- 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?
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
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Antworten (6)
Rick Rosson
am 25 Aug. 2011
Hi Cody,
Can you please answer the following questions?
Thanks!
Rick
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Sean de Wolski
am 25 Aug. 2011
for m=10:-1:1
b(m,:)=xcorr( a(m,:) );
end
5 Kommentare
Sean de Wolski
am 26 Aug. 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 :(
Rick Rosson
am 25 Aug. 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
Honglei Chen
am 25 Aug. 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 Kommentare
Rick Rosson
am 25 Aug. 2011
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 Kommentare
Chaowei Chen
am 27 Aug. 2011
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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