to. For example, letg =
Interpolation[{{1, 1/3}, {2, 2/3}, {3, 1}},
InterpolationOrder ->0]Then, g /@ {1, 2, 3} returns {2/3,
2/3, 1} instead of {1/3, 2/3, 1}.In addition, g is continuous
from the left rather than from the right.Obviously I am not
aware of the considerations that went intodetermining the
behavior of InterpolationFunction whenInterpolationOrder ->
0.So I have two questions: (1) Does anyone have any opinions
about how InterpolatingFunctionought to behave with
InterpolationOrder -> 0?(2) Does anyone have a faster way to
evaluate an empirical CDF than acompiled Which function?By
the way, herešs my current
version:CompileEmpiricalCDF[list_?(VectorQ[#, NumericQ] &)]
:= Block[{x}, Compile[{{x, _Real}}, Evaluate[ Which @@
Flatten[ Append[ Transpose[{ Thread[x < Sort[list]], Range[0,
1 - 1/#, 1/#] & @ Length[list] }], {True, 1}]]
]]]--Mark====Try deleting
...Mathematica4.2DocumentationEnglishMainBookBrowserIndex.nb>
Your machine information and OS EXACTLY matches mine (except
that mypageŪle> is NOT on a separate physical drive --
perhaps this is a clue as to whereto start),> and I had no
problem building the help Ūle. Hence, as you say, thismust>
be a problem for Wolfram to address. Perhaps, someone from
WRI willenlighten> us via this newsgroup.> ....Terry>> I too
have been unable to get the help browser functioning. I have>
exactly the same problem. When trying to open the help browser
it gets> to the point where it says scanning index Ūle and
freezes. I tried> all the suggestions in the faq and the
links provided here to no> avail. Išm running Winows 2000
with service pack 3. I have 512mb of> memory with a 768mb
pageŪle on a separate drive, so I donšt think> thatšs an
issue. Mathematica 4.1 ran just Ūne on this setup, so>
judging by the number of other people affected by this, Išd
say itšs> some kind of bug with 4.2. Hopefully Wolfram will
address this with> either a link that will provide a Ūx that
actually works, or a patch.>>What is your operating system?
How much memory do you have in yourmachine?>Also, how big is
your pageŪle? Sounds to me like the rebuild ran out>of room
to do its thing in.>> I just installed Mathematica 4.2, and
had no problem rebuildingthe help index.>Subsequently, I ahd
no problem using the Help browser either.>Hope that
helps!>....Terry>>>>Išd start with the following
FAQ.>>>http://support.wolfram.com/mathematica/interface/
helpbrowser/howrebuildindex.html>>>>-Dale>> Sorry that i
failed say it immediately in a Ūrst place, but ofcourse>> i
did try it FAQ at Ūrst, and both tried to delete cache and
rebuild>> index, but results where the same - whenever i try
to invoke help>> browser (or rebuild index, for that matter),
mathematica stops>> responding (i did read your answer to the
same question asked>> a week ago before - actually that is
why i turned to the FAQ).>====>Try deleting
...>Mathematica4.2DocumentationEnglishMainBookBrowserIndex.nb
>====Simplify:5nth sqrt (3)/sqrt (6)5nth Sqrt(3)------------
Sqrt(6)1st.. is this the correct notation for use on a
computer2nd.. how is the solution solved, step by step
please.Reply-To: kuska@informatik.uni-leipzig.de====>
Simplify:> 5nth sqrt (3)/sqrt (6)> 5nth Sqrt(3)>
------------> Sqrt(6)> 1st.. is this the correct notation for
use on a computerNo. May be you mean3^(1/5)/Sqrt[6]??> 2nd..
how is the solution solved, step by step
please.3^(1/5)/Sqrt[6] // FullSimplify Jens====I want to make
a list of all symbols in the Global context, as
inNames[Global`*]and compute a ByteCount for each symbolšs
OwnValues -- withoutevaluating the symbols.It seems possible
in principle, but I havenšt found a way.Bobby TreatReply-To:
kuska@informatik.uni-leipzig.de====whats wrong
withToExpression[#, StandardForm, Hold] & /@ Names[Global`*]
/. Hold[a_] :> ByteCount[OwnValues[a]] Jens> I want to make a
list of all symbols in the Global context, as in>
Names[Global`*]> and compute a ByteCount for each symbolšs
OwnValues -- without> evaluating the symbols.> It seems
possible in principle, but I havenšt found a way.> Bobby
Treat====with the following Java comands you can execute any
Mathematica commands:ml.putFunction(EnterTextPacket,
1);ml.put(string);See JLinkUserGuide p. 164.Hermann
Schmitt----- Original Message ----- > Išve checked the
documentation on Wolfram, but did not Ūnd the Œhookš > for
ReadList from java. Any help would be appreciated. thanks.>
Pete> ====Dear Mark,I suggest trying the following code
before you putfurther energy in speeding up your functions.
Mycode is very short and seems to be fast in my Ūrsttest with
a random array of 100000 integers.cdf[li_List]:=
FoldList[#1+Length[#2]&, 0.0,
Split[Sort[li]]]/Length[li]t=Table[Random[Integer,{1,1000}],{
100000}];Timing[cdf[t];]{0.44 Second,Null}Probably the speed
could be increased further generatinga compiled version. But
this would require additionalprogramming effort. Then you had
to write a function whichcounts the number ofequal data Œby
handš while scanning through the data list. Johannes> Išm
trying to write a fast empirical cummulative distribution
function> (CDF). Empirical CDFs are step functions that can
be expressed in> terms of a Which statement. For example,
given the list of> observations {1, 2, 3},> f = Which[# < 1,
0, # < 2, 1/3, # < 3, 2/3, True, 1]&> is the empirical CDF.
Note that f /@ {1, 2, 3} returns {1/3, 2/3, 1}> and f is
continuous from the right.> When the number of observations
is large, the Which statement> evaluates fairly slowly (even
if it has been Compiled). Since> InterpolationFunction
evaluates so much faster in general, Išve tried> to use
Interpolation with InterpolationOrder -> 0. The problem is
that> the resulting InterpolatingFunction doesnšt behave the
way (I think)> it ought to. For example, let> g =
Interpolation[{{1, 1/3}, {2, 2/3}, {3, 1}},
InterpolationOrder ->> 0]> Then, g /@ {1, 2, 3} returns {2/3,
2/3, 1} instead of {1/3, 2/3, 1}.> In addition, g is
continuous from the left rather than from the right.>
Obviously I am not aware of the considerations that went
into> determining the behavior of InterpolationFunction when>
InterpolationOrder -> 0.> So I have two questions: > (1) Does
anyone have any opinions about how InterpolatingFunction>
ought to behave with InterpolationOrder -> 0?> (2) Does
anyone have a faster way to evaluate an empirical CDF than a>
compiled Which function?> By the way, herešs my current
version:> CompileEmpiricalCDF[list_?(VectorQ[#, NumericQ] &)]
:=> Block[{x}, Compile[{{x, _Real}}, Evaluate[> Which @@
Flatten[> Append[> Transpose[{> Thread[x < Sort[list]],>
Range[0, 1 - 1/#, 1/#] & @ Length[list]> }],> {True, 1}]]>
]]]> --Mark> <><><><><><><><><><><><>Johannes
LudsteckEconomics DepartmentUniversity of
RegensburgUniversitaetsstrasse 3193053
Regensburg====>-----Original Message----->Sent: Wednesday,
September 11, 2002 7:28 PM>Dear MathGroup,>>To plot
FrameTicks outward with the same style as the Frame, I
use>ExtendGraphics in the following
way.>>Needs[ExtendGraphics`Ticks`];>>tf[min_,max_] :=
TickFunction[min, max, MajorStyle ->{Thickness[0.003]},>
MajorLength - > {0, 0.01}, MinorStyle ->
{Thickness[0.003]},>MinorLength -> {0, 0.005}]>>Plot [-2.4
(x+6) (x-8), {x, -6.2, 9.2}, FrameTicks -> {tf, tf, >None,
None},>AspectRatio -> 18/25, Frame -> True, DefaultFont
->{Helvetica-Bold, 12},>Axes -> None,>FrameStyle ->
Thickness[0.003], FrameLabel -> {x, y, None, None},>PlotStyle
-> Thickness[0.006],>ImageSize -> 504];>>(instead of the same
Plot statement with FrameTicks -> {Automatic,>Automatic,
None, None})>>This arrangement of major and minor ticks is
clearly unacceptable. The>minor ticks do not divide>
intervals of adjacent labelled major ticks into equal
parts.>>Are there other solutions? I have noticed that some
major statistical>(plotting) programs donšt do minor ticks at
all.>>Tom Aldenberg>RIVM>Bilthoven>Netherlands>>Several
alternatives (if the default ticks of Plot, Show that is,
wouldnštdo):(1) generate the ticks explicitly (to get control
over supplied min, maxvalues for TickFunction):yticks =
tf[-40., 120.]xticks = tf[-6.2, 9.2]Plot[-2.4 (x + 6) (x -
8), {x, -6.2, 9.2}, FrameTicks -> {xticks, yticks, None,
None}, ...](2) try the Option TickNumbers, e.g.tf[min_, max_]
:= TickFunction[min, max, MajorStyle -> {Thickness[0.005]},
MajorLength -> {0, 0.01}, MinorStyle -> {Thickness[0.003]},
MinorLength -> {0, 0.005}, TickNumbers -> {5, 25}]gives an
acceptable result, or TickNumbers -> {6, 18};with TickNumbers
-> {8, 8} youšll have no minor ticks.(3) redeŪne the tick
functionšs helperBegin[ExtendGraphics`Ticks`Private`]??
