Compress down a 1000x1000 matrix into a 100x100 matrix

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Ahmed Abdulla
Ahmed Abdulla am 7 Jun. 2020
Kommentiert: larasupernovae am 3 Mär. 2022
I have a 1000x1000 matrix that contains various values. I would like to compress it down into a 100 by 100 matrix by averaging each 10 by 10 cell's values into 1, but im not really sure how to go about doing this

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Ameer Hamza
Ameer Hamza am 7 Jun. 2020
Bearbeitet: Ameer Hamza am 7 Jun. 2020
If you have image processing toolbox
M = rand(1000, 1000);
M_new = blockproc(M, [10 10], @(x) mean(x.data, 'all'))
Alternative Solution 1: (surprisingly the fastest)
M_new = conv2(M, ones(10)/100, 'valid');
M_new = M_new(1:10:end, 1:10:end);
Alternative Solution 2:
M_C = mat2cell(M, 10*ones(1,100), 10*ones(1,100));
M_new = cellfun(@(x) mean(x, 'all'), M_C);
Alternative Solution 3:
M_new = zeros(size(M)/10);
for i=1:100
for j=1:100
M_new(i, j) = mean(M(10*(i-1)+1:10*i,10*(j-1)+1:10*j), 'all');
end
end
  2 Kommentare
Robert Jansen
Robert Jansen am 24 Mär. 2021
Amazing array of solutions. These all give the same results. The last one is surprisingly fast too. Thanks.

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David Hill
David Hill am 7 Jun. 2020
count=1;
for col=1:100:1000
for row=1:100:1000
newMatrix(count)=mean(yourMatrix(row:row+99,col:col+99),'all');
count=count+1;
end
end
newMatrix=reshape(newMatrix,10,[]);

Jan
Jan am 24 Mär. 2021
Bearbeitet: Jan am 24 Mär. 2021
X = rand(1000, 1000);
Y = reshape(X, [10, 100, 10, 100]);
Z = reshape(sum(sum(Y, 1), 3), [100, 100]) / 100;
Or with FEX: BlockMean :
Z = BlockMean(X, 10, 10)

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