Skip to main content

plotting - ListContourPlot has wrong colouring: workaround?


Bug introduced in 8 or earlier and persisting through 11.0.1 or later




ListContourPlot uses the wrong colouring here:


res = Import["https://dl.dropboxusercontent.com/u/38623/res.wdx", "WDX"];

ListContourPlot[res, Contours -> Range[0.66, 0.9, 0.02], ColorFunction -> "Rainbow"]

enter image description here



The contour plot shows that the function value decreases around the bottom middle, even though it reaches its maximum there in reality.


The tooltips for the contour values between the red and orange regions are correct however: one contour shows 0.84 and the other 0.86. It's just the colouring that's wrong.


Compare


ListDensityPlot[res, ColorFunction -> "Rainbow"]

Mathematica graphics


What is the simplest, most convenient workaround for this problem? Also, is this a bug or am I missing something? Can anyone explain why this happens?


I do need these specific contour values. Interpolation + ContourPlot gives the exact same result:


if = Interpolation[res]


ContourPlot[if[x, y], {x, 0.1, 1}, {y, 7, 9.4},
Contours -> Range[0.66, 9, 0.02], ColorFunction -> "Rainbow"]

Answer



This seems like a bug of some kind to me. But replacing the second argument to Range with the very last contour you want to see, as a workaround, gives the same contours but with proper coloring


Grid[{ListContourPlot[res, Contours -> Range[0.2, #, 0.02], 
ColorFunction -> "Rainbow",
PlotLegends -> Automatic] & /@ {.86, .90}}]

enter image description here


And the bug seems to go away if you increase the scale of your data by a factor of 10, but it shows a completely different wrong color if you decrease it by a factor of 10.



Grid@Partition[#, 3] &@
Table[ListContourPlot[{#1, #2, x #3} & @@@ res,
Contours -> Range[x .2, x .9, x 0.02], ColorFunction -> "Rainbow",
ImageSize -> 300], {x, {.001, .01, .1, 1, 10, 100}}]

enter image description here


So apparently ListContourPlot and ContourPlot have trouble with their ColorFunctionScaling when the data is smaller than 1.


Comments

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...

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}]