TickPositionTickPosition[ min, max, num] returns a list of at
most num nicely roundedpositions between min and max. These
can be used for tick mark positions.TickPosition[x0_Real,
x1_Real, (num_Integer)?Positive] := Block[{dist, scale, min,
max, i, delta, space}, space = {1., 2., 2.5, 5., 10.}; dist =
(x1 - x0)/num; scale = 10.^Floor[Log[10, dist]]; dist =
dist/scale; If[dist < 1., dist *= 10.; scale /= 10.]; If[dist
>= 10., dist /= 10.; scale *= 10.]; delta =
First[Select[space, #1 >= dist & ]]*scale; min =
Ceiling[x0/delta]*delta; Table[Floor[x/delta + 0.5]*delta,
{x, min, x1, delta}]]End[]There are some things to play with:
space, scale, the deŪnition for delta,to get at more nicely
rounded positions; or rewrite completely.(4) for best
control, deŪne the ticks all by yourself, e.g. yticks1 =
Table[{t, t, {0, 0.01`}, {Thickness[0.003`]}} , {t, -25, 100,
25}]yticks2 = DeleteCases[ Table[{t, , {0, 0.005`},
{Thickness[0.003`]}} , {t, -25, 100, 12.5}], {t_, __} /;
Mod[t, 25] == 0]xticks = Join[xticks1, xticks2]xticks1 =
Table[{t, t, {0, 0.01`}, {Thickness[0.003`]}} , {t, -5, 7.5,
2.5}]xticks2 = DeleteCases[ Table[{t, , {0, 0.005`},
{Thickness[0.003`]}} , {t, -7.5, 8.5, 1.25}], {t_, __} /;
Mod[t, 2.5] == 0]yticks = Join[yticks1, yticks2]--Hartmut
Wolf====> I need to create an adjacency matrix from my data,
which is currently in> the form of a .txt Ūle and is
basically a two column incidence list.> For example: > 1 A >
1 B > 2 B > 3 C > . . > . . > . . > m n Past the following
notebook into Mathematica.Tom
BurtonNotebook[{Cell[CellGroupData[{Cell[Preparing an
adjacency matrix,Section],Cell[<This notebook was prepared
with the default celloutput format type set to
TraditionalForm. Then the output cells wereremoved to saved
space. I recommend that execute this entire notebookbefore
editing
it.>,Text],Cell[CellGroupData[{Cell[Introduction,Subsection],
Cell[<Given the list of pairs of actors and events, itappears
that most of the work is in preparing the associated list
ofundirected edges. So if you want only an adjacency matrix,
it is perhapsmore trouble that itšs worth to prepare the
associated graph. But thegraph is nice, so I am showing both
methods. >, Text]}, Open]],Cell[CellGroupData[{Cell[The
structure of a graph,Subsection],Cell[<We need to investigate
the structure of a graph sowe can make our own graph.>,
Text],Cell[BoxData[ (<<DiscreteMath`Combinatorica`)],
Input],Cell[Start with the simplestgraph:,
Text],Cell[BoxData[ (aa =
CompleteGraph[2])],Input],Cell[BoxData[ (InputForm[aa])],
Input],Cell[<Itappears to be a list of undirected edges
followed by a list of vertices,each element of which is
wrapped in an extra pair of braces. Letšs tryit:>,
Text],Cell[BoxData[ (bb = Graph[{{{1, 2}}},{{{(-1. ), 0}},
{{1. , 0}}}])], Input],Cell[BoxData[(ToAdjacencyMatrix[aa])],
Input],Cell[BoxData[(ToAdjacencyMatrix[bb])], Input]},
Open]],Cell[CellGroupData[{Cell[Example of a list of
actor-event pairs,Subsection],Cell[BoxData[ (data = {{1,
<B>}, {4, <A>}, {3,<C>}, {1, <A>}, {1, <D>}, {3, <B>}, {2,
<C>}, {2,<B>}})], Input],Cell[<Get the number of actors,
which will bethe dimension of the matrix and the number of
vertices in the graph.>,Text],Cell[BoxData[ (d =
Max[data[([All, 1])]])], Input]},Open
]],Cell[CellGroupData[{Cell[Getting the list of
edges,Subsection],Cell[TextData[{ Given the list of actors
and actions,this is most of the work. Išll do it in baby
steps, as I would like to seewere I new to ,
StyleBox[Mathematica, FontSlant->Italic], . Itcould be done
more directly.}], Text],Cell[Put events
Ūrst.,Text],Cell[BoxData[ (data2 = Reverse /@
data)],Input],Cell[Group by event, with increasing actors
within each event.,Text],Cell[BoxData[ (data3 =
Sort[data2])],Input],Cell[BoxData[ (data4 = Split[data3,
First[#1] == First[#2]&])], Input],Cell[Remove isolated
pairs., Text],Cell[BoxData[(data5 = DeleteCases[data4, {{_,
_}}])], Input],Cell[Drop theactions., Text],Cell[BoxData[
(data6 = data5[LeftDoubleBracket]All, All,
2[RightDoubleBracket])],Input],Cell[<Edges are all subsets of
connected vertices taken twoat a time:>, Text],Cell[BoxData[
(data7 = (KSubsets[#, 2]&) /@ data6)], Input],Cell[Collect
edges into one list,Text],Cell[BoxData[ (data8 =
Flatten[data7, 1])],Input],Cell[and remove duplicates,
Text],Cell[BoxData[(onedir = Union[data8])], Input]},
Open]],Cell[CellGroupData[{Cell[Adjacency matrix via a list
of edges usingMapAt, Subsection],Cell[Convert to directed
edges,Text],Cell[BoxData[ (bothdir = Join[onedir, Reverse /@
onedir])],Input],Cell[<and Ūnally, construct the adjacency
matrix, putting1šs into elements corresponding to directed
edges:>,Text],Cell[BoxData[ (emptymat = Table[0, {d},
{d}])],Input],Cell[BoxData[ (adjmat = MapAt[1 &, emptymat,
bothdir])],Input]}, Open
]],Cell[CellGroupData[{Cell[Adjacency matric
viaToAdjacencyMatrix, Subsection],Cell[Construct a set of
vertices fordisplay, Text],Cell[BoxData[ (verts =
Table[{Cos[2 [Pi]j/d], Sin[2 [Pi] j/d]}, {j,
d}])],Input],Cell[<Prepare graph, adding extra braces and
making thecoordinates of the vertices real numbers (this
appears to benecessary)>, Text],Cell[BoxData[ (g = Graph[List
/@ onedir,N[List /@ verts]])], Input],Cell[Check our
handiwork:,Text],Cell[BoxData[ (InputForm[g])],
Input],Cell[BoxData[((ShowGraph[g];))],
Input],Cell[BoxData[(ToAdjacencyMatrix[g])], Input]}, Open
]]}, Open]]},ScreenRectangle->{{43, 1152},{0,
746}},WindowSize->{565, 624},WindowMargins->{{243,
Automatic}, {48,Automatic}}]====I have a notebook that will
become a web page. I want to have a logo GIF atthe top.My
Ūrst thought was that I could drag and drop a GIF into a cell
in thenotebook, but that does not work.I can Import a GIF and
Show it. This creates an input cell, the GIF, and anoutput
cell. I donšt want the input cell and the output cell to show
whenthe HTML code is created.How can I do this?Tom
Compton====Instead of using Legend in plots with multiple
series is there a wayto have arrows with a text at the end
identifying the differentseries? The reason Išd like this is
that the legend is a bit Œbulkyšlooking and Išd want
something tidier.I know there is an Arrow package but I think
youšd need to manuallyenter each start and end point of each
arrow - is there a simplecommand to do this?Also - is there a
way to adjust the legend font size? I.e. make itsmall so that
it doesnšt interfere with data series. _ | thankyouReply-To:
kuska@informatik.uni-leipzig.de====Needs[Graphics`Arrow`]Plot
[{Sin[x], Cos[x]}, {x, 0, 2Pi}, Epilog -> {{GrayLevel[0.9],
Rectangle[{3.75, 0.25}, {5.2, 0.9}]}, Arrow[{4, 0.75}, {2,
Sin[2]}], Text[Sin[x], {4, 0.75}, {-1, 0}], Arrow[{4, 0.5},
{2, Cos[2]}], Text[Cos[x], {4, 0.5}, {-1, 0}]}] Jens> Instead
of using Legend in plots with multiple series is there a way>
to have arrows with a text at the end identifying the
different> series? The reason Išd like this is that the
legend is a bit Œbulkyš> looking and Išd want something
tidier.> I know there is an Arrow package but I think youšd
need to manually> enter each start and end point of each
arrow - is there a simple> command to do this?> Also - is
there a way to adjust the legend font size? I.e. make it>
small so that it doesnšt interfere with data series.> _> |> >
> > > thankyou====>-----Original Message----->Išm trying to
write a fast empirical cummulative distribution
function>(CDF). Empirical CDFs are step functions that can be
expressed in>terms of a Which statement. For example, given
the list of>observations {1, 2, 3},>>f = Which[# < 1, 0, # <
2, 1/3, # < 3, 2/3, True, 1]&>>is the empirical CDF. Note
that f /@ {1, 2, 3} returns {1/3, 2/3, 1}>and f is continuous
from the right.>>When the number of observations is large, the
Which statement>evaluates fairly slowly (even if it has been
Compiled). Since>InterpolationFunction evaluates so much
faster in general, Išve tried>to use Interpolation with
InterpolationOrder -> 0. The problem is that>the resulting
InterpolatingFunction doesnšt behave the way (I think)>it
ought to. For example, let>>g = Interpolation[{{1, 1/3}, {2,
2/3}, {3, 1}}, InterpolationOrder ->>0]>>Then, g /@ {1, 2, 3}
returns {2/3, 2/3, 1} instead of {1/3, 2/3, 1}.>In addition, g
is continuous from the left rather than from the
right.>>Obviously I am not aware of the considerations that
went into>determining the behavior of InterpolationFunction
when>InterpolationOrder -> 0.>>So I have two questions: >>(1)
Does anyone have any opinions about how
InterpolatingFunction>ought to behave with InterpolationOrder
-> 0?>>(2) Does anyone have a faster way to evaluate an
empirical CDF than a>compiled Which function?>>By the way,
herešs my current
version:>>CompileEmpiricalCDF[list_?(VectorQ[#, NumericQ] &)]
:=> Block[{x}, Compile[{{x, _Real}}, Evaluate[> Which @@
Flatten[> Append[> Transpose[{> Thread[x < Sort[list]],>
Range[0, 1 - 1/#, 1/#] & @ Length[list]> }],> {True, 1}]]>
]]]>>--Mark>Mark,as to (1) I donšt know, but following your
observation, and regarding ff = CompileEmpiricalCDF[{1, 2,
3}]Plot[ff[x], {x, 0, 4}, PlotRange -> {All, {0, 1}}]as a
correct CDF, we get something quite similar withobs = {1, 2,
3}t = With[{n = Length[obs]}, Transpose[{Append[obs,
Last[obs] + 1], Range[0, 1, 1/n]}]]g = Interpolation[t,
InterpolationOrder ->
0]Off[InterpolatingFunction::dmval]Plot[g[x], {x, -1, 5},
PlotRange -> All]See that all values are shifted by one
place, zero must be deŪned and onepoint added to the
right.Now, comparingff /@ ({0, 1, 2, 3, 4}){0., 0.333333,
0.666667, 1., 1.}g /@ ({0, 1, 2, 3, 4}){0, 1/3, 1/3, 2/3,
1}We get something wrong for g, but forg /@ ({0, 1, 2, 3, 4}
+ 2$MachineEpsilon){0, 1/3, 2/3, 1, 1}obviously is all right.
