When plotting a function using Plot, I would like to obtain the set of data points plotted by the Plot command.
For instance, how can I obtain the list of points {t,f} Plot uses in the following simple example?
f = Sin[t]
Plot[f, {t, 0, 10}]
I tried using a method of appending values to a list, shown on page 4 of Numerical1.ps (Numerical Computation in Mathematica) by Jerry B. Keiper, http://library.wolfram.com/infocenter/Conferences/4687/ as follows:
f = Sin[t]
flist={}
Plot[f, {t, 0, 10}, AppendTo[flist,{t,f[t]}]]
but generate error messages no matter what I try.
Any suggestions would be greatly appreciated.
f = Sin[t];
plot = Plot[f, {t, 0, 10}]
One way to extract points is as follows:
points = Cases[
Cases[InputForm[plot], Line[___],
Infinity], {_?NumericQ, _?NumericQ}, Infinity];
ListPlot to 'take a look'
ListPlot[points]
giving the following:
EDIT
Brett Champion has pointed out that InputForm is superfluous.
ListPlot#Cases[
Cases[plot, Line[___], Infinity], {_?NumericQ, _?NumericQ},
Infinity]
will work.
It is also possible to paste in the plot graphic, and this is sometimes useful. If,say, I create a ListPlot of external data and then mislay the data file (so that I only have access to the generated graphic), I may regenerate the data by selecting the graphic cell bracket,copy and paste:
ListPlot#Transpose[{Range[10], 4 Range[10]}]
points = Cases[
Cases[** Paste_Grphic _Here **, Point[___],
Infinity], {_?NumericQ, _?NumericQ}, Infinity]
Edit 2.
I should also have cross-referenced and acknowledged this very nice answer by Yaroslav Bulatov.
Edit 3
Brett Champion has not only pointed out that FullForm is superfluous, but that in cases where a GraphicsComplex is generated, applying Normal will convert the complex into primitives. This can be very useful.
For example:
lp = ListPlot[Transpose[{Range[10], Range[10]}],
Filling -> Bottom]; Cases[
Cases[Normal#lp, Point[___],
Infinity], {_?NumericQ, _?NumericQ}, Infinity]
gives (correctly)
{{1., 1.}, {2., 2.}, {3., 3.}, {4., 4.}, {5., 5.}, {6., 6.}, {7.,
7.}, {8., 8.}, {9., 9.}, {10., 10.}}
Thanks to Brett Champion.
Finally, a neater way of using the general approach given in this answer, which I found here
The OP problem, in terms of a ListPlot, may be obtained as follows:
ListPlot#Cases[g, x_Line :> First#x, Infinity]
Edit 4
Even simpler
ListPlot#Cases[plot, Line[{x__}] -> x, Infinity]
or
ListPlot#Cases[** Paste_Grphic _Here **, Line[{x__}] -> x, Infinity]
or
ListPlot#plot[[1, 1, 3, 2, 1]]
This evaluates to True
plot[[1, 1, 3, 2, 1]] == Cases[plot, Line[{x__}] -> x, Infinity]
One way is to use EvaluationMonitor option with Reap and Sow, for example
In[4]:=
(points = Reap[Plot[Sin[x],{x,0,4Pi},EvaluationMonitor:>Sow[{x,Sin[x]}]]][[2,1]])//Short
Out[4]//Short= {{2.56457*10^-7,2.56457*10^-7},<<699>>,{12.5621,-<<21>>}}
In addition to the methods mentioned in Leonid's answer and my follow-up comment, to track plotting progress of slow functions in real time to see what's happening you could do the following (using the example of this recent question):
(* CPU intensive function *)
LogNormalStableCDF[{alpha_, beta_, gamma_, sigma_, delta_}, x_] :=
Block[{u},
NExpectation[
CDF[StableDistribution[alpha, beta, gamma, sigma], (x - delta)/u],
u \[Distributed] LogNormalDistribution[Log[gamma], sigma]]]
(* real time tracking of plot process *)
res = {};
ListLinePlot[res // Sort, Mesh -> All] // Dynamic
Plot[(AppendTo[res, {x, #}]; #) &#
LogNormalStableCDF[{1.5, 1, 1, 0.5, 1}, x], {x, -4, 6},
PlotRange -> All, PlotPoints -> 10, MaxRecursion -> 4]
etc.
