Optimizing: Getting rid of for loop

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Raphaël
Raphaël el 5 de Dic. de 2017
Editada: Matt J el 5 de Dic. de 2017
Hi, I've been scratching my head on this problem and keep seem to figureout a good way to do it.
I have a big matrix : size(A)=1024x1024x400 where the 3rd dimension is that of time and the two first of space. Here is an example of what I would want to do with it:
for x=1:1022
for y=1:1022
B(y,x,1) = mean2(A(y:y+2,x:x+2,1));
end
end
Doing it like that works but Matlab will do the calculations on one element at the time instead of all at the same time like in the following case:
B(:,:,1) = A(:,:,1).*A(:,:,2);
Is there a way to rewrite my initial code or use a function that would make it way faster? Thank you for any help.

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Guillaume
Guillaume el 5 de Dic. de 2017
Editada: Guillaume el 5 de Dic. de 2017
Not sure what the mean2d function do as it's not a standard matlab function. Assuming it's calculating a mean, then probably
kernel = ones(3, 3, size(A, 3));
kernel = kernel ./ numel(kernel); %convolution kernel for mean
B = convn(A, kernel, 'valid');
  2 comentarios
Raphaël
Raphaël el 5 de Dic. de 2017
Yes it works thanks. One question, do you know the difference between conv2 and convn for this? Thanks
Matt J
Matt J el 5 de Dic. de 2017
Editada: Matt J el 5 de Dic. de 2017
Because of the separable nature of the kernel, it may be more efficient to do this in separable passes,
B=convn(A,[1,1,1]/9, 'valid');
B=convn(B,[1;1;1],'valid');

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