large 3D Matrix calculation

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rahman
rahman 2015-1-20
评论： rahman 2015-1-25
Hi All I have two large matrix and I want to calculate an expression without for loop. this expression is something like matrix d as below
f=[f1 f2 f3]
r=[r1 r2 r3]
d=[f1-r1 f2-r1 f3-r1
f1-r2 f2-r2 f3-r2
f1-r3 f2-r2 f3-r3]
can any one help me?! consider that size(f)=1*1*400 and size(r)=50*50*900 and fii=f(1,1,ii) and rjj=r(:,:,jj)

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采纳的回答

Stephen23
Stephen23 2015-1-20
编辑：Stephen23 2015-1-21
You gave this code in a comment to my other answer:
size(f)==[1,350]
size(r)==[50,50,900]
for ii=1:350
for jj=1:900
s(ii,jj) = sum(sum(sum( repmat(f(ii),[50,50,1]) - r(:,:,jj) ))) ;
end
end
You can try this instead:
A = 50*50*reshape(f,1,[]);
B = reshape(sum(sum(r,1),2),[],1);
C = bsxfun(@minus,A,B);
It produces the same result as your nested loops.
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rahman
rahman 2015-1-25
tnx Stephen Cobeldick. your code was completely helpfull ;)

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更多回答（2 个）

Stephen23
Stephen23 2015-1-20
编辑：Stephen23 2015-1-20
This is exactly what bsxfun is for:
f = [f1,f2,f3];
r = [r1,r2,r3];
d = bsxfun(@minus,f,r.');
bsxfun calculates the output without requiring large intermediate arrays (eg using repmat). Although, depending on the size of d, you might still run out of memory...
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显示 1更早的评论
Stephen23
Stephen23 2015-1-20
编辑：Stephen23 2015-1-20
What you have now described is a different problem to the one that you posed in your original question. My code exactly solves your original question.
John D'Errico
John D'Errico 2015-1-20
The heartache of those who write code - shifting specs. The correct answer to the question asked but not the question intended. :)

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dpb
dpb 2015-1-20
d=repmat(f,size(r,2),1).-repmat(r.',1,size(f,2));
  1 个评论
rahman
rahman 2015-1-20
tnx dpb but this operation needs much RAMs according to what I said ( consider that size(f)=1*1*400 and size(r)=50*50*900 and fii=f(1,1,ii) and rjj=r(:,:,jj) )

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