Here is a very efficient way to get all the data points:
{plot, {points}} = Reap # Plot[Last#Sow#{x, Sin[x]}, {x, 0, 4 Pi}]
Based on the answer of Sjoerd C. de Vries, I've now written the following code which automates a plot preview (tested on Mathematica 8):
pairs[x_, y_List]:={x, #}& /# y
pairs[x_, y_]:={x, y}
condtranspose[x:{{_List ..}..}]:=Transpose # x
condtranspose[x_]:=x
Protect[SaveData]
MonitorPlot[f_, range_, options: OptionsPattern[]]:=
Module[{data={}, plot},
Module[{tmp=#},
If[FilterRules[{options},SaveData]!={},
ReleaseHold[Hold[SaveData=condtranspose[data]]/.FilterRules[{options},SaveData]];tmp]]&#
Monitor[Plot[(data=Union[data, {pairs[range[[1]], #]}]; #)& # f, range,
Evaluate[FilterRules[{options}, Options[Plot]]]],
plot=ListLinePlot[condtranspose[data], Mesh->All,
FilterRules[{options}, Options[ListLinePlot]]];
Show[plot, Module[{yrange=Options[plot, PlotRange][[1,2,2]]},
Graphics[Line[{{range[[1]], yrange[[1]]}, {range[[1]], yrange[[2]]}}]]]]]]
SetAttributes[MonitorPlot, HoldAll]
In addition to showing the progress of the plot, it also marks the x position where it currently calculates.
The main problem is that for multiple plots, Mathematica applies the same plot style for all curves in the final plot (interestingly, it doesn't on the temporary plots).
To get the data produced into the variable dest, use the option SaveData:>dest
Just another way, possibly implementation dependent:
ListPlot#Flatten[
Plot[Tan#t, {t, 0, 10}] /. Graphics[{{___, {_, y__}}}, ___] -> {y} /. Line -> List
, 2]
Just look into structure of plot (for different type of plots there would be a little bit different structure) and use something like that:
plt = Plot[Sin[x], {x, 0, 1}];
lstpoint = plt[[1, 1, 3, 2, 1]];
Related
First of all I would like to apologize for the newbie question.
I am just starting up with mathematica and I have 2 simple plots. What i want to do is have Mathematica automatically find the intersections, label them and show me the coordinates.
I have searched this forum and there are a lot of similar questions, but they are too advanced for me.
Can someone explain how i can do this the easiest way?
Solve for equality. Get values for the points using replacement : points = {x, x^2} /. sol would work just as well. Offset the labels and set as text in epilog.
sol = Solve[x^2 == x + 2, x];
points = {x, x + 2} /. sol;
offset = Map[# + {0, 3} &, points];
Plot[{x^2, x + 2}, {x, -6, 6},
Epilog -> {Thread[Text[points, offset]],
Blue, PointSize[0.02], Point[points]}]
I am trying to make an histogram without vertical lines. I'd like to have a plot which looks like a function. Like this:
The same question has been asked for R before ( histogram without vertical lines ) but I'm on Mathematica.
I have been looking into the ChartStyle options without success.
You could also use ListPlot with InterpolationOrder->0:
(* example data *)
data = RandomVariate[NormalDistribution[], 10^3];
hist = HistogramList[data, {.5}];
ListPlot[Transpose[{hist[[1]], ArrayPad[hist[[2]], {0, 1}, "Fixed"]}],
InterpolationOrder -> 0,
Joined -> True,
AxesOrigin -> {hist[[1, 1]], 0}]
There probably are ways to do this by fiddling with EdgeForm[] and FaceForm[] in Histogram, but I've found it simpler to roll one on my own, whenever I need it. Here's a very simple and quick example:
histPlot[data_, bins_, color_: Blue] := Module[{
countBorder =
Partition[Riffle[Riffle[#1, #1[[2 ;;]]], Riffle[#2, #2]], 2] & ##
HistogramList[data, bins, "PDF"]
},
ListLinePlot[countBorder, PlotStyle -> color]
]
Doing histPlot[RandomReal[NormalDistribution[],{1000}],{-3,3,0.1}] gives
You can then extend this to take any option instead of just "PDF", and for cases when you'd like to choose the bins automatically. I dislike automatic binning, because I like to control my bin widths and extents for predictability and easy comparison against other plots.
