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I ran into this problem when I try to solve a partial differential equation. Here is my code:
dd = NDSolve[{D[tes[t, x], t] ==D[tes[t, x], x, x] + Exp[-1/(tes[t, x])],
tes[t, 0] == 1, tes[t, -1] == 1, tes[0, x] == 1}, {tes[t, x]}, {t, 0, 5}, {x, -1, 0}]
f[t_, x_] = tes[t, x] /. dd
kkk = FunctionInterpolation[Integrate[Exp[-1.1/( Evaluate[f[t, x]])], {x, -1, 0}], {t, 0, 0.05}]
kkg[t_] = Integrate[Exp[-1.1/( Evaluate[f[t, x]])], {x, -1, 0}]
Plot[Evaluate[kkk[t]] - Evaluate[kkg[t]], {t, 0, 0.05}]
N[kkg[0.01] - kkk[0.01], 1]
It's strange that the deviation showed in the graph reaches up to more than 5*10^-7 around t=0.01, while it's only -3.88578*10^-16 when calculated by N[kkg[0.01] - kkk[0.01], 1], I wonder how this error comes out.
By the way, I feel it strange that the output of N[kkg[0.01] - kkk[0.01], 1] has so many decimal places, I've set the precision as 1, right?
Using Mathematica 7 the plot I get does not show a peak at 0.01:
Plot[kkk[t] - kkg[t], {t, 0, 0.05}, GridLines -> Automatic]
There is a peak at about 0.00754:
kkk[0.00754] - kkg[0.00754] // N
{6.50604*10^-7}
Regarding N, it does not change the precision of machine precision numbers as it does for exact or arbitrary precision ones:
N[{1.23456789, Pi, 1.23456789`50}, 2]
Precision /# %
{1.23457, 3.1, 1.2}
{MachinePrecision, 2., 2.}
Look at SetPrecision if you want to force (fake) a precision, and NumberForm if you want to print a number in a specific format.
I was initially attempting visualize a 4 parameter function with Plot3D and Manipulate sliders (with two params controlled by sliders and the other vary in the "x-y" plane). However, I'm not getting any output when my non-plotted parameters are Manipulate controlled?
The following 1d plot example replicates what I'm seeing in the more complex plot attempt:
Clear[g, mu]
g[ x_] = (x Sin[mu])^2
Manipulate[ Plot[ g[x], {x, -10, 10}], {{mu, 1}, 0, 2 \[Pi]}]
Plot[ g[x] /. mu -> 1, {x, -10, 10}]
The Plot with a fixed value of mu has the expected parabolic output in the {0,70} automatically selected plotrange, whereas the Manipulate plot is blank in the {0, 1} range.
I was suspecting that the PlotRange wasn't selected with good defaults when the mu slider control was used, but adding in a PlotRange manually also shows no output:
Manipulate[ Plot[ g[x], {x, -10, 10}, PlotRange -> {0, 70}], {{mu, 1}, 0, 2 \[Pi]}]
This is because the Manipulate parameters are local.
The mu in Manipulate[ Plot[ g[x], {x, -10, 10}], {{mu, 1}, 0, 2 \[Pi]}] is different from the global mu you clear on the previous line.
I suggest using
g[x_, mu_] := (x Sin[mu])^2
Manipulate[Plot[g[x, mu], {x, -10, 10}], {{mu, 1}, 0, 2 \[Pi]}]
The following works too, but it keeps changing the value of a global variable, which may cause surprises later unless you pay attention, so I don't recommend it:
g[x_] := (x Sin[mu])^2
Manipulate[
mu = mu2;
Plot[g[x], {x, -10, 10}],
{{mu2, 1}, 0, 2 \[Pi]}
]
It may happen that you Clear[mu], but find that it gets a value the moment the Manipulate object is scrolled into view.
