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Improving the performance of a for loop in Matlab

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Yodish
Yodish on 19 Jul 2017
Edited: Yodish on 20 Jul 2017
Hello,
I am trying to improve the speed of the following for loop in Matlab. As it is now is incredibly slow. Maybe vectorizing? But one vector slides on the other and constantly changes.
for m = 1:size(phi,1) - (constant)/2
phi(m) = phi(m).*(mean(conj(phi(1+m:(constant)/2+m))));
end
Thanks!

  2 Comments

dpb
dpb on 19 Jul 2017
What are phi and constant? Is phi actually complex; and even if so,
mean(conj(x)) --> conj(mean(x))
so can be outside the loop.
Yodish
Yodish on 20 Jul 2017
Hi and thanks for the answer. phi would be a complex vector whereas constant is a constant number (of elements) < than the length of the aforementioned vector phi
Michele

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Accepted Answer

David Goodmanson
David Goodmanson on 20 Jul 2017
Edited: David Goodmanson on 20 Jul 2017
Hello Yodish,
'one vector slides on the other'. This is basically convolution, and since you are finding a mean it is convolution with a vector of ones, of length constant/2. Since the code is not iterative, all the means can be found first.
N = 10000;
constant = 5000;
phi_orig = rand(1,N)+i*rand(1,N);
tic % first way
phi = phi_orig;
for m = 1:length(phi) - (constant)/2
phi(m) = phi(m).*(mean(conj(phi(1+m:(constant)/2+m))));
end
toc
tic % second way
phi2 = phi_orig;
c2 = constant/2;
A = conv(phi2,ones(1,c2),'valid')/c2; % means
A(1) = [];
sA = length(A);
phi2(1:sA) = phi2(1:sA).*conj(A);
toc
d = max(abs(phi2-phi)) % should be zero
In your code I used length(phi) rather than size(phi,1) so that phi could be either a row vector or a column vector.

  1 Comment

Yodish
Yodish on 20 Jul 2017
Yes thanks, it is indeed a convolution. It would also work with the movmean function. They both improve speed by a factor of 100.
Thanks for your precious help guys

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