returning the indexes of an image displayed with a specific colormap

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.

[Kevin Claytor]

Can you be more specific on what you want to do? You already have the numeric data - it's stored in IMAGE. If you want to look at that on the screen, you can print out a certain range; "IMAGE(1:10,1:10)" or get a single pixel value with the cursor + sticky note icon in the figure toolbar.

[Mario Trevino]

Suppose you have an IMAGE and display it using imagesc(IMAGE) using a colormap that switches between black and white 0,1,0,1... etc the displayed result is a black and white map composed by 0s and 1s.... I want to recover this matrix of 0s and 1s....

[Matt Fig]

What format is the IMAGE? m-by-n-by-3 or m-by-n or what?

[Sean de Wolski]

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

[Mario Trevino]

@ 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.

Answer 1

[Matt Fig]

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))

[Mario Trevino]

its getting closer but still not there... the problem is that my colormap is not a linear function. check it out

% building the colormap
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);

in this case how would you get your 'idx'????

[Matt Fig]

I think that INTERP1 will not work. But this seems to:

A = rand(3,3)
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 for your colormap colormap_MT, use:

idx = ceil(A*size(colormap_MT,1))

[Mario Trevino]

no, sorry, its not working. the outcome shoul be a colormap with 0s and 1s only... and its not being mapped correctly.... i still dont 'get what I see'...

[Matt Fig]

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))

[Mario Trevino]

You got it man! Its working for me now!

I dont know how the reshape function works... would it hold with zeros?

that you very much Matt

[Matt Fig]

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]

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.