Manipulation of matrix addition and multiplication.
1 vue (au cours des 30 derniers jours)
Afficher commentaires plus anciens
Hello Friends,
I have the following:
A = [1 2 3; 4 5 6; 7 8 9];
B = [10 11 12; 13 14 15];
[N1, D1] = size(A);
[N2, D2] = size(B);
A_sq = sum(A.^2, 2);
B_sq = sum(B.^2, 2)';
D = A_sq(:,ones(1,N2)) + B_sq(ones(1,N1),:) - 2.*(A*B');
where D is N1 x D1 matrix.
I want to write expression for D in one single step, i.e., something like this (this is for illustration purpose, but it should compute the same Euclidean distance as the code above):
D = sum(X - C).^2;
I will appreciate any advise.
3 commentaires
the cyclist
le 16 Août 2016
Also, this equivalent formulation seems closer to your prototype formula, but I still don't quite see a simpler set of matrix operations to get you there:
D = bsxfun(@plus,diag(A*A'),diag(B*B')') - 2.*(A*B')
(I think this version is likely more computationally intensive, but somewhat more elegant.)
James Tursa
le 16 Août 2016
Modifié(e) : James Tursa
le 16 Août 2016
How is this different from this earlier post which was already answered?
Réponse acceptée
Matt J
le 16 Août 2016
Modifié(e) : Matt J
le 16 Août 2016
Bp=permute(B,[3,2,1]);
D=reshape( sum(bsxfun(@minus, A, Bp).^2,2)) , N1,N2);
5 commentaires
Matt J
le 18 Août 2016
Well... permutes are expensive as compared to reshapes. I was seeking to minimize them. It is possible to do this entirely without permutes/transposes if the OP had organized the 3x1 vectors in matrix columns instead of matrix rows.
the cyclist
le 18 Août 2016
I repmat'ed his matrices to make them pretty huge, and found nearly identical timing for the reshape algorithm and the permute algorithm.
Interestingly, my original solution (in the comments)
D = bsxfun(@plus,sum(A.^2, 2),sum(B'.^2, 1)) - 2.*(A*B');
absolutely crushed both of these in timing.
So, as always, best to try to solve the problem (multiple ways if possible!), and then do optimization.
Plus de réponses (0)
Voir également
Catégories
En savoir plus sur Loops and Conditional Statements dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!