Quickest way to join images preserving each colormap in MATLAB
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I am a having problem finding a way to join images preserving each colormap. I basically want to read images from a folder, place them in a rectangular grid (image11, image12; image21, image22) each with its own colormap. I have tried imtile() but I get black boxes at times and also for some reason the colormap is a greyscale instead of colours.
What would be the quickest way to do this? maybe forming a supermatrix containing all pixels of all images and then saving that as an image?
My pseudocode is simple:
for namefiles
im = imread(namefile);
imshow(im); %this works
out=imtile(im,'GridSize', [ypos xpos]); %this shows the pictures in black and white
imshow(out)
end
2 个评论
Jonas
2022-7-4
can you provide a sample set and can you show an image of your problem/error.
Are all images of same size, is the position only defined by the file name and which colormap do you want to use? or do you mean the colors as they should be when you use imshow() on them?
采纳的回答
DGM
2022-7-4
编辑:DGM
2022-7-4
Read the images into a cell array in the loop. If the images are indexed color images, then convert them to RGB using ind2rgb() before storing them. Call imtile() or montage() on the cell array outside the loop.
nframes = 4;
C = cell(nframes,1);
for f = 1:nframes
[thisframe map] = imread(sprintf('pepframe%d.png',f));
if ~isempty(map)
thisframe = ind2rgb(thisframe,map);
end
C{f} = thisframe;
end
montage(C)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1054610/image.png)
5 个评论
DGM
2022-7-8
编辑:DGM
2022-7-8
Ah. I didn't think about that. By default, ind2rgb() produces an array of class 'double'. You can reduce the memory requirements by a factor of 8 by doing:
C{f,sim_case} = im2uint8(thisframe);
If they still won't fit in memory, then the problem gets more complicated. There may have to be some compromises made. If there are further memory issues, it would be helpful to know how large the images actually are (height, width) and how much memory is available. That way I can at least estimate what might fit. If the images are all the same size, the requirements can be reduced by a factor of about 2.
The memory constraints would also mean that using mimread()/imstacker() from MIMT would be out of the question.
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