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I'm trying to compare different calculation following different approaches and software programs. I basically plot a DensityPlot with the following legend color bars:

color bars

The bar on the left was generated by Python, while the bar on the right was generated by Mathematica via ColorData[{"RedBlueTones", "Reverse"}]. Although they are very similarly, the one in Mathematica doesn't have the 'white' color transition between blue and red. I'm wondering whether it would be possible to have this 'white' color transitions in Mathematica too. Can I create my own ColorData style using the colors from Python Colorbar?

Thanks!

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    $\begingroup$ "Can I create my own ColorData style using the colors from Python Colorbar?" - this is possible, but you hadn't specified which colormap (from matplotlib?) you're using. $\endgroup$
    – J. M.'s persistent exhaustion ♦
    Feb 24, 2021 at 7:13

2 Answers 2

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How about

French = Blend[{{0, Blue}, {1/2 - 0.1, White},
     {1/2 + 0.1, White}, {1, Red}}, #1] &;

Then

  French /@ (Range[15]/15.)

enter image description here

ContourPlot[x, {x, 0, 1}, {y, 0, 1}, ColorFunction -> French]

enter image description here

Any variation is possible: e.g.

French2 = Blend[{{0, Darker[Blue, 0.7]}, {0.15, Blue}, {1/2 - 0.05, 
      White},
     {1/2 + 0.05, White}, {0.9, Red}, {1, Darker[Red, 0.5]}}, #1] &;

enter image description here

Or alternatively,

col0 = Join[
   ColorData["SiennaTones"] /@ (Range[0, 15]/15), 
   {White},
   ColorData[{"DeepSeaColors", "Reverse"}] /@ (Range[0, 15]/15)];
nn = Length[col0] - 1;

col[x_] = Blend[Transpose[{Range[0, nn]/nn, col0}], x];

col /@ (Range[0,12]/12.)

enter image description here

Then

ContourPlot[x y, {x, -1, 1}, {y, -1, 1}, ColorFunction -> col,  Contours -> 36, 
ContourStyle -> None, PlotRange -> Full]

enter image description here

Update

If you want something which looks more closely to brewer-23

col0=List[List[0,RGBColor[List[Rational[103,255],0,Rational[31,255]]]],List[Rational[1,23],RGBColor[List[Rational[8,15],Rational[3,85],Rational[7,51]]]],List[Rational[2,23],RGBColor[List[Rational[167,255],Rational[19,255],Rational[8,51]]]],List[Rational[3,23],RGBColor[List[Rational[63,85],Rational[44,255],Rational[52,255]]]],List[Rational[4,23],RGBColor[List[Rational[41,51],Rational[74,255],Rational[67,255]]]],List[Rational[5,23],RGBColor[List[Rational[44,51],Rational[104,255],Rational[28,85]]]],List[Rational[6,23],RGBColor[List[Rational[232,255],Rational[133,255],Rational[7,17]]]],List[Rational[7,23],RGBColor[List[Rational[244,255],Rational[163,255],Rational[128,255]]]],List[Rational[8,23],RGBColor[List[Rational[50,51],Rational[62,85],Rational[52,85]]]],List[Rational[9,23],RGBColor[List[Rational[254,255],Rational[211,255],Rational[188,255]]]],List[Rational[10,23],RGBColor[List[Rational[254,255],Rational[227,255],Rational[212,255]]]],List[Rational[11,23],RGBColor[List[Rational[251,255],Rational[16,17],Rational[232,255]]]],List[Rational[12,23],RGBColor[List[Rational[82,85],Rational[247,255],Rational[83,85]]]],List[Rational[13,23],RGBColor[List[Rational[77,85],Rational[241,255],Rational[49,51]]]],List[Rational[14,23],RGBColor[List[Rational[43,51],Rational[233,255],Rational[242,255]]]],List[Rational[15,23],RGBColor[List[Rational[193,255],Rational[223,255],Rational[14,15]]]],List[Rational[16,23],RGBColor[List[Rational[167,255],Rational[208,255],Rational[229,255]]]],List[Rational[17,23],RGBColor[List[Rational[46,85],Rational[193,255],Rational[13,15]]]],List[Rational[18,23],RGBColor[List[Rational[104,255],Rational[172,255],Rational[209,255]]]],List[Rational[19,23],RGBColor[List[Rational[5,17],Rational[3,5],Rational[199,255]]]],List[Rational[20,23],RGBColor[List[Rational[56,255],Rational[134,255],Rational[63,85]]]],List[Rational[21,23],RGBColor[List[Rational[14,85],Rational[23,51],Rational[12,17]]]],List[Rational[22,23],RGBColor[List[Rational[29,255],Rational[97,255],Rational[11,17]]]],List[1,RGBColor[List[Rational[6,85],Rational[74,255],Rational[134,255]]]]];

So that

col0//Transpose

enter image description here

Then

col[x_] = Blend[col0, x]

Would work like this:

ContourPlot[Sin[ 10 x y] , {x, 0, 1}, {y, 0, 1}, ColorFunction -> col]

enter image description here

Or

ContourPlot[x y , {x, -1, 1}, {y, -1, 1}, ColorFunction -> col, 
 Contours -> 26, ContourStyle -> None, PlotRange -> Full]

enter image description here

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    $\begingroup$ Here's a more compact way to define your col: With[{col0 = RGBColor /@ {"#67001f", "#880923", "#a71328", "#bd2c34", "#cd4a43", "#dc6854", "#e88569", "#f4a380", "#faba9c", "#fed3bc", "#fee3d4", "#fbf0e8", "#f6f7f9", "#e7f1f5", "#d7e9f2", "#c1dfee", "#a7d0e5", "#8ac1dd", "#68acd1", "#4b99c7", "#3886bd", "#2a73b4", "#1d61a5", "#124a86"}}, col[x_] := Blend[col0, x]] $\endgroup$
    – J. M.'s persistent exhaustion ♦
    Feb 24, 2021 at 7:19
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    $\begingroup$ @J.M. Thank you! You may recall that you taught me this stuff! $\endgroup$
    – chris
    Feb 24, 2021 at 7:34
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A cheap way to get what you want is to use Lighter[] along with "RedBlueTones" and an appropriate bell-shaped curve:

LinearGradientImage[Function[x, Lighter[ColorData["RedBlueTones", x],
                                        Sech[5 (x - 1/2)]]], {300, 30}]

modified gradient

ContourPlot[x y, {x, -1, 1}, {y, -1, 1}, 
            ColorFunction -> (Lighter[ColorData["RedBlueTones", #], 
                                      Sech[7.2 (# - 0.5)]] &),
            Contours -> 26, ContourStyle -> None, PlotRange -> Full]

contour plot with modified gradient

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