Here are two methods that work in version 7, using post-processing:
rdat = RandomReal[NormalDistribution[0, 1], 200];
MapAt[
{Blue,
Line[# /. {{Rectangle[{x_, y_}, {X_, Y_}]}} :> Sequence[{x, Y}, {X, Y}]] } &,
Histogram[rdat, PerformanceGoal -> "Speed"],
{1, 2, 2, 2}
]
Cases[
Histogram[rdat, PerformanceGoal -> "Speed"],
Rectangle[{x_, y_}, {X_, Y_}] :> {{x, Y}, {X, Y}},
\[Infinity]
];
Graphics[Line[Join ## %], AspectRatio -> 1/GoldenRatio, Axes -> True]
Consider the following definition:
f[x_]=Piecewise[{{0,x<1/2},{Interval[{0,1}],x==1/2},{1,x>1/2}}];
Then when one does the Plot[f[x],{x,0,1}] of the function, the graph does not depict the interval value f[1/2] of the graph.
Any ideas on how to plot interval-valued functions in Mathematica would be much appreciated.
Update #1: I've found a hack:
Plot[ f[x], {x,0,1}, ExclusionsStyle->Opacity[1]];
The hack, however, does not work on a general interval-valued function, such as
f[x_]=Piecewise[{{0,x<1/2},{Interval[{0,1}],1/2<=x<=1}}];
which is the essence of the question.
Update #2:
As a followup to the neat example of #Heike below: it's only a partial solution. For if one tries the following:
f[x_] = Piecewise[{{0, x < 1/2}, {Interval[{x, 1}], 1/2 <= x <= 1}}];
Plot[ {f[x] /. Interval[a_] :> a[[1]], f[x] /. Interval[a_] :> a[[2]]},
{x, 0, 1}, Filling -> {1 -> {2}}]
then the graph depicts a segment at x=1/2 that is equal to the value [0,1] instead of [1/2,1].
Maybe you could do something like
f[x_]=Piecewise[{{0,x<1/2},{Interval[{0,1}],1/2<=x<=1}}];
Plot[{f[x] /. Interval[a_] :> a[[1]],
f[x] /. Interval[a_] :> a[[2]]}, {x, 0, 1}, Filling -> {1 -> {2}}]
Frequently I have to visualize multiple datasets simultaneously, usually in ListPlot or its Log-companions. Since the number of datasets is usually larger than the number of easily distinguishable line styles and creating large plot legends is still somewhat unintuitiv I am still searching for a good way to annotate the different lines/sets in my plots. Tooltip is nice when working on screen, but they don't help if I need to pritn the plot.
Recently, I played around with the Mesh option to enumerate my datasets and found some weird stuff
GraphicsGrid[Partition[Table[ListPlot[
Transpose#
Table[{Sin[x], Cos[x], Tan[x], Cot[x]}, {x, 0.01, 10, 0.1}],
PlotMarkers -> {"1", "2", "3", "4"}, Mesh -> i, Joined -> True,
PlotLabel -> "Mesh\[Rule]" <> ToString[i], ImageSize -> 180], {i,
1, 30}], 4]]
The result looks like this on my machine (Windows 7 x64, Mathematica 8.0.1):
Funnily, for Mesh->2, 8 , and 10 the result looks like I expected it, the rest does not. Either I don't understand the Mesh option, or it doesn't understand me.
Here are my questions:
is Mesh in ListPLot bugged or do I use it wrongly?
how could I x-shift the mesh points of successive sets to avoid overprinting?
do you have any other suggestions how to annotate/enumerate multiple datasets in a plot?
You could try something along these lines. Make each line into a button which, when clicked, identifies itself.
plot=Plot[{Sin[x],Cos[x]},{x,0,2*Pi}];
sinline=plot[[1,1,3,2]];
cosline=plot[[1,1,4,2]];
message="";
altplot=Append[plot,PlotLabel->Dynamic[message]];
altplot[[1,1,3,2]]=Button[sinline,message="Clicked on the Sin line"];
altplot[[1,1,4,2]]=Button[cosline,message="Clicked on the Cos line"];
altplot
If you add an EventHandler you can get the location where you clicked and add an Inset with the relevant positioned label to the plot. Wrap the plot in a Dynamic so it updates itself after each button click. It works fine.