So we ran into problems of numerical precision. Now,what is
the sense of the CDF at the point of jump? None Išd say, at
somepoint above the jump it must count for the fraction of
all events below inrespect to over all. And this is done
within the limits of numericalprecision. Conceptionally the
CDF is function used to integrate as aStieltjes Integral, so
only interpretations in this contex are possible, andyou have
to look at the limit from above (within the conŪnes of
numericalprecision), all ok, I think.To dig up a bigger
sample:<< Statistics`ContinuousDistributions`gdist =
GammaDistribution[3, 1]cdfunction = CDF[gdist, x]pTheory =
Plot[cdfunction, {x, 0, 10}, PlotStyle -> Hue[.7]]data =
RandomArray[gdist, 1000];t = With[{n = Length[data], obs =
Sort[data]}, Transpose[{Append[obs, Last[obs] + 1], Range[0,
1, 1/n]}]];g = Interpolation[t, InterpolationOrder ->
0]pExperiment = Plot[g[x], {x, 0, 10}, PlotPoints ->
200]Show[pExperiment, pTheory]Nice comparison!ff =
CompileEmpiricalCDF[Sort[data]];pExperiment2 = Plot[ff[x],
{x, 0, 10}, PlotStyle -> {Hue[.4, 1, .5], Thickness[.003]},
PlotPoints -> 200]Show[pTheory, pExperiment2, pExperiment]I
couldnšt see any differences between pExperiment and
pExperiment2enlarged.Timings:rx = Table[Random[Real, {0,
10}], {10000}];(s1 = g /@ rx); // Timing{1.082 Second,
Null}(s2 = ff /@ rx); // Timing{28.39 Second, Null}s1 ==
s2True So, I think, with deliberate use, your observation can
be well exploited as(2)!--Hartmut Wolf====First of all, Import
is not really meant for this sort of situation; it is better
to use ReadList. Suppose your Ūle is:1 A1 B2 B3 C4 B5 D5 E6
F7 Aand its name is adj.txt. Here is one way you can deal
with this case. First, read in the
Ūle:In[1]:=ss=ReadList[adj.txt,{Number,Word}]Out[1]={{1,A},{1
,B},{2,B},{3,C},{4,B},{5,D},{5,E},{6,F},{7,A}}Next, Ūnd how
many actors there
are:In[2]:=l=Length[Union[First[Transpose[ss]]]]Out[2]=
7Construct the adjacency
matrix:In[3]:=Table[If[Intersection[Cases[ss,{i,x_}->x,{1}],
Cases[ss,{j,x_}->x,{1}]]=!={},1,0](1-KroneckerDelta[i,j]),{i,
l},{j,l}]//MatrixFormOut[3]//MatrixForm=0 1 0 1 0 0 11 0 0 1 0
0 00 0 0 0 0 0 01 1 0 0 0 0 00 0 0 0 0 0 00 0 0 0 0 0 01 0 0 0
0 0 0The function ToAdjacencyMatrix form the Combinatorica
package constructs the adjacency matrix from a graph, so it
has no direct application to your problem.Andrzej
KozlowskiToyama International UniversityJAPAN> I need to
create an adjacency matrix from my data, which is currently >
in> the form of a .txt Ūle and is basically a two column
incidence list.> For example:>> 1 A> 1 B> 2 B> 3 C> . .> . .>
. .> m n>> Where 1 to m represent actors and A to n represent
events. My goal is > to> have an (m x m) matrix where cell
i,j equals 1 if two actors are> incident to the same event
(in the sample above, 1 and 2 are both> incident to B) and 0
otherwise (w/ zeros on the diagonal).>> Išm new to
Mathmatica, and so Išm on the steep part of the learning>
curve ... All Išve been able to Ūgure out so far is how to
get my> incidence list into the program using
Import[Ūlename.txt]. But then> what? How do I convert to the
adjacency matrix? Išve found the> ToAdjacencyMatrix[] command
in DiscreteMath`Combinatorica`, but I canšt> seem to get it to
work ...> Tom>>
**********************************************> Thomas P.
Moliterno> Graduate School of Management> University of
California, Irvine> tmoliter@uci.edu>
**********************************************>====When I
click on The Mathematica Book, I see no chapters, no
chapteralready fully expanded and there were chapter numbers
next to thechapter names.Anyway, what is Floor doing in a
chapter on Advanced Mathematics? Whyisnšt there a chapter on
notation? I suppose thatšs why an unexpandedIf you Go To
notation or notations, therešs no link to Mathematicaland
Other Notations. notation and notations yield different
links,too. Very strange.As for looking for LeftFloor -- that
makes sense only if you alreadyknow about LeftFloor. (Now I
do.) Links from Floor to LeftFloor andRightFloor would make a
lot of sense, but thatšs probably why theyšrenot there.very
kindly helping to make up for it.Bobby Treat-----Original
Message-----Išm reaching it by clicking on The Mathematica
Book and then clickingdie chapter and subchapters as
mentioned above. Itšs chapter 3.10.4 (inYou can also Ūnd it
by looking for [LeftFloor] in the Master index.Yours,
Alexander Dreyer-- / Alexander Dreyer, Dipl.-Math. -
Abteilung Adaptive Systeme / Fraunhofer Institut fuer Techno-
und Wirtschaftsmathematik (ITWM) Gottlieb-Daimler-Strasse,
Geb. 7^2=49/313 D-67663 Kaiserslautern /====One can make my
original code (below) much faster by adding an auxiliary
function and using dynamic
programming:In[1]:=ss=ReadList[adj.txt,{Number,Word}]Out[1]={
{1,A},{1,B},{2,B},{3,C},{4,B},{5,D},{5,E},{6,F},{7,A}}In[2]:=
l=Length[Union[First[Transpose[ss]]]]Out[2]=7a[i_] := a[i] =
Cases[ss, {i, x_} -> x,
{1}]In[4]:=Table[If[Intersection[a[i],a[j]]=!={},1,0](1-
KroneckerDelta[i,j]),{i,l},{j,
l}]//MatrixFormOut[4]//MatrixForm=0 1 0 1 0 0 11 0 0 1 0 0 00
0 0 0 0 0 01 1 0 0 0 0 00 0 0 0 0 0 00 0 0 0 0 0 01 0 0 0 0 0
0For large lists you will see a large difference in
speed.Andrzej KozlowskiToyama International UniversityJAPAN>
First of all, Import is not really meant for this sort of
situation; > it is better to use ReadList. Suppose your Ūle
is:>> 1 A> 1 B> 2 B> 3 C> 4 B> 5 D> 5 E> 6 F> 7 A>> and its
name is adj.txt. Here is one way you can deal with this >
case. First, read in the Ūle:> In[1]:=>
ss=ReadList[adj.txt,{Number,Word}]>> Out[1]=>
{{1,A},{1,B},{2,B},{3,C},{4,B},{5,D},{5,E},{6,F},{7,A}}>>
Next, Ūnd how many actors there are:>> In[2]:=>
l=Length[Union[First[Transpose[ss]]]]>> Out[2]=> 7>>
Construct the adjacency matrix:>> In[3]:=>
Table[If[Intersection[Cases[ss,{i,x_}->x,{1}],> >
Cases[ss,{j,x_}->x,{1}]]=!={},1,0](1-KroneckerDelta[i,j]),{i,
> l},{j,l}]//MatrixForm>> Out[3]//MatrixForm=> 0 1 0 1 0 0
1>> 1 0 0 1 0 0 0>> 0 0 0 0 0 0 0>> 1 1 0 0 0 0 0>> 0 0 0 0 0
0 0>> 0 0 0 0 0 0 0>> 1 0 0 0 0 0 0>> The function
ToAdjacencyMatrix form the Combinatorica package > constructs
the adjacency matrix from a graph, so it has no direct >
application to your problem.>> Andrzej Kozlowski> Toyama
International University> JAPAN>>> I need to create an
adjacency matrix from my data, which is currently >> in>> the
form of a .txt Ūle and is basically a two column incidence
list.>> For example:>> 1 A>> 1 B>> 2 B>> 3 C>> . .>> . .>> .
.>> m n>> Where 1 to m represent actors and A to n represent
events. My goal is >> to>> have an (m x m) matrix where cell
i,j equals 1 if two actors are>> incident to the same event
(in the sample above, 1 and 2 are both>> incident to B) and 0
otherwise (w/ zeros on the diagonal).>> Išm new to Mathmatica,
and so Išm on the steep part of the learning>> curve ... All
Išve been able to Ūgure out so far is how to get my>>
incidence list into the program using Import[Ūlename.txt].
But then>> what? How do I convert to the adjacency matrix?
Išve found the>> ToAdjacencyMatrix[] command in
DiscreteMath`Combinatorica`, but I >> canšt>> seem to get it
to work ...>> Tom>>
**********************************************>> Thomas P.
Moliterno>> Graduate School of Management>> University of
California, Irvine>> tmoliter@uci.edu>>
**********************************************>>>>====
script.Unfortunately when I start Mathreader I get the error
message No Ūle exists for resource 318 while the programme is
initialising the text resource manager. It is not possible to
click this message away.Can anybody help me with this
problem? The only hint I found was the which I didnšt Ūnd
very useful.Best,M.====I have written the following in a
JLink
program:drawArea=JavaNew[com.wolfram.jlink.MathCanvas];then I
perform an analysis of some data with output of this
form....plot =
DisplayTogether[ListPlot[],ListPlot[].etc...];Unfortunately,
at this point I dont know how to associate Œplotš
withdrawArea so that draw.Area@RepaintNow[]; repaints the
math screen to show the plots Išvecomputed...Išve tried
draw.Area@update[plot] and draw.Area@setImage[plot] but
thesedont work........how to Ūx?.........thanks.....jerry
blimbaum NSWC panama city, ū====Somehow, we at WRI seem to
have completely neglected to make any slide shows with
internal hyperlinks (probably because slide show content, by
its very nature, is just less likely to need internal
hyperlinks), and so we never saw this problem.Wešll Ūx it in
a future version of Mathematica, but in the mean time, herešs
a workaround. The workaround is not exactly perfect...it can
cause ūashing on screen and may not put the scrollbar in
exactly the right place...but it does work.With the slide
show youšre editing open, perform the following steps:*
Format->Edit Style Sheet... Choose Import Private Copy
(unless you want to modify the global style sheet, which I
wouldnšt recommend unless you need this in a lot of
notebooks).* In the stylesheet, open the Hyperlinks group.
Open the group for the Hyperlink style with that.* Place the
cell insertion point after the prototype for the Hyperlink
style and copy/paste the following in (pressing Yes when
asked whether to interpret the
cell):Cell[StyleData[Hyperlink, SlideShow],
ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ {
CompoundExpression[ FrontEnd`SetOptions[
FrontEnd`ButtonNotebook[ ], ScrollingOptions -> {
PagewiseDisplay -> False}], FrontEnd`NotebookLocate[ #2],
FrontEnd`SetOptions[ FrontEnd`ButtonNotebook[ ],
ScrollingOptions -> { PagewiseDisplay -> Inherited}]]}]&)}]*
Close the style sheet, then save your slideshow notebook.
Internal hyperlinks should now work.Sincerely,John
Fultzjfultz@wolfram.comUser Interface GroupWolfram Research,
Inc.>I am interested in creating a slide presentation using
slide view>mode in Mathematica 4.2. I would like to have
hyperlinks between>different slides ( e.g., a hyperlink in
say slide 6 takes me back to>slide 3).>>I have no difŪculty
in creating such a hyperlink (using tags) when>I am in the
author mode. But the hyperlink does not work when a>revert
back to slide view mode.>>I guess it has somthing to do with
the fact that in author view>mode your slides are a subset of
a single notebook but in slide view>mode each slide becomes a
distinct notebook.>>So my question: Is there a way to
reference the individual slides>using the hyperlink
command?>Brian====BRIEFMathematica seems to support 3 levels
of nesting:section, subsection, and subsubsection.Q: how can
I obtain more levels of nesting?I typically go 6-7 levels
deep.DETAIL= ConŪg =I am using Mathematica 4 on a Sun Sparc
at work.= Good Style or Not =I *know*, good style guides say
that you should nevergo more than 3 levels of section deep. I
disagree, anddonšt care. After editting my documents may only
endup with three levels of subsection nesting, but whileI am
editting, moving stuff around, I frequently want more.(Often,
I move a whole subtree of the document around,so that all
sections become subsubsections, subsectionsbecome
subsubsections, etc.) E.g. HTML originally had only H1, H2,
H3,but now it goes all the way up to H10.= How are Sections
Indicated? =I had hoped to deŪne my own styles, section,
subsection, sub2section, sub3section, etc.(ideally,
section(level) if styles can be parameterized).But, I cannot
see how to indicate how deeply nested a style is.In
particular, I want to use the folding or hidingfeature, so
that after I have deŪned sub6section, etc.I can hide
everything below a sub4section.== Nesting Level or Bracket?
==It appears that Mathematicašs sections have the level
hardwiredinto them. section is at level 0 subsection at level
1 subsubsection at level 2This means that the next section at
the same level terminatesa nesting group.The alternative
implementation is to use brackets, e.g. XML <section> level 1
<section> level 2 <section> level3 </section> level 2
</section> level 1 </section>This does not require the
section level to be hardwired into the style.Obviously,
Mathematica supports the Ūrst, hardwired or explicit
levelscheme. But, I cannot Ūgure out how to extend it to
multiple levels.I woulds be especially happy if the second
approach were available,but, again, do not know how to create
a Mathematica notebook stylethat accomplishes it.APOLOGIESIšm
sorry if this is a frequently asked question. I have searched
the FAQand the MathGroup archives to no avail. I did Ūnd stuff
on section numbering,which is related - I usually like Dewey
decimal section numbers extended to the samearbitrary depth -
but none on deeper nesting than subsubsection.====Perhaps, it
is a stupid question, but somehow, I was not able to Ūnd any
help on the following:I have a notebook, which does some
calculations and saves results onto disk. All I need is to
execute this notebook somewhere within another notebook.
