autocorrelate rows of matrix without using a for loop
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Hello, can anyone help me to write 'vectorized' code to compute the autocorrelation of each row of a matrix without using a for loop? My current code computes the autocorrelation of the rows like this:
for m=1:10
b(m,:)=xcorr( a(m,:) );
end
This is slow and inefficient. There must be a better way? Thanks, -Cody
Réponses (6)
Rick Rosson
le 25 Août 2011
0 votes
Hi Cody,
Can you please answer the following questions?
- Did you pre-allocate the output array b ?
- Also, have you considered that MATLAB stores arrays column-major, so that your code might be faster and more efficient if you used the transpose of a, even though it is still in a for loop?
Thanks!
Rick
Sean de Wolski
le 25 Août 2011
for m=10:-1:1
b(m,:)=xcorr( a(m,:) );
end
5 commentaires
Rick Rosson
le 25 Août 2011
Hi Sean,
I am confused: why is this code faster or more efficient than the original code?
Thanks!
Rick
Cody
le 25 Août 2011
Honglei Chen
le 25 Août 2011
@Rick, @Cody
When you write it as Sean does, MATLAB knows that it should preallocate the memory for 10 rows at the first iteration, sort of like a preallocation. Otherwise, MATLAB increases memory allocation at each iteration.
Honglei
Sean de Wolski
le 26 Août 2011
Honglei explained it well, it dynamically preallocates b to have m rows of length(xcorr(a(m,:))) so basically getting you the speed gain of calling zeros before hand
b = zeros(m,length_of_2nd_dim);
for ii = 1:m
%do stuff
end
A few months ago Matt Fig showed dynamically preallocating, as I did a above, to sometimes be faster than preallocating the conventional way. I can't seem to find that question though :(
Cody
le 26 Août 2011
Rick Rosson
le 25 Août 2011
Also, please check the documentation for xcorr:
>> doc xcorr
It looks like you may be able to accomplish what you want without a for loop if you first transpose a and then consider each column of a as an independent channel.
HTH.
Rick
1 commentaire
Cody
le 25 Août 2011
Honglei Chen
le 25 Août 2011
You can use FFT if your data is large, e.g.,
ffta = fft(a,NFFT,2);
b = fftshift(ifft(ffta.*conj(ffta),[],2),2)
Choose your NFFT as 2*size(a,2)-1 to match xcorr behavior. You may gain even more if you can transpose your data first to make these function working along columns.
HTH
2 commentaires
Cody
le 26 Août 2011
Sebastian Holmqvist
le 21 Mar 2013
This is actually what xcorr does. Just my two cents.
Rick Rosson
le 25 Août 2011
0 votes
Is the issue here that you really need to make this code run faster, or is it that you are just hoping to find a more elegant (e.g. vectorized) way to accomplish this task without resorting to a for loop?
How much time is it taking to run this code now? How fast do you need it to be?
2 commentaires
Cody
le 26 Août 2011
Sean de Wolski
le 26 Août 2011
0.29 seconds for how big a matrix?
Chaowei Chen
le 27 Août 2011
0 votes
The idea is to convert mat to cell since each row is independent. After processing, convert cell back to mat.
a2=mat2cell(a,ones(1,size(a,1)),size(a,2));
b2=cellfun(@xcorr,a2,'uniformoutput',false);
b2=cell2mat(b2);
isequal(b,b2)
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