large 3D Matrix calculation

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)

Answer 1

[Stephen23]

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.

[rahman]

tnx Stephen Cobeldick. your code was completely helpfull ;)

Answer 2

[Stephen23]

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...

[rahman]

Dear friends let me say exactly what I need ! suppose that size(f) = 1*350 ; size(r)=50*50*900; I want to calculate the S matrix as below

 for ii=1:350
  for jj=1:900
   s(ii,jj) = sum(sum(sum( repmat(f(ii),[50,50,1]) - r(:,:,jj) ))) ;
  end
 end
 Now, I want to calculate this matrix without for loops and by faster matrix calculation method !

[Stephen23]

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]

The heartache of those who write code - shifting specs. The correct answer to the question asked but not the question intended. :)

Answer 3

[dpb]

d=repmat(f,size(r,2),1).-repmat(r.',1,size(f,2));

[rahman]

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) )