Sometimes it is nice to keep some parts of the code separated
in different notebook Ūles, which depending on the
circumstances could be run manually via FrontEnd or within an
automated loop from a notebook.To a certain extent this
philosophy resembles, for example, what they have for scripts
in another system.Please, reply directly to
akolzine@uiuc.eduSincerely,Alexei.====Jens-Peer Kuska tells
me that he gets no crash with Mathematica 4.2, however after
workingwith this program for more than a month, trying every
possible permutation andmanipulation, and having it still
crash, I am still worried that upgrading to 4.2wonšt solve my
problem. Can anyone else with 4.2 try running this program for
me agood number of times, say 3 or 4, and see if they get
similar good results?Bernard GressDear Group,I have a program
to take the local polynomial non parametric regressionof two
variables. It uses Compile, unfortunately, and regularly,
butinconsistently, causes Mathematica to crash (in Win2K,
with I forgetwhat error, and in Win98/Mathematica4.0 with an
invalid memory access fromMathDLL.dll). I am running
Mathematica 4.1, and have this problem consistentlyon 4
different machines.Here is the code, if it doesnšt crash the
Ūrst time, it will the secondor third:(*w = Compile[{{xj,
_Real, 0}, {XX, _Real, 1}, {YY, _Real, 1}, {h, _Real,0}, {nn,
_Integer, 0}, {ord, _Integer,
0}},First[Inverse[Sum[Outer[Times, Table[If[Positive[q],
(XX[[i]] - xj)^q,1], {q, 0, ord}],Table[If[Positive[q],
(XX[[i]] - xj)^q, 1.], {q, 0,ord}]*E^(-0.5*((XX[[i]] -
xj)/h)^2)], {i, nn}]] .Sum[(Table[If[Positive[q], (XX[[i]] -
xj)^q, 1.], {q, 0, ord}]*YY[[i]])*E^(-0.5*((XX[[i]] -
xj)/h)^2), {i, nn}]]];*)(*!(tt = MemoryInUse[];
ListPlot[Table[{((i + 1))^2, (Timing[ nn = ((i + 1))^2;
[Epsilon] = Table[Random[]*3, {i,nn}]; testX = Table[i* .3 -
Random[]*2, {i, nn}]; testY = Table[Sin[i/3] - 2, {i, nn}] +
[Epsilon]; Map[w[#, testX, testY, .1 + Random[], nn, 0] &,
testX];])[([1])]/Second}, {i, 15}], PlotJoined ->True,
PlotLabel -> <Calculation Times as a function of n>];
MemoryInUse[] - tt)*)====I ran it 10 times on 4.2 with no
crashes.> Jens-Peer Kuska tells me that he gets no crash with
Mathematica 4.2,however after working> with this program for
more than a month, trying every possible permutationand>
manipulation, and having it still crash, I am still worried
that upgradingto 4.2> wonšt solve my problem. Can anyone else
with 4.2 try running this programfor me a> good number of
times, say 3 or 4, and see if they get similar goodresults?>
Bernard Gress>> Dear Group,>> I have a program to take the
local polynomial non parametric regression> of two variables.
It uses Compile, unfortunately, and regularly, but>
inconsistently, causes Mathematica to crash (in Win2K, with I
forget> what error, and in Win98/Mathematica4.0 with an
invalid memory access from> MathDLL.dll). I am running
Mathematica 4.1, and have this problemconsistently> on 4
different machines.>> Here is the code, if it doesnšt crash
the Ūrst time, it will the second> or third:>> (*> w =
Compile[{{xj, _Real, 0}, {XX, _Real, 1}, {YY, _Real, 1}, {h,
_Real,> 0},> {nn, _Integer, 0}, {ord, _Integer, 0}},>
First[Inverse[Sum[Outer[Times, Table[If[Positive[q], (XX[[i]]
- xj)^q,> 1], {q, 0, ord}],> Table[If[Positive[q], (XX[[i]] -
xj)^q, 1.], {q, 0,> ord}]*E^(-0.5*((XX[[i]] - xj)/h)^2)], {i,
nn}]] .> Sum[(Table[If[Positive[q], (XX[[i]] - xj)^q, 1.], {q,
0, ord}]*YY[[i]])*> E^(-0.5*((XX[[i]] - xj)/h)^2), {i,
nn}]]];> *)>> (*> !(tt = MemoryInUse[];> ListPlot[Table[{((i
+ 1))^2, (Timing[> nn = ((i + 1))^2; [Epsilon] =
Table[Random[]*3, {i,> nn}];> testX = Table[i* .3 -
Random[]*2, {i, nn}];> testY = Table[Sin[i/3] - 2, {i, nn}] +
[Epsilon];> Map[w[#, testX, testY, .1 + Random[], nn, 0] &,>
testX];])[([1])]/Second}, {i, 15}], PlotJoined ->> True,>
PlotLabel -> <Calculation Times as a function of n>];>
MemoryInUse[] - tt)> *)>Reply-To:
kuska@informatik.uni-leipzig.de====my Mathematica 4.2 does
not crash and you may upgradeyour version. Jens> Dear Group,>
I have a program to take the local polynomial non parametric
regression> of two variables. It uses Compile, unfortunately,
and regularly, but> inconsistently, causes Mathematica to
crash (in Win2K, with I forget> what error, and in
Win98/Mathematica4.0 with an invalid memory access from>
MathDLL.dll). I am running Mathematica 4.1, and have this
problem consistently> on 4 different machines.> Here is the
code, if it doesnšt crash the Ūrst time, it will the second>
or third:> (*> w = Compile[{{xj, _Real, 0}, {XX, _Real, 1},
{YY, _Real, 1}, {h, _Real,> 0},> {nn, _Integer, 0}, {ord,
_Integer, 0}},> First[Inverse[Sum[Outer[Times,
Table[If[Positive[q], (XX[[i]] - xj)^q,> 1], {q, 0, ord}],>
Table[If[Positive[q], (XX[[i]] - xj)^q, 1.], {q, 0,>
ord}]*E^(-0.5*((XX[[i]] - xj)/h)^2)], {i, nn}]] .>
Sum[(Table[If[Positive[q], (XX[[i]] - xj)^q, 1.], {q, 0,
ord}]*YY[[i]])*> E^(-0.5*((XX[[i]] - xj)/h)^2), {i, nn}]]];>
*)> (*> !(tt = MemoryInUse[];> ListPlot[Table[{((i + 1))^2,
(Timing[> nn = ((i + 1))^2; [Epsilon] = Table[Random[]*3,
{i,> nn}];> testX = Table[i* .3 - Random[]*2, {i, nn}];>
testY = Table[Sin[i/3] - 2, {i, nn}] + [Epsilon];> Map[w[#,
testX, testY, .1 + Random[], nn, 0] &,>
testX];])[([1])]/Second}, {i, 15}], PlotJoined ->> True,>
PlotLabel -> <Calculation Times as a function of n>];>
MemoryInUse[] - tt)> *)> I have followed Ted Ersekšs tips and
tricks>
(http://www.verbeia.com/mathematica/tips/tip_index.html)
about Compile,> and have tried changing all variable names
(works sometimes), removing> any hidden spaces, restructuring
the formulas, changing all 0šs to 0. Œs> and all 1šs to 1. Œs,
etcetera etcetera, but I still canšt comprehend> what the
problem might be. Doing w[[-2]] shows a list of op-code>
numbers, and one function name, Inverse[#1]&, so it seems to
me that> there are no problems with the use of Compile here.
Would anyone have> any thoughts???> Bernard Gress>
burnthebiscuit@netscape.net====Mr Kuska,4.2, I thought I had
solved this problem so many times and still it would
suddenlystart crashing again our of the blue, after 20 or 30
executions.Bernard>> my Mathematica 4.2 does not crash and
you may upgrade> your version.>====>Is it possible to
conŪgure a style sheet so that functions or data shown
in>graphs are coloured (colored :)) in working but black in
printing...without>actually re-executing the function that
created the graphics?>>thanks>>MikeThere is no way do this,
but with a couple of tricks you can create the same
effect.First, we generate 2 graphics. One is the normal
graphic. The other uses all black colors. This process is
automated by changing $DisplayFunction.$DisplayFunction = (
(* Ūrst the normal graphic *) Display[$Display, #1]; (* then
a new B&W graphic *) CellPrint[ Cell[GraphicsData[PostScript,
DisplayString[#1 /. {_RGBColor | _CMYKColor ->
GrayLevel[0]}]], GraphicsPrintout]]; #1 ) &;If you set this
and do a plot you will see the two graphics. The second
graphic will look strange because itšs in a different cell
style. Wešre going to use the cell styles to control which
graphic is shown.Next, open the style sheet (menu item
Format>Edit Style Sheet). Replace the Graphics/Printout style
with the following cell.Cell[StyleData[Graphics, Printout],
ShowCellBracket->False, CellMargins->{{0, 0}, {0, 0}},
CellElementSpacings->{CellMinHeight->0},
CellGroupingRules->NormalGrouping, CellFrameMargins->False,
CellSize->{Inherited, 0}, Background->None]It hides the
colored cell in the Printout environment. Then, add the
following styles.Cell[StyleData[GraphicsPrintout],
CellMargins->{{4, Inherited}, {Inherited, Inherited}},
CellGroupingRules->GraphicsGrouping,
CellHorizontalScrolling->True, PageBreakWithin->False,
GeneratedCell->True, CellAutoOverwrite->True,
ShowCellLabel->False,
DefaultFormatType->DefaultOutputFormatType,
LanguageCategory->None, FormatType->InputForm,
ImageMargins->{{43, Inherited}, {Inherited, 0}},
StyleMenuListing->None, FontFamily->Courier,
FontSize->10]Cell[StyleData[GraphicsPrintout, Printout],
ImageMargins->{{30, Inherited}, {Inherited, 0}},
MagniŪcation->0.8]Cell[StyleData[GraphicsPrintout, Working],
ShowCellBracket->False, CellMargins->{{0, 0}, {0, 0}},
CellElementSpacings->{CellMinHeight->0},
CellGroupingRules->NormalGrouping, CellFrameMargins->False,
CellSize->{Inherited, 0}, Background->None]It hides the B&W
cell in the Working environment and shows the cell in the
Printout environment.Finally, try it out.Plot[x, {x, 0, 1},
PlotStyle -> RGBColor[1, 0, 0], TextStyle -> {FontColor ->
RGBColor[0, 0,
1]}]---------------------------------------------------------
-----Omega ConsultingThe Ūnal answer to your Mathematica
needsSpend less time searching and more time
Ūnding.http://www.wz.com/internet/Mathematica.html====I was
completely wrong about the pattern list_?(VectorQ[#,
NumericQ] &).I had forgotten about that notation and didnšt
think to look it up.Sorry!Bobby-----Original
Message-----evaluates fairly slowly (even if it has been
Compiled). SinceInterpolationFunction evaluates so much
faster in general, Išve triedto use Interpolation with
InterpolationOrder -> 0. The problem is thatthe resulting
InterpolatingFunction doesnšt behave the way (I think)it
ought to. For example, letg = Interpolation[{{1, 1/3}, {2,
2/3}, {3, 1}}, InterpolationOrder ->0]Then, g /@ {1, 2, 3}
returns {2/3, 2/3, 1} instead of {1/3, 2/3, 1}.In addition, g
is continuous from the left rather than from the
right.Obviously I am not aware of the considerations that
went intodetermining the behavior of InterpolationFunction
whenInterpolationOrder -> 0.So I have two questions: (1) Does
anyone have any opinions about how InterpolatingFunctionought
to behave with InterpolationOrder -> 0?(2) Does anyone have a
faster way to evaluate an empirical CDF than acompiled Which
function?By the way, herešs my current
version:CompileEmpiricalCDF[list_?(VectorQ[#, NumericQ] &)]
:= Block[{x}, Compile[{{x, _Real}}, Evaluate[ Which @@
Flatten[ Append[ Transpose[{ Thread[x < Sort[list]], Range[0,
1 - 1/#, 1/#] & @ Length[list] }], {True, 1}]] ]]]--Mark====I
was wondering if a way exists to move the main menu bar of
the front end.I am talking about the bar that occupies the
top of the screen and extends,by default, from the left end
of the screen to the right. It has a lot ofblank space to the
right of the menus. Making it narrower frees up someimportant
screen real estate, if you want to reach icons under it. How
couldI make Mathematica do that automatically? I cannot seem
to Ūnd it in thePreferences or in the sections on
manipulating notebooks.Nicholas====The various
Intro/Mathematica Programming books Išve seen (e.g.
