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How to change default values and limits within a colormap

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Hi, How are you?.
I want to change the default values in a colormap, In order to plot a map of "diopter powers", like in this image:
However each value is related with a value (or a range of values). For example: if a "diopter power" has a value 39 to 40, the color has to be green. (It is a standard).
Someone could help to know, how I can choose the default values to each color in the colormap.
Thanks for your help.
Rewards.

Answers (1)

Image Analyst
Image Analyst on 27 Oct 2016
Looks kind of like jet upside down. Try this
imshow(grayImage, []);
myColorMap = flipud(jet(256));
% Top row black
myColorMap(end, :) = 0;
% Bottom row white
myColorMap(1, :) = 1;
colormap(myColorMap);
colorbar;
  2 Comments
Jose abel De la Fuente
Jose abel De la Fuente on 28 Oct 2016
Edited: Jose abel De la Fuente on 28 Oct 2016
It is a grean idea!!
However I have tried to do it of different way (I know that is not the best idea), but It was the only way to solve my problem, like in this link:
This is my code:
supDiop = Dioptrica surface power; supAlt = surface height; pxEval = x axis; pyEval = y axis;
Each value of subAlt will be associated a color value which is related with the subDio.
surfc(pxEval,pyEval,(1./(supDiop>=1 & supDiop<=9)).*supAlt,'facecolor',[0/256,1/256,48/256]);
surfc(pxEval,pyEval,(1./(supDiop>=9 & supDiop<=14)).*supAlt,'facecolor', [0/256,0/256,82/256]);
surfc(pxEval,pyEval,(1./(supDiop>=14 & supDiop<19)).*supAlt,'facecolor', [0/256,0/256,115/256]);
surfc(pxEval,pyEval,(1./(supDiop>=19 & supDiop<24)).*supAlt,'facecolor', [0/256,0/256,163/256]);
surfc(pxEval,pyEval,(1./(supDiop>=24 & supDiop<29)).*supAlt,'facecolor', [0/256,0/256,198/256]);
surfc(pxEval,pyEval,(1./(supDiop>=29 & supDiop<35.5)).*supAlt,'facecolor', [0/256,0/256,254/256]);
Everything seemed fine, but I found other problem. when I coupled the surf one after the other (with hold on command) the images in the suft appear with omitted points like in this imagen:
can someone help me to know the cause?... and the solution.

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