Fast computation of diagonal elements

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Saurabh
Saurabh am 5 Nov. 2013
Kommentiert: Matt J am 5 Nov. 2013
I have two matrices, A of size m-by-n and B of size m-by-m. I need to quickly compute the digonal elements of (A'*B*A). Here m is of order 100 while n is of order 10000.
How would I do it ? Doing diag(A'*B*A) will be slow.
Neither B nor A is diagonal. they are full matrices.

Akzeptierte Antwort

Matt J
Matt J am 5 Nov. 2013
Bearbeitet: Matt J am 5 Nov. 2013
diagonal = sum(A.*(B*A),1);
  2 Kommentare
Azzi Abdelmalek
Azzi Abdelmalek am 5 Nov. 2013
or
sum((A'*B).*A',2)
Matt J
Matt J am 5 Nov. 2013
It works. But you'll have overhead from the transpositions.

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Weitere Antworten (2)

Sean de Wolski
Sean de Wolski am 5 Nov. 2013
m = 100;
n = 10000;
A = rand(m,n);
B = rand(m);
timeit(@()diag(A'*B*A)) % R2013b
ans = 0.8128
So it takes slightly less than a second on my Win7x64 laptop. How many times do you need to do this computation?
  2 Kommentare
Saurabh
Saurabh am 5 Nov. 2013
I need to compute this for possibly 200 iterations.
Sean de Wolski
Sean de Wolski am 5 Nov. 2013
Bearbeitet: Sean de Wolski am 5 Nov. 2013
I.e. less than three minutes total? I would just do the above 200x and go get a cup of coffee while MATLAB is crunchin'

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Azzi Abdelmalek
Azzi Abdelmalek am 5 Nov. 2013
Bearbeitet: Azzi Abdelmalek am 5 Nov. 2013
This will reduce the time
S=A'*B;
d1=arrayfun(@(x) S(x,:)*A(:,x),1:n);
  1 Kommentar
Saurabh
Saurabh am 5 Nov. 2013
This solution is slightly slower than Matt's.

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