5
$\begingroup$
Plot3D[Exp[-x^2 - y^2], {x, -2, 2}, {y, -2, 2},   ColorFunction -> (ColorData["VisibleSpectrum"][
     Rescale[#3, {0, 1}, {380, 750}]] &), PlotLegends -> Automatic]

enter image description here

There is no legend in this picture.

Plot3D[Exp[-x^2 - y^2], {x, -2, 2}, {y, -2, 2}, 
 ColorFunction -> (ColorData["VisibleSpectrum"][Rescale[#3, {0, 1}, {380, 750}]] &), 
 PlotLegends -> BarLegend[{"VisibleSpectrum", {0, 1}}]]

enter image description here

BarLegend["VisibleSpectrum"]

enter image description here

The color of lengend is wrong.

Mathematica 10.0.1. Is this a bug?

$\endgroup$
3
  • $\begingroup$ I think the correct usage of the second argument should be BarLegend[{"VisibleSpectrum", {380, 750}}] but no, that doesn't work either... $\endgroup$
    – C. E.
    Nov 24, 2014 at 5:19
  • $\begingroup$ related: 38895. @Pickett almost :) see my answer. $\endgroup$
    – Kuba
    Nov 24, 2014 at 7:22
  • $\begingroup$ @Kuba wow, I only tried BarLegend on its own. You can't even set the scaling then as far I as I know. I count this as a bug.. $\endgroup$
    – C. E.
    Nov 24, 2014 at 12:09

1 Answer 1

9
$\begingroup$
f@x_ := ColorData["VisibleSpectrum"][Rescale[x, {0, 1}, {380, 750}]]; 
Plot3D[
 Exp[-x^2 - y^2], {x, -2, 2}, {y, -2, 2},
 ColorFunction -> f, PlotLegends -> BarLegend[{f@# &, {0, 1}}]]

Mathematica graphics

$\endgroup$
1
  • $\begingroup$ I'm trying to perform your suggestion in a similar problem. However, here I'm using StreamDensityPlot instead of Plot3D. Therefore, my bar refers to the absolute value of the vector, i.e., ColorFunction -> (ColorData["VisibleSpectrum"][ Rescale[#5, {0, 1.2}, {380, 750}]] &). How could I adjust your code accordingly? $\endgroup$
    – sined
    Nov 2, 2020 at 20:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.