Gray:Mastering Mathematica -- now out of print but have a
copy on order) andMathematical Navigator and other books Išve
seen are based on V3. Is thereany Intro/Programming book based
on V4 (or even updated for 4.2) that isavailable besides the
on-line book?Does someone have notes that update the V3
books?====I puzzled over this and came up with an
explanation. The phenomenonsomebody noticed is that the
fractional part of(Sqrt[3] + Sqrt[2])^(2*n)approaches 1 from
below as n (integer) gets large. Hence, for instance,Mod[
N[(Sqrt[2] + Sqrt[3])^2002, 2000], 100]has a string of 996
nines after the decimal point. Well, herešs thereason for
that.Let x and y be the square roots of any integers and let
n be an eveninteger power. Then (x + y)^n == Sum[Binomial[n,
m]*y^m*x^(n - m), {m, 0, n}];and (x - y)^n ==
Sum[(-1)^m*Binomial[n, m]*y^m*x^(n - m), {m, 0, n}];so,
adding the two equations, we have (x + y)^n + (x - y)^n ==
Sum[Binomial[n, m]*y^m*x^(n - m)* (1 + (-1)^m), {m, 0, n}];1
+ (-1)^m is 0 when m is odd and 2 when m is even, so (x +
y)^n + (x - y)^n == 2*Sum[Binomial[n, 2*k]*y^(2*k)*x^(n -
2*k), {k, 0, n/2}];Since n is even, all powers of x and of y
are even in this expression.Since x and y are the square
roots of integers, it follows that all theterms are integers.
Hence(x + y)^n + (x - y)^nis integer. Now we assume, in
addition to our other conditions, that0 < x - y < 1In that
case, (x - y)^n approaches 0 from above as n gets large,
andFractionalPart[(x + y)^n] == 1 - (x - y)^nwhich approaches
one from below as n gets large. The number of ninesafter the
decimal point can be calculated asFloor[(-n)*Log[10, Sqrt[3]
- Sqrt[2]]]For n=2002 (the example we started with, this is
indeed)Floor[-2002*Log[10, Sqrt[3] - Sqrt[2]]]996Bobby
Treat-----Original Message-----> Ūnally take the result
modulo 10^n.> Mod[Floor[ N[(Sqrt[2] + Sqrt[3])^2002 , 1000]],
10^2]> results in 9, so the last two digits before the decimal
point are 09.> With a slight modiŪcation we can Ūnd the Ūrst n
digits after thedecimal> point. Simply Ūnd the last n digits
before the decimal point of 10^ntimes> the number.>
Mod[Floor[ N[10^2 (Sqrt[2] + Sqrt[3])^2002 , 1000]], 10^2]>
results in 99, so these are the digits you are interested
in.> But there is something curious about this number.>
Mod[Floor[N[10^1000 (Sqrt[2] + Sqrt[3])^2002 , 2000]],
10^1000]> results in 996 digits 9 followed by 7405.> You can
also play with the following command, resulting in the
digitsaround> the decimal point:> Mod[ N[(Sqrt[2] +
Sqrt[3])^2002, 2300], 10^6]> The decimal expansion of
(Sqrt[2]+Sqrt[3])^2002 contains a sequence of> 997
consecutive digits 9. Do you have any idea why?Not unusual.
Try other even exponents and there should be large blocks of
9s . Smaller number for small exponents. Some kind of
propagation of 9s as the exponent grows. Try other odd values
for 3 and the same thinghappens for some values. Also for
other values for 2.Larry> Fred Simons> Eindhoven University
of Technology> ====>I have written the following in a JLink
program:>drawArea=JavaNew[com.wolfram.jlink.MathCanvas];>>
then I perform an analysis of some data with output of this
form....>>plot =
DisplayTogether[ListPlot[],ListPlot[].etc...];>Unfortunately,
at this point I dont know how to associate Œplotš
with>drawArea so that>>draw.Area@RepaintNow[]; repaints the
math screen to show the plots Išve>computed...>Išve tried
draw.Area@update[plot] and draw.Area@setImage[plot] but
these>dont work........how to
Ūx?.........thanks.....Jerry,You use the
MathCanvas.setMathCommand() method to specify the Mathematica
command that is used to generate the image to display:
drawArea.setMathCommand(plot);You will Ūnd a very simple
example of this in section 1.2.8.2 of the J/Link User Guide
(in the Help Browser hierarchy, this is found at JLink/Part
1. Installable Java/Drawing and Displaying Images in Java
Windows/Showing Graphics and Typeset Expressions).You do not
need to use the repaintNow() method unless you are updating
the image continuously in response to a user action, like
dragging a slider.Todd GayleyWolfram
Research====try:(((((3^(1/2))^(1/2))^(1/2))^(1/2))^(1/2))/(6^
(1/2))Out:1/(Sqrt[2]*3^(15/32))You can see what happened:1st:
6^(1/2) = 2^(1/2)*3^(1/2);2nd: subtraction of the 3šs
exponents (1/2)^5 - 1/2 = 1/32 - 16/32 = -15/32.MATHEMATICA
does these straightforward simpliŪcations without you having
toinvoke Simplify[].Matthias BodeSal. Oppenheim jr. & Cie.
KGaAKoenigsberger Strasse 29D-60487 Frankfurt am
MainGERMANYMobile: +49(0)172 6 74 95 77Internet:
http://www.oppenheim.de-----UrsprĖ.b9ngliche
Nachricht-----Gesendet: Freitag, 13. September 2002 07:14An:
mathgroup@smc.vnet.netBetreff: nth rootsSimplify:5nth sqrt
(3)/sqrt (6)5nth Sqrt(3)------------ Sqrt(6)1st.. is this the
correct notation for use on a computer2nd.. how is the
solution solved, step by step please.====Dear all,I want to
include some Mathematica code in some lecture notes I am
writing.For that purpose, I use the LaTeX style provided with
Mathematica. It seemsto be working Ūne when I produce a
document that is just Mathematica code.The problem - as
expected - is that when I include the Mathematica code asa
part of a larger document, there are (1) clashes with
existing packages(2) some LaTeX elements, such as the
equations, start to use theMathematica fonts too. My
objective would be to typeset a document instandard LaTeX,
where only the Mathematica code has the Mathematica
feeling.Does anyone know of any style available that is
completely seperated fromall other LaTeX
deŪnitions?BestKyriakos_____+**+____+**+___+**+__+**+_
Kyriakos ChourdakisLecturer in Financial EconomicsURL:
http://www.qmw.ac.uk/~te9001tel: (++44) (+20) 7882 5086Dept
of EconomicsUniversity of London, QMLondon E1
4NSU.K.Reply-To: kuska@informatik.uni-leipzig.de====a) you
must change the fonts, because the Mathematica symbols are
designed to Ūt to Times/Helvetica/Courier and not to Computer
Modern ! Mixing Computer Modern and the Mathematica fonts look
not very nice because Comuter Modern has smaler strokes. You
need atleast wrisym.sty, i.e.
usepackage[monospacemath]{wrisym} but this will switch the
main document font to Times and the mono-space font to
courier. With Mathematica 4.2 you can use Adobe-Garamond and
Janson as document fonts too, with
usepackage[monospacemath,garamond]{wrisym} but this are
commercial fonts and you have to buy it from Adobe.b) If you
like the narrow monotype fonts similar to Computer Modern you
can use the CMTT fonts, now included in Mathematica 4.2 Just
use the package option usepackage[cmtt]{wrisym}c) I have a
TeX frontend that use only the wrisym package and a
preprocessor. It run Ūne with the most style Ūles But you
have to use TeX notebooks that are essential TeX Ūles with
mathinput environments. The TeX frontend send the contents of
the mathinput environments to the kernel and paste the output
into the Ūnal TeX Ūle. It also does some fancy formating with
the Mathematica input and replace -> with Rule, :> with
RuleDelayed .. You can have the program and the style Ūles if
you like. Jens> Dear all,> I want to include some Mathematica
code in some lecture notes I am writing.> For that purpose, I
use the LaTeX style provided with Mathematica. It seems> to be
working Ūne when I produce a document that is just Mathematica
code.> The problem - as expected - is that when I include the
Mathematica code as> a part of a larger document, there are>
(1) clashes with existing packages> (2) some LaTeX elements,
such as the equations, start to use the> Mathematica fonts
too. My objective would be to typeset a document in> standard
LaTeX, where only the Mathematica code has the Mathematica
feeling.> Does anyone know of any style available that is
completely seperated from> all other LaTeX deŪnitions?> Best>
Kyriakos> _____+**+____+**+___+**+__+**+_> Kyriakos
Chourdakis> Lecturer in Financial Economics> URL:
http://www.qmw.ac.uk/~te9001> tel: (++44) (+20) 7882 5086>
Dept of Economics> University of London, QM> London E1 4NS>
U.K.====What about this problem? I have tried the previous
suggestions but none ofthem worked. Will Wolfram help us in
this problem?====Have you tried deleting
...
Mathematica4.2DocumentationEnglishMainBookBrowserIndex.nbThere
are two BrowserIndex Ūles. Only delete the .nb one (or just
move itsomewhere else).> What about this problem? I have
tried the previous suggestions but none of> them worked. Will
Wolfram help us in this problem?>>>====First a thank you to
all those who pointed out to me (and othersapparently) about
the various ways to input the symbolic form of Floor.On a
larger note, it surprises me that no one has yet published a
book onhow to use Mathematica to write a technical document.
The number of questionsthat appear on this board concerning
the details of how to do itsuggests the need for such
documentation. Išm speculating when I say thatmany users give
up in frustration and go back to (or learn) Latex. Atleast
there are bookcases full of learning Latex. I myself have 5!I
donšt mean this to sound like a rant, but using Mathematica to
publish adocument is, for most of us, not worth the
effort.Jack====I often run into this difŪculty: when
designing a program, say as amodule, and testing it for
various inputs, I get wrong answers. What todo? I use a
method that works for me but may not be the best available.I
want to show my method and then ask a question about how it
can beimporved. (Oh yes, I abandoned Trace a long time
ago!)myFunction[f_] := Module[ {L1,L2,L3},L1 = ... ;L2 = ...
;l3 = ... ; Ūnal step ]To see what went wrong, I use (* *)
selectively as follows:Stage 1myFunction[f_] := Module[
{L1,L2,L3},L1 = ... (*;L2 = ... ;L3 = ... ; Ūnal step *)
]Thus I see if L1 worked as expected. The next step is to put
(* after L2and see if this works. I continue this til the
bitter end and I usuallyŪnd my errors.My question; The
process of moving (* *) step by step through theprogram is
quite tedious when the code has lots more lines. What I
wouldlike is a meta-program which (like FoldList) does this
job for me. Theoutput of this meta-program is the list of
outputs of each line in themodule, probably best printed as a
column.This sounds like Trace but my problem with Trace is it
is terriblydifŪcult to read. For the not-so-subtle
programming I do, the only thingI need is what expression is
returned line by line.Any advice? All remarks are
appreciated!JackReply-To:
kuska@informatik.uni-leipzig.de====does help you ? Jens> I
often run into this difŪculty: when designing a program, say
as a> module, and testing it for various inputs, I get wrong
answers. What to> do? I use a method that works for me but
may not be the best available.> I want to show my method and
then ask a question about how it can be> imporved. (Oh yes, I
abandoned Trace a long time ago!)> myFunction[f_] := Module[
{L1,L2,L3},> L1 = ... ;> L2 = ... ;> l3 = ... ;> Ūnal step>
]> To see what went wrong, I use (* *) selectively as
follows:> Stage 1> myFunction[f_] := Module[ {L1,L2,L3},> L1
= ... (*;> L2 = ... ;> L3 = ... ;> Ūnal step *)> ]> Thus I
see if L1 worked as expected. The next step is to put (*
after L2> and see if this works. I continue this til the
bitter end and I usually> Ūnd my errors.> My question; The
process of moving (* *) step by step through the> program is
quite tedious when the code has lots more lines. What I
would> like is a meta-program which (like FoldList) does this
job for me. The> output of this meta-program is the list of
outputs of each line in the> module, probably best printed as
a column.> This sounds like Trace but my problem with Trace is
it is terribly> difŪcult to read. For the not-so-subtle
programming I do, the only thing> I need is what expression
is returned line by line.> Any advice? All remarks are
appreciated!> Jack====Išve enhanced the code for
PartialEvaluation (see my post, which Iincorrectly attaced to
someone elsešs reply). It is somewhat moreūexible now and does
more error checking. It Ūnds the rule inDownValues that
matches f[args] (allowing for more than one rule
inDownValues); it allows the user to specify the position of
the mainCompoundExpression; and it allows the user to specify
an expressionthat gets appended to the truncated
CompoundExpression (and henceevaluated and returned). The
package is just a bit too large to posthere. It can be
downloaded from my web site
athttp://www.markŪsher.net/~meŪsher/mma/
mathematica.htmlNevertheless, I can give an outline of the
code here (absent the errorchecking
stuff).PartialEvaluation[f[args], n, expr]:(* get the
DownValues and turn them off *)dv =
DownValues[f];DownValues[f] = {};(* Ūnd the rule that matches
*)matches = Position[MatchQ[f[args], #]& /@ dv[[All, 1]],
True];match = dv[[ matches[[1, 1]] ]];(* Ūnd the main
CompoundExpression *)ppos = Position[match,
HoldPattern[CompoundExpression[__]]];pos =
First[Sort[ppos]];(* extract, truncate, append to, and
reinsert it *)held = Extract[match, pos, Hold];held =
ReplacePart[held, Sequence, {1, 0}];held = Take[held, n];held
= Join[held, Hold[expr]];held =
ReplacePart[Hold[Evaluate[held]], CompoundExpression, {1,
0}];match = ReplacePart[match, held, pos, 1];(* apply the
modiŪed rule and restore DownValues *)result = f[args] /.
