Multiply then sum elements of two matrices

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What's the best way to 1) multiply all the elements between two similar sized matrices then 2) sum them into one single number/outcome?

Accepted Answer

Walter Roberson
Walter Roberson on 28 Feb 2011
It is not clear from your question whether corresponding elements are to be multiplied or if you are wanting to do a matrix multiplication.
If corresponding elements are to be multiplied, then you could calculate the sum-of-products using
dot(A(:),B(:))
  1 Comment
Edoardo Bezzeccheri
Edoardo Bezzeccheri on 25 May 2016
The mathematical name of the operation is called Frobenius Product, by the way.

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More Answers (3)

Bruno Luong
Bruno Luong on 28 Feb 2011
I believe
sum(A(:).*B(:))
would be faster than dot(...) if that matter.
  2 Comments
Walter Roberson
Walter Roberson on 28 Feb 2011
Heh. I just looked at the source for dot. It does sum(conj(a).*b) so yes, your suggestion should be slightly faster due to have a small bit less overhead.
F
F on 1 Mar 2011
Hi Bruno and Walter, thanks for the answers. I used them both and they worked! I'm new to Matlab so I don't know yet how to clock the processing time....I will learn. Great user community feedback!

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James Tursa
James Tursa on 28 Feb 2011
  2 Comments
James Tursa
James Tursa on 28 Feb 2011
Or if you prefer to stay with MATLAB built-in functions:
A(:).'*B(:)
F
F on 1 Mar 2011
James, thanks for the answer as well!

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Matt Fig
Matt Fig on 28 Feb 2011
Another (this should be faster than calling SUM):
reshape(A,1,[])*reshape(B,[],1)
  9 Comments
James Tursa
James Tursa on 28 Feb 2011
e.g., 32-bit WinXP Intel Core 2 Duo:
>> A = rand(3000) + rand(3000)*1i;
>> B = rand(3000) + rand(3000)*1i;
>>
>> tic;dot(A(:),B(:));toc
Elapsed time is 0.158458 seconds.
>> tic;dot(A(:),B(:));toc
Elapsed time is 0.165634 seconds.
>>
>> tic;A(:)'*B(:);toc
Elapsed time is 0.101896 seconds.
>> tic;A(:)'*B(:);toc
Elapsed time is 0.101962 seconds.
>>
>> mtimesx('speedomp')
ans =
SPEEDOMP
>> tic;mtimesx(A(:),'C',B(:));toc
Elapsed time is 0.061016 seconds.
>> tic;mtimesx(A(:),'C',B(:));toc
Elapsed time is 0.061558 seconds.
The symmetric case difference is even more dramatic:
>> tic;dot(A(:),A(:));toc
Elapsed time is 0.146473 seconds.
>> tic;dot(A(:),A(:));toc
Elapsed time is 0.146098 seconds.
>>
>> tic;A(:)'*A(:);toc
Elapsed time is 0.084071 seconds.
>> tic;A(:)'*A(:);toc
Elapsed time is 0.082746 seconds.
>>
>> tic;mtimesx(A(:),'C',A(:));toc
Elapsed time is 0.032382 seconds.
>> tic;mtimesx(A(:),'C',A(:));toc
Elapsed time is 0.032521 seconds.
F
F on 1 Mar 2011
Matt, thanks for your feedback as well. Yours works too - but I never would have figured that out...something to think about. Again, great user community here.

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