Another way to overcome Manipulate's localization is to bring the function inside the Manipulate[]:
Manipulate[Module[{x,g},
g[x_]=(x Sin[mu])^2;
Plot[g[x], {x, -10, 10}]], {{mu, 1}, 0, 2 \[Pi]}]
or even
Manipulate[Module[{x,g},
g=(x Sin[mu])^2;
Plot[g, {x, -10, 10}]], {{mu, 1}, 0, 2 \[Pi]}]
Both of which give
Module[{x,g},...] prevents unwanted side-effects from the global context. This enables a simple definition of g: I've had Manipulate[]ed plots with dozens of adjustable parameters, which can be cumbersome when passing all those parameters as arguments to the function.
I tried this, but failed.
fig3D = ContourPlot3D[ x^2 + y^3 - z^2 == 0, {x, -2, 2}, {y, -2, 2}, {z, -2, 2},
PlotPoints -> 100]
pts = (InputForm#fig3D)[[1, 1, 1]];
ListSurfacePlot3D[pts]
The regenerated surface is very poor. Any suggestions? thanks!
Not too bad if you specify MaxPlotPoints
ListSurfacePlot3D[pts, Mesh -> None, MaxPlotPoints -> 100]
Compare with
ListSurfacePlot3D[pts]
Edit
Regarding the export to Excel, please consider that the Excel surface plot is a very basic construction and requires a matrix whose first file and column are the XY values with the Z values in the inner cells. Example:
So, exporting a working dataset to Excel may require (an unspecified amount of) data massaging.
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]];
I have data in the form { {x,y,z,f}...} I am using ListContourPlot3D but all I get is an empty box with dimensions -1 to 1 in each direction. Here is my code:
ListContourPlot3D[data5, PlotRange -> All,
AxesLabel -> {"[Beta]", "[Omega]", "Vo"}, Contours -> {1500}].
These are the first 5 points of my data:( the whole set has 55 points)
{{200, 20000 10^(1/3), 2000, 1226},
{200, 20000 10^(1/3), 2600, 1422},
{200, 20000 10^(1/3), 3200, 1581},
{200, 20000 10^(1/3), 3800, 1761},
{200, 20000 10^(1/3), 4400, 1872}}
Dimensions[data5] returns {55,4}
If I do IntegerPart[data5] it does it correctly so it must recognize the numbers in my data.
I appreciate any ideas.
Thank you.
It's hard to tell without having the entire dataset, but I am betting there is a problem with your Contours -> {1500} setting. What happens if you omit it altogether or use a different value?
Contours -> num
Plots num equally spaced levels contours.
Contours -> {num}
Plots the f[x,y,z] = num contour.
Did you mean the former? I doubt ListContourPlot3D can plot your data if it is too sparse or to localized. For the data sample you gave us x and y do not vary at all. Does x and y vary enough in you final data set to well populate coordinate space?
#Davorak's suggestion that the data set, as written, does not seem to vary may be the cause of the problem. Assuming that is not the case, try rotating the resulting graphic, and if you see a black plane appear, then it is the color scheme that is off. By default, ListContourPlot3D produces an opaque white surface, and I've had issues where it did not seem to produce anything, but it was just invisible. The solution: add a ContourStyle option, and set it to something like Red.
The problem is using the {x,y,z,f} form of ListContourPlot3D at low resolution.
I stumbled over this a few weeks ago as well, here is a minimal example of the bug:
xyzfdata[r_] := Flatten[#, 2] &#Table[{x, y, z, x^2 + y^2 + z^2 - 1},
{x, -2, 2, r}, {y, -2, 2, r}, {z, -2, 2, r}];
(* Low resolution {x,y,z,f} fails *)
ListContourPlot3D[xyzfdata[1], Contours -> {0}]
The solution in my case (I had my data on a grid) was to use the grid form and DataRange:
fdata[r_] := Table[x^2 + y^2 + z^2 - 1,
{z, -2, 2, r}, {y, -2, 2, r}, {x, -2, 2, r}];
(* Low resolution works ok for array data *)
ListContourPlot3D[fdata[1], Contours -> {0},
DataRange -> 2 {{-1, 1}, {-1, 1}, {-1, 1}}]
I think the issue is that for the {x,y,z,f} form, the implementation uses interpolation in a way that fails at low resolution. Upping the resolution in the first example, everything works:
(* Higher resolution {x,y,z,f} works *)
ListContourPlot3D[xyzfdata[.2], Contours -> {0}]