match;DownValues[f] = dv;result--Mark> I often run into this
difŪculty: when designing a program, say as a> module, and
testing it for various inputs, I get wrong answers. What to>
do? I use a method that works for me but may not be the best
available.> I want to show my method and then ask a question
about how it can be> imporved. (Oh yes, I abandoned Trace a
long time ago!)> myFunction[f_] := Module[ {L1,L2,L3},> L1 =
... ;> L2 = ... ;> l3 = ... ;> Ūnal step> ]> To see what went
wrong, I use (* *) selectively as follows:> Stage 1>
myFunction[f_] := Module[ {L1,L2,L3},> L1 = ... (*;> L2 = ...
;> L3 = ... ;> Ūnal step *)> ]> Thus I see if L1 worked as
expected. The next step is to put (* after L2> and see if
this works. I continue this til the bitter end and I usually>
Ūnd my errors.> My question; The process of moving (* *) step
by step through the> program is quite tedious when the code
has lots more lines. What I would> like is a meta-program
which (like FoldList) does this job for me. The> output of
this meta-program is the list of outputs of each line in the>
module, probably best printed as a column.> This sounds like
Trace but my problem with Trace is it is terribly> difŪcult
to read. For the not-so-subtle programming I do, the only
thing> I need is what expression is returned line by line.>
Any advice? All remarks are appreciated!> Jack====JM,You are
certainly correct in wanting to make a tidier plot! Legends
areoften poor because they actually distract from the message
of the data.However, the best solution will depend on the
particular nature of yourdata. If there are not too many
curves you could perhaps put Text labelsright on top of each
curve. If there are many curves, or some of them areclose
together, use arrows for some of them. If you have a really
largenumber of curves, then maybe a different approach is
needed.It is probably not possible to make a useful general
routine for labeledarrows because the best placement would
depend upon the particular nature ofthe graph. So, to make a
nice graphic you will have to do some hand work,specifying
each Arrow and Text label. You can actually click the
coordinatesoff the graph to put into the Arrow and Text
statements.If you want to actually show me your plot, I could
try to make somesuggestions.David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/ _ |
thankyou====>I am interested in creating a slide presentation
using slide view>mode in Mathematica 4.2. I would like to have
hyperlinks between>different slides ( e.g., a hyperlink in say
slide 6 takes me back to>slide 3).>>I have no difŪculty in
creating such a hyperlink (using tags) when I>am in the
author mode. But the hyperlink does not work when a
revert>back to slide view mode.>>I guess it has somthing to
do with the fact that in author view mode>your slides are a
subset of a single notebook but in slide view mode>each slide
becomes a distinct notebook.>>So my question: Is there a way
to reference the individual slides>using the hyperlink
command?>BrianIf you set PagewiseScrolling->True, then your
hyperlink will work. Go to the Option Inspector and set the
scope to Notebook. Then go to the
optionNotebookOptions|Window
Properties|ScrollingOptions|PagewiseScrollingAnd set it to
True. The disadvantage to this setting is that you can now
only have one screenšs worth of contents per slide. (But then
it is a slide show after all, so maybe thatšs not so
bad.)-Dale====Does anybody know how to calculate in
Mathematica:a)empirical CDF,b)empirical PDF,c)normal
QQ-plot;d)QQ-plot two different random
samples?!Swidrygiello.====DeclarePackage[OtherPackage`,
{BigVariable} ]. The idea was to preventthe loading of a
large Ūle in routine cases. However, when ThisPackage`deŪnes
its functions, inside the Private` area, it includes a
conditionalcall to BigVariable. It turns out that
OtherPackage is loaded when thatfunction is deŪned. I was
wondering if there is a way to avoid this.Roughly speaking,
here is the setting:BeginPackage[ OtherPackage`
]BigVariable::usage=exampleBegin[Private`]BigVariable=Table[x
y,{x,1000},{y,1000}]End[ ]EndPackage[ ]BeginPackage[
ThisPackage`
]function::usage=exampleBegin[Private`]function[x_]:=Module[{
y},y=If[x>1000,BigVariable[[x]],x] ]End[ ]EndPackage[ ]I
donšt want OtherPakage to be loaded unless function[x] is
called withx>1000 but it loads when function is deŪned. Any
ideas?NicholasReply-To:
kuska@informatik.uni-leipzig.de====BeginPackage[
OtherPackage`
]BigVariable::usage=exampleBegin[`Private`]BigVariable=Table[
x y,{x,1000},{y,1000}]End[ ]EndPackage[ ]Print[Loading
...];andBeginPackage[ ThisPackage`
]function::usage=exampleBegin[`Private`]function[x_]:=Module[
{y}, y=If[x>1000, Needs[OtherPackage`];
OtherPackage`BigVariable[[x]], x] ]End[ ]EndPackage[ ]should
do that. Jens> DeclarePackage[OtherPackage`, {BigVariable} ].
The idea was to prevent> the loading of a large Ūle in routine
cases. However, when ThisPackage`> deŪnes its functions,
inside the Private` area, it includes a conditional> call to
BigVariable. It turns out that OtherPackage is loaded when
that> function is deŪned. I was wondering if there is a way
to avoid this.> Roughly speaking, here is the setting:>
BeginPackage[ OtherPackage` ]> BigVariable::usage=example>
Begin[Private`]> BigVariable=Table[x y,{x,1000},{y,1000}]>
End[ ]> EndPackage[ ]> BeginPackage[ ThisPackage` ]>
function::usage=example> Begin[Private`]>
function[x_]:=Module[{y},> y=If[x>1000,BigVariable[[x]],x] ]>
End[ ]> EndPackage[ ]> I donšt want OtherPakage to be loaded
unless function[x] is called with> x>1000 but it loads when
function is deŪned. Any ideas?> Nicholas==== Dynamic
programming might help: replace the line deŪning
BigVariablewithBigVariable:=BigVariable=Table[x
y,{x,1000},{y,1000}]Then the table will not be calculated
unless it is needed.(See the section on Functions that
Remember Values They Have Found, 2.4.9in the Mathematica
Book).John Jowett> DeclarePackage[OtherPackage`,
{BigVariable} ]. The idea was to prevent> the loading of a
large Ūle in routine cases. However, when ThisPackage`>
deŪnes its functions, inside the Private` area, it includes a
conditional> call to BigVariable. It turns out that
OtherPackage is loaded when that> function is deŪned. I was
wondering if there is a way to avoid this.> Roughly speaking,
here is the setting:>> BeginPackage[ OtherPackage` ]>
BigVariable::usage=example> Begin[Private`]>
BigVariable=Table[x y,{x,1000},{y,1000}]> End[ ]> EndPackage[
]>> BeginPackage[ ThisPackage` ]> function::usage=example>
Begin[Private`]> function[x_]:=Module[{y},>
y=If[x>1000,BigVariable[[x]],x] ]> End[ ]> EndPackage[ ]> I
donšt want OtherPakage to be loaded unless function[x] is
called with> x>1000 but it loads when function is deŪned. Any
ideas?>> Nicholas>>>====Could you send a sample example of
your Ūle -- just a few records.Tomas GarzaMexico City-----
Original Message -----> This appears to pull in a line. Now I
want to take characters 25 to 110 toget just the stuff I
want:> y1=StringTake[y ,{25,110}];>> Herešs the output.
StringTake doesnšt seem to work.>321. 317. 367. -115. 126.
146. -410. -426.000000EF 75.},{25, 110}]>> It doesnšt take
loading another package as far> as I can tell from the help.
Išm thinking that it doesnšt work becauseitšs trying to work
on a list> rather than a string. Išve tried Flatten, and
other stuff to try to get tojust a string and not a list but>
nothing has worked so far. Išm a long way from getting to
those numbersin there but heck, I> canšt even get to the
string. Can anyone point me in the right direction?>>====Išve
got to extract some numbers from a Ūle that are in lines of
text. Since the line contents are not numbers, I presume I
must pull the line out as a string. Here I start by pulling
out just one line:inFile = OpenRead[197-tst.txt]y =
ReadList[inFile, String, 1, RecordLists ->
True]Close[inFile];This appears to pull in a line. Now I want
to take characters 25 to 110 to get just the stuff I
want:y1=StringTake[y ,{25,110}];Herešs the output. StringTake
doesnšt seem to work.It doesnšt take loading another package
as faras I can tell from the help. Išm thinking that it
doesnšt work because itšs trying to work on a listrather than
a string. Išve tried Flatten, and other stuff to try to get to
just a string and not a list butnothing has worked so far. Išm
a long way from getting to those numbers in there but heck,
Icanšt even get to the string. Can anyone point me in the
right direction?====this thing. Here is what Išm using
now:In[1]:=a=ReadList[197-tst.txt,Word,RecordLists->True,
WordSeparators->None];In[2]:=ūds={{25,31},
{32,39},{40,46},{47,54},{55,62},{63,70},{71,78},{79,86},{87,
94},{95,102},{103,110}};DeŪne this
function:In[3]:=f[x_]:=ToExpression[StringTake[x[[1]],#]]&/@
ūdsNow, map this function onto
a:In[8]:=f/@aOut[8]={{5935.8,5946.66,27.06,-1281.9,-229.,321.
,317.,367.,-115.,126.,146.},{5935.8....************The data
is now in a matrix of numbers and each variable is in a
column.Now I just transpose this to get each variable in a
row for easy access (apparentlyMathematica has no way to
directly access a column in a matrix, you gotta transpose I
think)Rob> Išve got to extract some numbers from a Ūle that
are in lines of text. Since the line contents are not
numbers, I presume I must pull the line out as a string. Here
I start by pulling out just one line:>> inFile =
OpenRead[197-tst.txt]> y = ReadList[inFile, String, 1,
RecordLists -> True]> Close[inFile];>> This appears to pull
in a line. Now I want to take characters 25 to 110 to get
just the stuff I want:> y1=StringTake[y ,{25,110}];>> Herešs
the output. StringTake doesnšt seem to work.> It doesnšt take
loading another package as far> as I can tell from the help.
Išm thinking that it doesnšt work because itšs trying to work
on a list> rather than a string. Išve tried Flatten, and other
stuff to try to get to just a string and not a list but>
nothing has worked so far. Išm a long way from getting to
those numbers in there but heck, I> canšt even get to the
string. Can anyone point me in the right direction?>====> The
data is now in a matrix of numbers and each variable is in a
column.> Now I just transpose this to get each variable in a
row for easy access> (apparently Mathematica has no way to
directly access a column in a matrix,> you gotta transpose I
think)You can use All:In[1]:= mat = {{a,b,c},{d,e,f}};In[2]:=
mat[[All,2]]Out[2]= {b, e}--Bhuvanesh,Wolfram Research.====>
inFile = OpenRead[197-tst.txt]> y = ReadList[inFile, String,
1, RecordLists -> True]> Close[inFile];> This appears to pull
in a line. Now I want to take characters > 25 to 110 to get
just the stuff I want:> y1=StringTake[y ,{25,110}];> Herešs
the output. StringTake doesnšt seem to work.> -1281.9 -229.
