If you have an image of a colormap from some other figure, you can fairly easily just extract the map without having to guess about what path it makes in some colorspace.
% read image and process it
margins = [7 7 15 11]; % N S E W
A = imread('https://www.mathworks.com/matlabcentral/answers/uploaded_files/853000/image.png');
A = A(1+margins(1):end-margins(2),1+margins(4):end-margins(3),:);
CT0 = permute(mean(im2double(A),1),[2 3 1]);
CT0 = CT0([true; ~all(diff(CT0,1,1)==0,2)],:); % remove duplicate rows
% CT0 is now a fixed-length color table
% make it whatever length we want
N = 32; % specify the number of colors in table
na = size(CT0,1);
CT = interp1(linspace(0,1,na),CT0,linspace(0,1,N));
x = linspace(-pi/2,pi/2,30);
y = x.';
z = sin(x).*sin(y);
surf(x,y,z)
colormap(CT)
colorbar
view(-17,50)
The extracted color table (CT0) can be stored as a .mat file or used by a simple function containing the interpolation routine above. In that manner, you can have a convenient function that can be called like jet() or parula() to return a color table of any desired length. Attached is such a function that will return the above colormap for any user-specified length.
C = mycolortable(16)
% show color tables of different lengths
CT = [];
for n = [16 32 64 128 256]
thisct = permute(mycolortable(n),[1 3 2]);
CT = [CT imresize(thisct,[256 16],'nearest')];
end
CT = rot90(imresize(CT,[512 256],'nearest'),1);
imshow(CT)
The given map appears to be done simply in RGB, and could be reconstructed as a path. The problem is that there are quite a few breakpoints near the LHS of the image, and the compression artifacts pretty much turn any attempt to reconstruct it into an exercise in guessing. If exact replication of the breakpoints isn't an option, then simple interpolation is as good as any guess.
