Fast computation of diagonal elements
3 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
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.
0 Kommentare
Akzeptierte Antwort
Weitere Antworten (2)
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
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'
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);
Siehe auch
Kategorien
Mehr zu Operating on Diagonal Matrices finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!