321. 317. 367. -115. 126. > 146. -410. -426.000000EF 75.},
{25, 110}]> It doesnšt take loading another package as far>
as I can tell from the help. Išm thinking that it doesnšt >
work because itšs trying to work on a list> rather than a
string. Išve tried Flatten, and other stuff to > try to get
to just a string and not a list but> nothing has worked so
far. Išm a long way from getting to > those numbers in there
but heck, I> canšt even get to the string. Can anyone point
me in the > right direction?You might try lst =
Read[StringToStream[y1], {Word, Table[Number, {16}],
Word,Number}]//Flatten;You can then pick the numbers (or
words) from the list lst.A quicker alternative might be as
follows:-iŪle = ReadList[197-tst.txt, {Word, Table[Number,
{16}], Word,
Number}];Dave.========================================== Dr.
David Annetts EM Modelling Analyst Australia
David.Annetts@csiro.au=======================================
========First of all, I used ReadList directly, with Word
instead of String, andwith the option WordSeparators -> None,
like this (I presume your Ūle isadequately located, so that
there is no problem in Ūnding it):In[1]:=a =
ReadList[197-tst.txt, Word, RecordLists -> True,
WordSeparators ->None];This allowed me to examine your
records and I found out that in this wayeach record comes out
as a list of length 1:
In[2]:=Head[a[[1]]]Out[2]=ListIn[3]:=Length[a[[1]]]Out[3]=
1That is,In[3]:=a[[1]]Out[3]=317. 367. -115. 126. 146. -410.
-426.000000EF 75.}In[4]:=StringLength[a[[1,1]]]Out[4]=140and
the characters you want areIn[5]:=StringTake[a[[1,1]], {25,
110}]Out[5]=5935.80 5946.66 27.06 -1281.9 -229. 321. 317.
367. -115. 126. 146.So far, so good. It seems that you want
these 11 numbers, OK? The problemnow, I think, is that this
is just a string and I can think of no easy wayto convert it
precisely into a list of 11 real numbers. Then, I suggest
youread the Ūle in a different way, without the
WordSeparators
option:In[6]:=b=ReadList[197-tst.txt,Word,RecordLists ->
True];In[7]:=b[[1]]Out[7]=115.,126.,146.,-410.,-426.000000EF,
75.}In[8]:=Head[b[[1]]]Out[8]=ListIn[9]:=Length[b[[1]]]Out[9]
=19so that each record is now a list of 19 strings. What you
want is strings 6to 16, but converted to reals (unless Išm
being presumptuous). This willachieve
that:In[10]:=ToExpression[Take[b[[1]],{6,16}]]Out[10]={
5935.8,5946.66,27.06,-1281.9,-229.,321.,317.,367.,-115.,126.,
146.} Now you have a nice list of real numbers to work with.
You can do this forthe whole Ūle like
this:In[11]:=ToExpression[Take[#,{6,16}]&/@b]; I hope this
will solve your problem.Tomas GarzaMexico City> -----
Original Message -----> Sent: Friday, September 13, 2002
12:14 AM>> Išve got to extract some numbers from a Ūle that
are in lines of text.> Since the line contents are not
numbers, I presume I must pull the lineout> as a string. Here
I start by pulling out just one line:>> inFile =
OpenRead[197-tst.txt]> y = ReadList[inFile, String, 1,
RecordLists -> True]> Close[inFile];>> This appears to pull
in a line. Now I want to take characters 25 to 110to> get
just the stuff I want:> y1=StringTake[y ,{25,110}];>> Herešs
the output. StringTake doesnšt seem to work.>-229.> 321. 317.
367. -115. 126. 146. -410. -426.000000EF 75.},> {25, 110}]>>
It doesnšt take loading another package as far> as I can tell
from the help. Išm thinking that it doesnšt work because> itšs
trying to work on a list> rather than a string. Išve tried
Flatten, and other stuff to try to getto> just a string and
not a list but> nothing has worked so far. Išm a long way
from getting to those numbers> in there but heck, I> canšt
even get to the string. Can anyone point me in the
rightdirection?>>--------------------------------------------
------------------------====> BRIEF> Mathematica seems to
support 3 levels of nesting:> section, subsection, and
subsubsection.> Q: how can I obtain more levels of nesting?>
I typically go 6-7 levels deep.Well, I started by editting
the notebook in a text editor(having used Edit Style Sheet to
unshare),and I created stylesSub3section ... Sub9section,with
Dewey decimal numbering 1.2.3.4.5.6.7.8.9.10I got collapsable
group working, by mucking with
CellGroupingRules->{SectionGrouping, 100},changing the
number.I am not sure what the number is, but Išm guessingit
may be a pixel count for the nesting boxesat the left of the
screen.Išm a bit worried about whether this is fragile.===Was
there any way to accomplish this from the GUI?Or was using a
text editor mandatory?===Related: this exercise makes it
obvious that acommon operation is to push down or pull upa
whole subtree - e.g. a Subsection becomes aSubsubsection, a
Subsubsection becomes aSub3section, etc. The sort of thing
that MicrosoftWord does with Outline mode.Q: has anyone got
something like Outline modefor Mathematica?Išm tempted to go
the other way, and ask if anyonehas a front end that allows
Mathematica to be embeddedin Word documents. But that might
be a hassle,since Word runs on Windows, and the
Mathematicalicence I have access to runs on Suns.====> inFile
= OpenRead[197-tst.txt]> y = ReadList[inFile, String, 1,
RecordLists -> True]> Close[inFile];> This appears to pull in
a line. Now I want to take characters > 25 to 110 to get just
the stuff I want:> y1=StringTake[y ,{25,110}];> Herešs the
output. StringTake doesnšt seem to work.> -1281.9 -229. 321.
317. 367. -115. 126. > 146. -410. -426.000000EF 75.}, {25,
110}]> It doesnšt take loading another package as far> as I
can tell from the help. Išm thinking that it doesnšt > work
because itšs trying to work on a list> rather than a string.
Išve tried Flatten, and other stuff to > try to get to just a
string and not a list but> nothing has worked so far. Išm a
long way from getting to > those numbers in there but heck,
I> canšt even get to the string. Can anyone point me in the >
right direction?You might try lst = Read[StringToStream[y1],
{Word, Table[Number, {16}], Word,Number}]//Flatten;You can
then pick the numbers (or words) from the list lst.A quicker
alternative might be as follows:-iŪle = ReadList[197-tst.txt,
{Word, Table[Number, {16}], Word,
Number}];Dave.========================================== Dr.
David Annetts EM Modelling Analyst Australia
David.Annetts@csiro.au=======================================
========I was and do have problems reading Ūles Exported Ūles
in AdobeIllustrator format (.ai) from Adobe Illustrator
10.0.3.With an upgrade to Adobe Illustrator 10.0.3, I can
read EPS Ūlescreated by Mathematica 4.0.1.A message from tech
support at Wolfram implies that Adobe Illustratorformat Ūles
can be read ONLY if you are using the correct Adobe
andWolfram programs. Additionally, Mathematica is going to
favor the exportof EPS Ūles (is that clear enough?).My
approach to using Mathematica as a source of images for
AdobeIllustrator 10.0.3 is:Create the image in
MathematicaExport[ImageFileName.eps,
theMathematicaImageCreated, EPS]Open the Ūle
ImageFileName.eps within Adobe Illustrator 10.0.3Save the Ūle
as a .ai ŪleBe sure to check the ImageSize as it is not often
what you toldMathematica.====In his not-so-subtle way,
Jens-Peer is trying to tell you that since yousee z in the
error message, thatšs what NIntegrate saw too. z is asymbol,
and NIntegrate needs numbers. Now Išll have to search
onlinefor your original problem, since I deleted my copy long
ago... (Excuseme a few minutes...) Hmm... OK, Išm back.Well...
therešs still a missing (or extra) parenthesis in the
deŪnitionof f and a missing bracket in the deŪnition of F,
but I also canšt seethat youšve given z a numerical value.
The error shouldnšt occur untilyou use F though, since you
used SetDelayed. So, the error message goeswith a line you
didnšt include in your post. However, in your latestpost, the
linesNumericQ[z]Falsetell me z is NOT a real number. For
instance,NumericQ[1.2]TrueSo, proceeding with the theory that
z needs to be numeric, letšs tryF[1.2]NIntegrate::itraw:Raw
object 1.2` cannot be used as an iterator.NIntegrate[f[y,
1.2], {1.2, 0, InŪnity}]NIntegrate::itraw:Raw object !(1.2`)
cannot be used as aniterator.As you see, in the deŪnition of
F, youšve used z as an iterator. Nomatter what the argument z
to F is -- number, symbol, whatever -- youcanšt use it as an
iterator, because it already has an identity, anditerators
are stand-ins -- temporary variables that donšt exist
outside(in this case) NIntegrate. So, as Jens-Peer pointed
out, y should havebeen your iterator. (Probably. We donšt
actually know what you weretrying to do. We only know for
sure that z couldnšt be the iterator.)So, if Išve made the
right guesses, the deŪnition for F should beF[z_] :=
NIntegrate[f[y, z], {y, 0, InŪnity}]But... F[z] will still
give you an error if z doesnšt have a numericalvalue, so itšs
even better to deŪne F this way:ClearAll[F]F[z_?NumericQ] :=
NIntegrate[f[y, z], {y, 0, InŪnity}]That way, F[z] is left
unevaluated if z isnšt numeric.I canšt go any further, since
I donšt know the value of f -- because Idonšt know where to
add or subtract that pesky parenthesis. (I askedstill
isnšt.)Bobby Treat-----Original Message-----> F[z_?NumericQ]:
= NIntegrate[f[y,z], {y, 0, InŪnity}]>Išm a newbie, bat I
mean:*************************In[1]=
NumericQ[z]Out[1]=False*************************BTW, z is a
real number.--Rob_jack====Išm learning the Ūne art of
patterns and Hold one tricky example at
atime.Bobby-----Original Message-----> I want to make a list
of all symbols in the Global context, as in> Names[Global`*]>
and compute a ByteCount for each symbolšs OwnValues --
without> evaluating the symbols.> It seems possible in
principle, but I havenšt found a way.> Bobby Treat====Herešs
the fairly useful bit of code I came up with, using
Jens-Peeršsbrilliant solution:names = Names[Global`*];counts
= ToExpression[#, StandardForm, Hold] & /@ names /. Hold[a_]
:> ByteCount[OwnValues[a]];Select[Transpose[{names, counts}],
Last@# > 16 &] // TableFormIt tells me how some of my memory
is being spent.Bobby Treat-----Original Message-----> I want
to make a list of all symbols in the Global context, as in>
Names[Global`*]> and compute a ByteCount for each symbolšs
OwnValues -- without> evaluating the symbols.> It seems
possible in principle, but I havenšt found a way.> Bobby
Treat====Jack,Often, what I do when developing a slightly
complicated module is to Ūrsttest it after I add each
statement L1, L2, etc. But then often I want tomake changes
after I have all the statements in. Then to debug I just
addtemporary Print statements. For example...myFunction[f_]
:= Module[ {L1,L2,L3},L1 = ... ;L2 = ... ;Print[{L1, L2}];l3
= ... ;]Sometimes I use multiple Print statements. The only
problem with thisapproach is that sometimes the difŪculty
might be in a subexpression of alonger expression.This forces
me to temporarily break out the longer expression into
multiplestatements, or perhaps duplicate the subexpression in
the Print statement.But I Ūnd that the easiest method to track
down errors.David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/l3 =
... ; Ūnal step ]To see what went wrong, I use (* *)
selectively as follows:Stage 1myFunction[f_] := Module[
{L1,L2,L3},L1 = ... (*;L2 = ... ;L3 = ... ; Ūnal step *)
]Thus I see if L1 worked as expected. The next step is to put
(* after L2and see if this works. I continue this til the
bitter end and I usuallyŪnd my errors.My question; The
process of moving (* *) step by step through theprogram is
quite tedious when the code has lots more lines. What I
wouldlike is a meta-program which (like FoldList) does this
job for me. Theoutput of this meta-program is the list of
outputs of each line in themodule, probably best printed as a
column.This sounds like Trace but my problem with Trace is it
is terriblydifŪcult to read. For the not-so-subtle
programming I do, the only thingI need is what expression is
returned line by line.Any advice? All remarks are
appreciated!Jack====The following code is my attempt to
provide the functionality yourequested. It does some
rudimentary error checking, but I havenšttried very hard to
fool it. The code manipulates the DownValues,locating the
highest CompoundExpression (see the lines in whichpos is
deŪned), keeping only those expressions speciŪed.DownValues
are restored after the expression is evaluated. The
trickypart is keeping the CompoundExpression from evaluating
while it isbeing manipulated. (See the lines in which held
and new aredeŪned.) Išm not sure Išm doing this part the most
elegant way, butit seems to work.(* code starts here
*)PartialEvaluation::usage = PartialEvaluation[f[args], n]
returnsf[args] where only the Ūrst n expressions are
evaluated in themainCompoundExpression in DownValues[f]. For
example:ntClear[f]ntf[x_] := Module[{a,b}, a=3; b=4; a b
x]ntPartialEvaluation[f, 2]n returns
4.PartialEvaluation::dvprob = DownValues[``] is either empty
or hasmorethan one element.PartialEvaluation::toobig = There
are only `1` expressions inthe CompoundExpression in
`2`.PartialEvaluation::noce = There are no
CompoundExpressions
inDownValues[``].SetAttributes[PartialEvaluation,
HoldFirst]PartialEvaluation[f_[args__], n_Integer] :=
Module[{dv, pos, held, new, eval}, Catch[ dv = DownValues[f];
If[Length[dv] != 1, Message[PartialEvaluation::dvprob, f];
Throw[HoldForm[f[args]]]]; pos = Sort[Position[dv,
CompoundExpression]]; If[pos == {},
Message[PartialEvaluation::noce, f];
Throw[HoldForm[f[args]]]]; pos = Drop[First @ pos, -1]; held
= Extract[dv, pos, Hold] /. CompoundExpression -> Sequence;
If[Abs[n] > Length[held], Message[PartialEvaluation::toobig,
Length[held], HoldForm[f[args]]]; Throw[HoldForm[f[args]]]];
new = ReplacePart[dv, Take[held, n] /. Hold[x__] :>
Hold[CompoundExpression[x]], pos, 1]; DownValues[f] = new;
eval = f[args]; DownValues[f] = dv; eval ]](* code ends here
*)--Mark.P.S. The version of this message I posted directly
from my newsreaderdidnšt show up, so Išm reposting (a
slightly improved version) fromGoogle.> Jack,> Often, what I
do when developing a slightly complicated module is to Ūrst>
test it after I add each statement L1, L2, etc. But then
often I want to> make changes after I have all the statements
in. Then to debug I just add> temporary Print statements. For
example...> myFunction[f_] := Module[ {L1,L2,L3},> L1 = ...
