Compress down a 1000x1000 matrix into a 100x100 matrix

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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

Respuesta aceptada

Ameer Hamza
Ameer Hamza el 7 de Jun. de 2020
Editada: Ameer Hamza el 7 de Jun. de 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 comentarios
Robert Jansen
Robert Jansen el 24 de Mzo. de 2021
Amazing array of solutions. These all give the same results. The last one is surprisingly fast too. Thanks.
larasupernovae
larasupernovae el 3 de Mzo. de 2022
thank you so much!

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Más respuestas (2)

David Hill
David Hill el 7 de Jun. de 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 el 24 de Mzo. de 2021
Editada: Jan el 24 de Mzo. de 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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