bsxfun vs for loop. Code Optimization.

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Hello everyone.
I'm trying to optimize some code that is running in a for loop. The code is the following:
for k=...
n=20;
u = rand(4386,21,n);
v = rand(size(u(:,:,1)));
F = zeros(size(u,1),n);
for m=1:n
F(:,m) = squeeze(max(u(:,:,m).*v,[],2));
end
output(k) = function(F);
end
I've tried replacing the inner for loop by bsxfun , this way:
for k=...
n=20;
u = rand(4386,21,n);
v = rand(size(u(:,:,1)));
F = squeeze(max(bsxfun(@times,u,v),[],2));
output(k) = function(F);
end
But bsxfun is taking more time than the previous solution.
Why is that happening?
Any other ideas to optimize this code?
I'm working with the specified dimensions, and the bottleneck of the problem is located in the creation of array F .
Thanks in advance.
Dani

Accepted Answer

Star Strider
Star Strider on 28 Apr 2014
If I understand bsxfun correctly, it’s taking longer because it’s expanding v to multiply elements of v by every element of u, not simply the 3rd dimension.
From the documentation:
Whenever a dimension of A or B is singleton (equal to one), bsxfun virtually replicates the array along that dimension to match the other array.
  2 Comments
Daniel Pereira
Daniel Pereira on 29 Apr 2014
The thing is that v is [4836×21], so I understand that bsxfun is only expanding the third dimension, that is the first singleton dimension that does not match the dimension of u.
I think the two algorithms are doing the same, since both give the same result for F.
Thank you anyway.
Star Strider
Star Strider on 29 Apr 2014
Edited: Star Strider on 29 Apr 2014
They’re both doing the same on the dimensions of interest to you, with bsxfun doing it on the third dimension as well, making it slower. They would give you the same result, because you’re only looking at some of the dimensions of F. The bsxfun function is just doing what you asked it to.
My pleasure!

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