pdist2 equivalent in MATLAB version 7

2020-01-25 11:25发布

I need to calculate the euclidean distance between 2 matrices in matlab. Currently I am using bsxfun and calculating the distance as below( i am attaching a snippet of the code ):

for i=1:4754
test_data=fea_test(i,:);
d=sqrt(sum(bsxfun(@minus, test_data, fea_train).^2, 2));
end

Size of fea_test is 4754x1024 and fea_train is 6800x1024 , using his for loop is causing the execution of the for to take approximately 12 minutes which I think is too high. Is there a way to calculate the euclidean distance between both the matrices faster?

I was told that by removing unnecessary for loops I can reduce the execution time. I also know that pdist2 can help reduce the time for calculation but since I am using version 7. of matlab I do not have the pdist2 function. Upgrade is not an option.

Any help.

Regards,

Bhavya

3条回答
等我变得足够好
2楼-- · 2020-01-25 11:29

Here is vectorized implementation for computing the euclidean distance that is much faster than what you have (even significantly faster than PDIST2 on my machine):

D = sqrt( bsxfun(@plus,sum(A.^2,2),sum(B.^2,2)') - 2*(A*B') );

It is based on the fact that: ||u-v||^2 = ||u||^2 + ||v||^2 - 2*u.v


Consider below a crude comparison between the two methods:

A = rand(4754,1024);
B = rand(6800,1024);

tic
D = pdist2(A,B,'euclidean');
toc

tic
DD = sqrt( bsxfun(@plus,sum(A.^2,2),sum(B.^2,2)') - 2*(A*B') );
toc

On my WinXP laptop running R2011b, we can see a 10x times improvement in time:

Elapsed time is 70.939146 seconds.        %# PDIST2
Elapsed time is 7.879438 seconds.         %# vectorized solution

You should be aware that it does not give exactly the same results as PDIST2 down to the smallest precision.. By comparing the results, you will see small differences (usually close to eps the floating-point relative accuracy):

>> max( abs(D(:)-DD(:)) )
ans =
  1.0658e-013

On a side note, I've collected around 10 different implementations (some are just small variations of each other) for this distance computation, and have been comparing them. You would be surprised how fast simple loops can be (thanks to the JIT), compared to other vectorized solutions...

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▲ chillily
3楼-- · 2020-01-25 11:30

You could fully vectorize the calculation by repeating the rows of fea_test 6800 times, and of fea_train 4754 times, like this:

rA = size(fea_test,1);
rB = size(fea_train,1);

[I,J]=ndgrid(1:rA,1:rB);

d = zeros(rA,rB);

d(:) = sqrt(sum(fea_test(J(:),:)-fea_train(I(:),:)).^2,2));

However, this would lead to intermediary arrays of size 6800x4754x1024 (*8 bytes for doubles), which will take up ~250GB of RAM. Thus, the full vectorization won't work.

You can, however, reduce the time of the distance calculation by preallocation, and by not calculating the square root before it's necessary:

rA = size(fea_test,1);
rB = size(fea_train,1);
d = zeros(rA,rB);

for i = 1:rA
    test_data=fea_test(i,:);
    d(i,:)=sum( (test_data(ones(nB,1),:) -  fea_train).^2, 2))';
end

d = sqrt(d);
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Rolldiameter
4楼-- · 2020-01-25 11:36

Try this vectorized version, it should be pretty efficient. Edit: just noticed that my answer is similar to @Amro's.

function K = calculateEuclideanDist(P,Q)
% Vectorized method to compute pairwise Euclidean distance
% Returns K(i,j) = sqrt((P(i,:) - Q(j,:))'*(P(i,:) - Q(j,:)))

[nP, d] = size(P);
[nQ, d] = size(Q);

pmag = sum(P .* P, 2);
qmag = sum(Q .* Q, 2);

K = sqrt(ones(nP,1)*qmag' + pmag*ones(1,nQ) - 2*P*Q');

end
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