;> L2 = ... ;> Print[{L1, L2}];> l3 = ... ;> ]> Sometimes I
use multiple Print statements. The only problem with this>
approach is that sometimes the difŪculty might be in a
subexpression of a> longer expression.> This forces me to
temporarily break out the longer expression into multiple>
statements, or perhaps duplicate the subexpression in the
Print statement.> But I Ūnd that the easiest method to track
down errors.> David Park> djmp@earthlink.net>
http://home.earthlink.net/~djmp/> I often run into this
difŪculty: when designing a program, say as a> module, and
testing it for various inputs, I get wrong answers. What to>
do? I use a method that works for me but may not be the best
available.> I want to show my method and then ask a question
about how it can be> imporved. (Oh yes, I abandoned Trace a
long time ago!)> myFunction[f_] := Module[ {L1,L2,L3},> L1 =
... ;> L2 = ... ;> l3 = ... ;> Ūnal step> ]> To see what went
wrong, I use (* *) selectively as follows:> Stage 1>
myFunction[f_] := Module[ {L1,L2,L3},> L1 = ... (*;> L2 = ...
;> L3 = ... ;> Ūnal step *)> ]> Thus I see if L1 worked as
expected. The next step is to put (* after L2> and see if
this works. I continue this til the bitter end and I usually>
Ūnd my errors.> My question; The process of moving (* *) step
by step through the> program is quite tedious when the code
has lots more lines. What I would> like is a meta-program
which (like FoldList) does this job for me. The> output of
this meta-program is the list of outputs of each line in the>
module, probably best printed as a column.> This sounds like
Trace but my problem with Trace is it is terribly> difŪcult
to read. For the not-so-subtle programming I do, the only
thing> I need is what expression is returned line by line.>
Any advice? All remarks are appreciated!> Jack====Mathematica
often produces complex resluts. But the problem is not
complex:TrySolve[FR == Pi r1^2 - Pi r2^2, r1]It is simple to
eliminate it in this case. Just square it.Does someone give
me a hint, how to write ra rule for simplify, lik Ted Ersek
did
inhttp://www.verbeia.com/mathematica/tips/Links/Tricks_lnk_41
.htmlI thin, one has to to replace the patterni
Sqrt[-a_]bySqrt[(i Sqrt[-a_])^2]and then one has to simplify
it?Peter Klamser====There has been an interesting discussion
about using InterpolatingFunctionfor an empirical CDF.If one
is willing to sacriŪce some theoretical soundness, an
empirical CDFmay be computed in a simpliŪed
form:simpleEmpiricalCDF[distr_/;VectorQ[distr,NumberQ]]
:=With[{nonu=Sort[distr],
u=Union[distr]},Transpose[{u,FoldList[Plus,0,(Count[nonu,#]&/
@u)/Length[nonu]]//Rest}]]This code does not return a
function, and the result needs someinterpretation.
Nevertheless, it may be of some use.data = {0.59, 0.72, 0.47,
0.43, 0.31, 0.56, 0.22, 0.9, 0.96, 0.78, 0.66,0.18, 0.73,
0.43, 0.58, 0.11}(the tie [0.43] will be accounted for in a
way that is consistent with
sometextbooks)simpleEmpiricalCDF[data]//N
returns{{0.11,0.0625},{0.18,0.125},{0.22,0.1875},{0.31,0.25},
{0.43,0.375},{0.47,0.4375},{0.56,0.5},{0.58,0.5625},{
0.59,0.625},{0.66,0.6875},{0.72,0.75},{0.73,0..8125},{
0.78,0.875},{0.9,0.9375},{0.96,1.}}A plot may be given
with:simpleEmpiricalCDFPlot[distr_/;VectorQ[distr,NumberQ]]
:=Module[{res,x,p},res=simpleEmpiricalCDF[distr];x=({#,#}& /@
Transpose[res][[1]])//Flatten;p=({#,#}& /@
Transpose[res][[2]])//Flatten;p=
Join[{0},Drop[p,-2],{1}];ListPlot[Transpose[{x,p}],c]]Again,
this plot does not reūect theoretical considerations at the
lowerand upper end.With regardHermann Meier====Caution: where
right- and left-continuity are concerned, Danielšsrelying on
undocumented behavior that may change in the next
version.Bobby-----Original Message-----that evaluates the
empirical CDF given the observations in the list.The function
is deŪned on the entire real
line.MakeEmpiricalCDF[list_?(VectorQ[#, NumericQ]&)] :=
Module[{n, s, a, r, idata}, n = Length[list]; s = Sort[list];
a = Append[s, s[[-1]] + 1]; (* phantom obs. *) r = Range[1/n,
1 + 1/n, 1/n]; (* phantom value 1 + 1/n *) idata = Last /@
Split[Transpose[{-a, r}], #1[[1]] == #2[[1]]&]; (* -a is the
Ūrst sign change *) Block[{x}, Function @@ {x, Which @@ { x <
s[[ 1]], 0., x > s[[-1]], 1., True, Interpolation[idata,
InterpolationOrder -> 0][-x] (* -x is the second sign change
*) }}] ]The construction Last /@ Split[ ... ] accounts for
duplicate values.Here are two examples.
Needs[Statistics`ContinuousDistributions`]list1 =
RandomArray[NormalDistribution[0, 1], 100];f1 =
MakeEmpiricalCDF[list1];Plot[f1[x], {x, -4, 4}]list2 =
Table[Random[Integer, {1, 10}], {10}];f2 =
MakeEmpiricalCDF[list2];Plot[f2[x], {x, 0, 11}]--Mark> Išm
trying to write a fast empirical cummulative distribution
function> (CDF). Empirical CDFs are step functions that can
be expressed in> terms of a Which statement. For example,
given the list of> observations {1, 2, 3},> f = Which[# < 1,
0, # < 2, 1/3, # < 3, 2/3, True, 1]&> is the empirical CDF.
Note that f /@ {1, 2, 3} returns {1/3, 2/3, 1}> and f is
continuous from the right.> When the number of observations
is large, the Which statement> evaluates fairly slowly (even
if it has been Compiled). Since> InterpolationFunction
evaluates so much faster in general, Išve tried> to use
Interpolation with InterpolationOrder -> 0. The problem is
that> the resulting InterpolatingFunction doesnšt behave the
way (I think)> it ought to. For example, let> g =
Interpolation[{{1, 1/3}, {2, 2/3}, {3, 1}},
InterpolationOrder ->> 0]> Then, g /@ {1, 2, 3} returns {2/3,
2/3, 1} instead of {1/3, 2/3, 1}.> In addition, g is
continuous from the left rather than from the right.>
Obviously I am not aware of the considerations that went
into> determining the behavior of InterpolationFunction when>
InterpolationOrder -> 0.> So I have two questions: > (1) Does
anyone have any opinions about how InterpolatingFunction>
ought to behave with InterpolationOrder -> 0?> (2) Does
anyone have a faster way to evaluate an empirical CDF than a>
compiled Which function?> By the way, herešs my current
version:> CompileEmpiricalCDF[list_?(VectorQ[#, NumericQ] &)]
:=> Block[{x}, Compile[{{x, _Real}}, Evaluate[> Which @@
Flatten[> Append[> Transpose[{> Thread[x < Sort[list]],>
Range[0, 1 - 1/#, 1/#] & @ Length[list]> }],> {True, 1}]]>
]]]> --Mark====Hey,How can I duplicate those nifty ŒMoreš
hyperlinks in my usagestatements? That is the hyperlinks that
appears at the end of text thatis generated after invoking a
Œ?š to get more information about a
function.Lawrence====>Išve got to extract some numbers from a
Ūle that are in lines of>text. Since the line contents are not
numbers, I presume I must pull>the line out as a string. Here
I start by pulling out just one line:>>inFile =
OpenRead[197-tst.txt] >y = ReadList[inFile, String, 1,
RecordLists -> True] >Close[inFile];>>This appears to pull in
a line. Now I want to take characters 25 to>110 to get just
the stuff I want: y1=StringTake[y ,{25,110}];>>Herešs the
output. StringTake doesnšt seem to work.ReadList returns a
List not a String. So, y is a List and StringTake fails since
it expects a String.Also, you do not need the options
RecordLists->True when reading Strings. Nor do you need the
OpenRead/Close statements with ReadListTryy =
First[readList[197-tst.txt,String,1];y1 = StringTake[y,
{25,110}];====>Does anybody know how to calculate in
Mathematica: >a)empirical CDF,>b)empirical PDF, >c)normal
QQ-plot; >d)QQ-plot two different random samples?!Yes, but
there are a number of issues particularly with an empirical
PDF. A very nice package that does all of the above and more
is mathStatica. See http://www.mathstatica.com for
details.Obviously, it is less expensive to write your own
functions.Just recently in message Mark Fisher posted code
that addresses the empirical CDF. However, in this code you
may want to replace 1/n with 1/(n+1) or (j-0.5)/n depending
on your application. Note, these will have no signiŪcant
effect for large data sets.====> I am trying to implement a
very simple sorted tree to quickly store some> real numbers I
need. I have written an add, delete,