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3

ent is an Entity of "Color" you can investigate its properties. EntityProperties["Color"] Of the many properties the "Value" property (strangely it has CommonName "Wolfram Language") will return the RGB color you are seeking. ent["Value"] (* RGBColor[1., 0.96, 0.93] *) Hope this helps.


2

I was unable to locate the function R[t] so I made one up. I also changed the BoxRatios to make it a bit easier to see. R[t_] := 4 + Cos[\[Pi] t] r[t_, ϕ0_] = {R[t] Cos[ϕ0], R[t] Sin[ϕ0], t} ParametricPlot3D[r[t, ϕ0], {ϕ0, 0, 2 π}, {t, 0, 2}, MeshFunctions -> {#4 &}, Axes -> None, Boxed -> False, BoxRatios -> {1, 1, 1}, ImageSize -&...


0

This bug seems to affect different versions differently, so more than one workaround is needed. Here is another workaround that does not work in version 10.3.1, but it does work on other versions. GeoRegionValuePlot[vals, GeoRange -> {{-60, 75}, {-130, 165}}, GeoBackground -> None, ColorRules -> Flatten@Union[({# -> colfoo@#} & /@ vals[[...


1

That's a funky bug. As JasonB points out in the comments, this can be reduced to a two-country example: GeoRegionValuePlot[ { Entity["Country", "UnitedStates"] -> -1, Entity["Country", "India"] -> 0 } , GeoRange -> {{-60, 75}, {-130, 165}} , ColorFunctionScaling -> False , ColorFunction -> (Blend[{{-1, Red}, {0, Yellow}}, #] &) ...


3

One could override the setting for each polygon group (or GraphicsGroup[]): cp /. p_Polygon :> {Lighting -> {{"Ambient", White}}, p} cp /. gg_GraphicsGroup :> {Lighting -> {{"Ambient", White}}, gg} Update: Addendum. While Lighting shows up in Options@ChromaticityPlot3D, it is not listed among the options in the docs for ChromaticityPlot3D....


5

If you look at the InputForm of ChromaticityPlot3D[{"WideGamutRGB", "sRGB"}] you'll find several spots where it says Lighting -> "Neutral" So if you want to change that, you'll have to modify the output of ChromaticityPlot3D using a replacement rule. Here is an extreme example, one that totally ruins the plot but shows how to change the lighting, ...


2

No legend for this slight simplification of Kuba's proposal, but it can be easily added if desired: RegionPlot[True, {x, -3, 3}, {y, -3, 3}, Background -> Black, BoundaryStyle -> None, ColorFunction -> Function[{x, y}, Hue[(Arg[x + I y] - Pi/2)/Pi, 1, 1, (x^2 + y^2) Exp[1 - x^2 - y^2]]], ...


4

Edit Table Case Kuba, can you add also a solution for tables instead of functions, because in fact I have something like this: table1 = Table[(x^2 + y^2) Exp[-x^2 - y^2], {x, -3, 3}, {y, -3, 3}], table2 = Table[ArcTan[x, y], {x, -2.999, 3}, {y, -3., 3}] – Mushegh { table1 = Table[(x^2 + y^2) Exp[-x^2 - y^2], {x, ##}, {y, ##}], table2 = ...


7

Here is a modification of Karsten's answer attempting to get the behavior you described in a comment. The selected value is assigned to Global`$color (as shown by Dynamic) I made the picker wider to get rid of the annoying bottom scroll bar Code: Button["Pick a color", Module[{picker, box = EvaluationBox[]}, picker = FrontEndResource["...


10

Do you mean Button[ Tooltip["click", Column[{"Click this button to select a ColorScheme.", "This button will be replaced with ColorData"}]], With[{box = EvaluationBox[]}, SelectionMove[box, All, Expression]; FrontEndExecute@ FrontEnd`AttachCell[box, FrontEndResource["ColorSchemeSelector"], {1, {Right, Top}}, {Left, Top}, "...


4

Actually I think what you need is ColorFunction and ColorFunctionScaling. In ColorFunction you can set the color in different regions according to the points' {x,y,z} coordination. And I think what you need is just setting some part in a color (in my code, Bed) and the other in another color(in my code, Blue). Then, simply create a function discribing this ...


3

A recommendation to the question poser: Pose your question in the absolute simplest terms, limiting to the minimal example that addresses your point. There is no need here, for instance, for the community to have to download a complicated data set in order to see how to color one part of a plot differently from others. Why do we need to incorporate text ...


2

One simple option: ListDensityPlot[mydata, ColorFunction -> (Blend[{{-10, Blue}, {0, White}, {5, Red}}, #] &), ColorFunctionScaling -> False, PlotLegends -> Automatic] A simple way to control blend: minLegend = Min[mydata[[;; , -1]]]; maxLegend = Max[mydata[[;; , -1]]]; spdLegend = 2; cf = Blend[{{minLegend , Blue}, {minLegend / ...


5

Here is a solution via Epilog and graphic primitive Rectangle[] Plot[{i, 4, 6}, {i, 0, 10}, Epilog -> {Opacity[0.3], Lighter@Red, Rectangle[{0, 0}, {10, 4}], Lighter@Blue, Rectangle[{0, 4}, {10, 6}]}]


6

Check out Plot for reference. Use {} to define multiple functions to be plotted and Filling for the shading. Plot[{i, 4, 6}, {i, 0, 10}, Filling -> {3 -> {2}, 2 -> Bottom}]


2

You can density plot a transformation of the function and subsequently inverse transform the bar labels using the Ticks options: concentration2D[x_, y_] = 1/x^2 + y; trans[x_] = E^x/(1 + E^x); inverse[y_] = Log[y/(1 - y)]; With[{ticks = 18}, DensityPlot[trans[concentration2D[x, y]], {x, -1, 1}, {y, -1, 1}, ColorFunction -> "Rainbow", PlotLegends ...


2

It is because MinimumVertexColoring is from the old Combinatorica package, and the data format of Graph is different from the new Graph in V10. To fix this problem, you will need to convert the new atomic graph object to the old representation. Needs["Combinatorica`"]; Needs["GraphUtilities`"]; g = System`RandomGraph[{5, 6}]; cg = ToCombinatoricaGraph[g]; ...


8

FromEntity and ToEntity can do the conversion. FromEntity[StarData["Sirius", "Color"]] (* RGBColor[0.73, 0.8, 1.] *)


2

What is the bug? It seems that GeoRegionValuePlot will not work correctly when two or more entities have the exact same value. Consider these examples (and ignore the legend, which is always wrong unless you give an explicit ColorFunction as below): GeoRegionValuePlot[{Entity[ "AdministrativeDivision", {"Arkansas", "UnitedStates"}] -> 1, Entity["...


3

First Method If you just want to get different color of your ball,but don't want to name the color.I will recommend you this solution,which makes your life easier.In the meantime,it will change other place's color,not just the ball,like CSP's answer img = Import["http://i.stack.imgur.com/Qr7Tx.jpg"]; Select[ColorCombine /@ Tuples[ColorSeparate[img], {3}], ...


1

ArrayPlot[a /. 0 -> White, ColorFunction -> "Rainbow"]



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