Improving the performance of a for loop in Matlab

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Yodish
Yodish el 19 de Jul. de 2017
Editada: Yodish el 20 de Jul. de 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 comentarios
dpb
dpb el 19 de Jul. de 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 el 20 de Jul. de 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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David Goodmanson
David Goodmanson el 20 de Jul. de 2017
Editada: David Goodmanson el 20 de Jul. de 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 comentario
Yodish
Yodish el 20 de Jul. de 2017
Editada: Yodish el 20 de Jul. de 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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