In response to comments, here is a fuller version:
plot = Plot[{Sin[x], Cos[x]}, {x, 0, 2*Pi}];
sinline = plot[[1, 1, 3, 2]];
cosline = plot[[1, 1, 4, 2]];
AddLabel[label_] := (AppendTo[plot[[1]],
Inset[Framed[label, Background -> White], pt]];
(* Remove buttons for final plot *)
plainplot = plot;
plainplot[[1, 1, 3, 2]] = plainplot[[1, 1, 3, 2, 1]];
plainplot[[1, 1, 4, 2]] = plainplot[[1, 1, 4, 2, 1]]);
plot[[1, 1, 3, 2]] = Button[sinline, AddLabel["Sin"]];
plot[[1, 1, 4, 2]] = Button[cosline, AddLabel["Cos"]];
Dynamic[EventHandler[plot,
"MouseDown" :> (pt = MousePosition["Graphics"])]]
To add a label click on the line. The final annotated chart, set to 'plainplot', is printable and copyable, and contains no dynamic elements.
[Later in the day] Another version, this time generic, and based on the initial chart. (With parts of Mark McClure's solution used.) For different plots 'ff' and 'spec' can be edited as desired.
ff = {Sin, Cos, Tan, Cot};
spec = Range[0.1, 10, 0.1];
(* Plot functions separately to obtain line counts *)
plots = Array[ListLinePlot[ff[[#]] /# spec] &, Length#ff];
plots = DeleteCases[plots, Line[_?(Length[#] < 3 &)], Infinity];
numlines = Array[Length#Cases[plots[[#]], Line[_], Infinity] &,
Length#ff];
(* Plot functions together for annotation plot *)
plot = ListLinePlot[##spec & /# ff];
plot = DeleteCases[plot, Line[_?(Length[#] < 3 &)], Infinity];
lbl = Flatten#Array[ConstantArray[ToString#ff[[#]],
numlines[[#]]] &, Length#ff];
(* Line positions to substitute with buttons *)
linepos = Position[plot, Line, Infinity];
Clear[line];
(* Copy all the lines to line[n] *)
Array[(line[#] = plot[[Sequence ## Most#linepos[[#]]]]) &,
Total#numlines];
(* Button function *)
AddLabel[label_] := (AppendTo[plot[[1]],
Inset[Framed[label, Background -> White], pt]];
(* Remove buttons for final plain plot *)
plainplot = plot;
bpos = Position[plainplot, Button, Infinity];
Array[(plainplot[[Sequence ## Most#bpos[[#]]]] =
plainplot[[Sequence ## Append[Most#bpos[[#]], 1]]]) &,
Length#bpos]);
(* Substitute all the lines with line buttons *)
Array[(plot[[Sequence ## Most#linepos[[#]]]] = Button[line[#],
AddLabel[lbl[[#]]]]) &, Total#numlines];
Dynamic[EventHandler[plot,
"MouseDown" :> (pt = MousePosition["Graphics"])]]
Here's how it looks. After annotation the plain graphics object can be found set to the 'plainplot' variable.
One approach is to generate the plots separately and then show them together. This yields code that is more like yours than the other post, since PlotMarkers seems to play the way we expect when dealing with one data set. We can get the same coloring using ColorData with PlotStyle. Here's the result:
ff = {Sin, Cos, Tan, Cot};
plots = Table[ListLinePlot[ff[[i]] /# Range[0.1, 10, 0.1],
PlotStyle -> {ColorData[1, i]},
PlotMarkers -> i, Mesh -> 22], {i, 1, Length[ff]}];
(* Delete the spurious asymptote looking thingies. *)
plots = DeleteCases[plots, Line[ll_?(Length[#] < 4 &)], Infinity];
Show[plots, PlotRange -> {-4, 4}]
Are you going to be plotting computable curves or actual data?
If it's computable curves, then it's common to use a plot legend (key).
You can use different dashings and thicknesses to differentiate between the lines on a grayscale printer. There are many examples in the PlotLegends documentation.
If it's real data, then normally the data is sparse enough that you can use PlotMarkers for the actual data points (i.e. don't specify Mesh). You can use automatic PlotMarkers, or you can use custom PlotMarkers including BoxWhisker markers to indicate the various uncertainties.
