returning the indexes of an image displayed with a specific colormap

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Mario Trevino 2012-8-14

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采纳的回答： Matt Fig

hi, does anyone know hot to get the numeric indexes of a displayed image with a specific user defined colormap???

for example imagesc(IMAGE) colormap(user_colormap)

displays an image mapped into the colormap....and I want the numeric representation of the displayed image.

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Sean de Wolski 2012-8-14

That's what IMAGE is.... It's just a 2d matrix of indices into the colormap.

Mario Trevino 2012-8-14

@ Matt: m-by-n only. @ Sean: I know! yet the colormap is transforming the matrix, and I want THAT transformation (or mapping), and not the original one. I want to recover the transformed 2d matrix after using imagesc with a specific user defined colormap.

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Matt Fig 2012-8-14

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编辑：Matt Fig 2012-8-14

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If I understand you correctly, you want (for example):

A = rand(4);
C = [0 0 0;.5 .5 .5;1 1 1];
imagesc(A);
colormap(C);
%  idx = interp1([0 .5 1],[0 .5 1],A,'nearest')
idx = ceil(A*size(C,1))

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Matt Fig 2012-8-14

编辑：Matt Fig 2012-8-14

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Oh, you said you want the indexes into the colormap, which is what I gave you. Now you want to return the values of the colormap using those indices. That's just another step.

Look at idx here:

A = rand(3)
C = [0 0 0;.5 .5 .5;1 1 1];
imagesc(A,[0 1]);  % Note the second arg.
colormap(C)
idx = ceil(A*size(C,1))

So if you want to look at the colors for each pixel of A, index into the colormap:

C(idx(1,1),:)  % The color of pixel A(1,1)
C(idx(2,1),:)  % The color of pixel A(2,1),  etc.

So for your task, do the same.

Now since you only have black and white, the first column of idx can denote the color. Take a look:

A = rand(3)
k=32; % cycles
x=(0:2*pi/63:2*pi)';
y=(-cos(k.*x)+1)./2; % sinusoidal waveform between 0 and 1
% variant (black and white)
yy=y;
yy(find(y>=0.5))=1;
yy(find(y<0.5))=0;
% colormap_MT=horzcat(y,y,y);
colormap_MT=horzcat(yy,yy,yy);
imagesc(A,[0 1])
colormap(colormap_MT)
idx = ceil(A*size(colormap_MT,1));
% Finally, we index into the map with idx.  Compare to the image.
CI = reshape(colormap_MT(idx,1),size(A))

Matt Fig 2012-8-14

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I don't know what you mean by holding with zeros. But the reshape function does pretty much what the name indicates. Take a look:

X = [1 0 2 0 3 0 4 0 5 0 6 0]
reshape(X,4,3)
reshape(X,3,4)
reshape(X,6,2)
reshape(X,2,6)
reshape(X,12,1)

Mario Trevino 2012-8-14

you are absolutely right Matt. What I saw is that none of the elements from A (or idx) can be 0 for the reshape function to work.

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