How to remove for-loop in computing a vector of sums?

How to compute rcross vector without the for-loop?

   rcross = zeros( length(elementProduct) - prefixLen , 1 );
   for ii = 1: length(elementProduct) - prefixLen
      rcross(ii) = sum( elementProduct(ii : ii + prefixLen - 1) );
   end

In one scenario prefixLen is 8. length(elementProduct) can vary from a few hundred to a few million.

[KSSV]

It looks like a moving window.....have a look on movsum.

[Paul Hoffrichter]

Thank you. Very close. prefixLen is 8. M is shifted to right by 4 points. Any way to adjust so that M is equal to rcross?

M = movsum( elementProduct, prefixLen  );

I will check for odd prefixLen. May not be as good.

[Image Analyst]

Another way is to use conv() and there you have options for how to handle end effects, like 'full', 'valid', or 'same'.

windowWidth = 9; % Usually an odd number but doesn't have to be.
kernel = ones(windowWidth);
outputSum = conv(elementProduct, kernel, 'same');

[Paul Hoffrichter]

Had some issues with odd vs even prefixLen, but then found this one that worked. Thanks.

M = movsum( elementProduct, prefixLen, 'Endpoints','discard');

rcross and M were then off by extremely small amounts.

[Paul Hoffrichter]

@KSSV,

Thank you for your comment. If you move it to an answer, I will accept it. Thanks again!

~Paul

[Paul Hoffrichter]

@Image Analyst,

I think your approach will also work. But I believe (without testing) that using conv would be slower than movsum.