Skip to main content

plotting - ArrayPlot and non-integer PlotRange


There seems to weird bug regarding non-integer PlotRange in ArrayPlot. For example...


ArrayPlot[
Table[Sin[x/50 π] Tanh[y/50 π], {y, 100}, {x, 100}],
DataRange -> {{0, 1}, {0, 1}},
PlotRange -> {{0, 1}, {0.5, .9}}
]


gives an error:



Value of option PlotRange -> {{0,1},{0.5,0.9}} is not All, Full, Automatic, a positive machine number, or an appropriate list of range specifications.



The same code works, if one of the plot range parameters for y-axes is turned into an integer.


Did I do smth wrong? Is there a workaround?


I need to use array plot as ListContourPlot is too slow for large datasets and MatrixPlot has the same problem.


I am using Mathematica 8 64 bit version on linux.


Edit


After some more testing I see, that it works whenever PlotRange includes at least one integer. For example PlotRange -> {{0, 1}, {0.5, 1.2}} works as it includes 1. So one possible workaround is just to scale the range, such that max-min would be larger than 1.



But I am still looking forward, if anyone finds a way to have shorter than 1 range on the axes. Dirty way would be just using manual 'Ticks'.


Edit 2


Possible workaround.


As I wrote in the previous edit, it works, then the range includes at least one integer. So one could just scale the range. Here is a naive hard-coded example how it might be done.


ArrayPlot[
Table[Sin[x/50 π] Tanh[y/50 π], {y, 100}, {x, 100}],
DataRange -> {{0, 1}, {0, 1} 10},
PlotRange -> {{0, 1}, {0.1, 0.8} 10},
FrameTicks -> {Table[{y, ToString[y/10 // N]}, {y, 1, 8, 1}],
Table[{x, ToString[x // N]}, {x, 0, 1, 0.25}]},

AspectRatio -> 1
]

This means that I have bypassed the only side-effect of scaling the range, which is messing up the tick labels.


I would submit it as a solution, when I have waited my 8 hours and there is no nicer one proposed.



Answer



Here is a workaround that works in V7,8,9,10 (the underlying problem persists):


Show[
ArrayPlot[
Table[Sin[x/50 π] Tanh[y/50 π], {y, 100}, {x, 100}],

DataRange -> {{0, 1}, {0, 1}}],
PlotRange -> {{0, 1}, {0.5, .9}}, FrameTicks -> All]

Mathematica graphics


Comments

Popular posts from this blog

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1.