Vectorizing bsxfun

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Brendan
Brendan am 16 Aug. 2011
Hi,
I have two matrices (which are really lists of vectors) and would like a matrix of the pair-wise squared distances between all of them. The following code does what I want, but I'm curious if there's any way to vectorize this.
Thank you
rX=rand(nTrain,numDim);
rXClass=rand(nClass,numDim);
dists=zeros(nTrain,nClass);
for ii=1:nTrain
thisX=rX(ii,:);
dists(ii,:)=sum(bsxfun(@minus,thisX,rXClass).^2,2)/D;
end

Antworten (3)

the cyclist
the cyclist am 16 Aug. 2011
rX2 = permute(rX,[1 3 2]);
rXClass2 = permute(rXClass,[3 1 2]);
dists = sum(bsxfun(@minus,rX2,rXClass2).^2,3)/D;

Sean de Wolski
Sean de Wolski am 16 Aug. 2011
dists2 = squeeze(sum(bsxfun(@minus,rX,reshape(rXClass',[1 numDim nClass])).^2,2))/D;
With all three sizes equaling 150, I have your elementary for-loop running the fastest:
nTrain = 150;
numDim = 150;
nClass = 150;
D = 1;
rX=rand(nTrain,numDim);
rXClass=rand(nClass,numDim);
t1 = 0;
t2 = 0;
t3 = 0;
for jj = 1:50
tic
dists=zeros(nTrain,nClass);
for ii=1:nTrain
thisX=rX(ii,:);
dists(ii,:)=sum(bsxfun(@minus,thisX,rXClass).^2,2)/D;
end
t1 = t1+toc;
tic
dists2 = squeeze(sum(bsxfun(@minus,rX,reshape(rXClass',[1 numDim nClass])).^2,2))/D;
t2 = t2+toc;
tic
rX2 = permute(rX,[1 3 2]);
rXClass2 = permute(rXClass,[3 1 2]);
dists3 = sum(bsxfun(@minus,rX2,rXClass2).^2,3)/D;
t3 =t3+toc;
end
isequal(dists,dists2,dists3)
[t1 t2 t3]
ans =
1
ans =
3.1505 4.0336 4.0368
  4 Kommentare
Sean de Wolski
Sean de Wolski am 16 Aug. 2011
Nope, probably not. The only hope I can think of for that one is maybe with James' mtimesx on File Exchange.
the cyclist
the cyclist am 17 Aug. 2011
If you prefer the vectorized code, they can both be sped up a fair amount by pulling apart the one-liner calculation of "dists" into separate lines for the bsxfun call, the squaring, and the sum

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Andrei Bobrov
Andrei Bobrov am 16 Aug. 2011
my small contribution:
[a b] = meshgrid(1:nTrain,1:nClass);
dists = reshape(sum((rX(a(:),:) - rXClass(b(:),:)).^2,2),[],nTrain)'/D

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