Manipulation of matrix addition and multiplication.
1 visualizzazione (ultimi 30 giorni)
Mostra commenti meno recenti
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 Commenti
the cyclist
il 16 Ago 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
il 16 Ago 2016
Modificato: James Tursa
il 16 Ago 2016
How is this different from this earlier post which was already answered?
Risposta accettata
Matt J
il 16 Ago 2016
Modificato: Matt J
il 16 Ago 2016
Bp=permute(B,[3,2,1]);
D=reshape( sum(bsxfun(@minus, A, Bp).^2,2)) , N1,N2);
5 Commenti
Matt J
il 18 Ago 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
il 18 Ago 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.
Più risposte (0)
Vedere anche
Categorie
Scopri di più su Loops and Conditional Statements in Help Center e File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!