Multiply then sum elements of two matrices

10 vues (au cours des 30 derniers jours)
F
F le 28 Fév 2011
Commenté : Ron Fredericks le 12 Déc 2020
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?
  1 commentaire
Ron Fredericks
Ron Fredericks le 12 Déc 2020
Another way to calculate Frobenious dot product:
trace( A'*B );

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Réponse acceptée

Walter Roberson
Walter Roberson le 28 Fév 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 commentaire
Edoardo Bezzeccheri
Edoardo Bezzeccheri le 25 Mai 2016
The mathematical name of the operation is called Frobenius Product, by the way.

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Plus de réponses (3)

Bruno Luong
Bruno Luong le 28 Fév 2011
I believe
sum(A(:).*B(:))
would be faster than dot(...) if that matter.
  2 commentaires
Walter Roberson
Walter Roberson le 28 Fév 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 le 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 le 28 Fév 2011
  2 commentaires
James Tursa
James Tursa le 28 Fév 2011
Or if you prefer to stay with MATLAB built-in functions:
A(:).'*B(:)
F
F le 1 Mar 2011
James, thanks for the answer as well!

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Matt Fig
Matt Fig le 28 Fév 2011
Another (this should be faster than calling SUM):
reshape(A,1,[])*reshape(B,[],1)
  9 commentaires
James Tursa
James Tursa le 28 Fév 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 le 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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