I am looking to plot something like the whispering gallery modes -- a 2D cylindrically symmetric plot in polar coordinates. Something like this:
I found the following code snippet in Trott's symbolics guidebook. Tried running it on a very small data set; it ate 4 GB of memory and hosed my kernel:
(* add points to get smooth curves *)
addPoints[lp_][points_, \[Delta]\[CurlyEpsilon]_] :=
Module[{n, l}, Join ## (Function[pair,
If[(* additional points needed? *)
(l = Sqrt[#. #]&[Subtract ## pair]) < \[Delta]\[CurlyEpsilon], pair,
n = Floor[l/\[Delta]\[CurlyEpsilon]] + 1;
Table[# + i/n (#2 - #1), {i, 0, n - 1}]& ## pair]] /#
Partition[If[lp === Polygon,
Append[#, First[#]], #]&[points], 2, 1])]
(* Make the plot circular *)
With[{\[Delta]\[CurlyEpsilon] = 0.1, R = 10},
Show[{gr /. (lp : (Polygon | Line))[l_] :>
lp[{#2 Cos[#1], #2 Sin[#1]} & ###(* add points *)
addPoints[lp][l, \[Delta]\[CurlyEpsilon]]],
Graphics[{Thickness[0.01], GrayLevel[0], Circle[{0, 0}, R]}]},
DisplayFunction -> $DisplayFunction, Frame -> False]]
Here, gr is a rectangular 2D ListContourPlot, generated using something like this (for example):
data = With[{eth = 2, er = 2, wc = 1, m = 4},
Table[Re[
BesselJ[(Sqrt[eth] m)/Sqrt[er], Sqrt[eth] r wc] Exp[
I m phi]], {r, 0, 10, .2}, {phi, 0, 2 Pi, 0.1}]];
gr = ListContourPlot[data, Contours -> 50, ContourLines -> False,
DataRange -> {{0, 2 Pi}, {0, 10}}, DisplayFunction -> Identity,
ContourStyle -> {Thickness[0.002]}, PlotRange -> All,
ColorFunctionScaling -> False]
Is there a straightforward way to do cylindrical plots like this?.. I find it hard to believe that I would have to turn to Matlab for my curvilinear coordinate needs :)
Previous snippets deleted, since this is clearly the best answer I came up with:
With[{eth = 2, er = 2, wc = 1, m = 4},
ContourPlot[
Re[BesselJ[(Sqrt[eth] m)/Sqrt[er], Sqrt[eth] r wc] Exp[I phi m]]/.
{r ->Norm[{x, y}], phi ->ArcTan[x, y]},
{x, -10, 10}, {y, -10, 10},
Contours -> 50, ContourLines -> False,
RegionFunction -> (#1^2 + #2^2 < 100 &),
ColorFunction -> "SunsetColors"
]
]
Edit
Replacing ContourPlot by Plot3D and removing the unsupported options you get:
This is a relatively straightforward problem. The key is that if you can parametrize it, you can plot it. According to the documentation both ListContourPlot and ListDensityPlot accept data in two forms: an array of height values or a list of coordinates plus function value ({{x, y, f} ..}). The second form is easier to deal with, such that even if your data is in the first form, we'll transform it into the second form.
Simply, to transform data of the form {{r, t, f} ..} into data of the form {{x, y, f} ..} you doN[{#[[1]] Cos[ #[[2]] ], #[[1]] Sin[ #[[2]] ], #[[3]]}]& /# data, when applied to data taken from BesselJ[1, r/2] Cos[3 t] you get
What about when you just have an array of data, like this guy? In that case, you have a 2D array where each point in the array has known location, and in order to plot it, you have to turn it into the second form. I'm partial to MapIndexed, but there are other ways of doing it. Let's say your data is stored in an array where the rows correspond to the radial coordinate and the columns are the angular coordinate. Then to transform it, I'd use
R = 0.01; (*radial increment*)
T = 0.05 Pi; (*angular increment*)
xformed = MapIndexed[
With[{r = #2[[1]]*R, t = #2[[1]]*t, f = #1},
{r Cos[t], r Sin[t], f}]&, data, {2}]//Flatten[#,1]&
which gives the same result.
If you have an analytic solution, then you need to transform it to Cartesian coordinates, like above, but you use replacement rules, instead. For instance,
ContourPlot[ Evaluate[
BesselJ[1, r/2]*Cos[3 t ] /. {r -> Sqrt[x^2 + y^2], t -> ArcTan[x, y]}],
{x, -5, 5}, {y, -5, 5}, PlotPoints -> 50,
ColorFunction -> ColorData["DarkRainbow"], Contours -> 25]
gives
Two things to note: 1) Evaluate is needed to ensure that the replacement is performed correctly, and 2) ArcTan[x, y] takes into account the quadrant that the point {x,y} is found in.