Azzera filtri
Azzera filtri

Minimise memory requirements when importing many images

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I am using imread to read in 430 monochrome (16-bit) images of resolution 1024x1280. I apply a common mask to each image, which is a 1024x1280 logical array. I also dispose of any pixels with values outside the range 50-60000.
I'm storing the processed images as layers of a 3-d array:
cube1 = zeros(1024,1280,430);
for k=1:430
image = double(imread (myImageFiles(k).name)));
image(~mask) = NaN;
image(image<50 | image>60000) = NaN; % replace 'bad' pixels outside limits with NaN
cube1(:,:,k) = image;
end
The image is cast as a double because I want to replace certain pixels with NaNs. If i just read in as uint16 then these pixels are stored as zeros. I also want to create a second cube in the same way and divide the elements in cube1 by the corresponding elements in cube2.
newCube = cube1./cube2;
However, I am coming up against 'out of memory' issues (system has 8GB RAM). I can reduce memory by not casting as a double but lose the NaN feature. Please could someone advise on a way to tackle this? I've tried saving as a .mat file, but each cube array is over 4GB (before any compression).
  2 Commenti
Adam Danz
Adam Danz il 18 Lug 2022
Modificato: Adam Danz il 18 Lug 2022
Some ideas
  1. Use single instead of double.
  2. Since you're doing element-wise division, do you need to load all 430 images into the same array? Can you load image 1 from cube1 and then load image 1 from cube2, and then store the result of division, then get rid of the two image data sets and repeat for the next pair of images?

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Risposte (1)

Catalytic
Catalytic il 19 Lug 2022
Modificato: Catalytic il 19 Lug 2022
Why is it better to use NaNs instead of zeros? A value of 0 doesn't conflict with anything because all of your nontrivial values lie between 50 and 60000.
cube = zeros(1024,1280,430,'single');
for k=1:430
image1 = imread (myFiles1(k).name));
image2 = imread (myFiles2(k).name));
image1(image1<50 | image1>60000) = 0;
image2(image2<50 | image2>60000) = 0;
image3 = single(image1)./single(image2);
image3(~isfinite(image3))=0;
cube(:,:,k)=image3;
end

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R2020b

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