Faster way to calculate the euclidean distance of a matrix than pdist and squareform?
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Dennis_Pana
am 21 Jul. 2016
Beantwortet: Walter Roberson
am 21 Jul. 2016
design=[0.625 0.875;0.25 0.75;0.375 0.125;0.875 0.375;0.125 0.625;
0 0;0.5 0.5;0.75 0.25;0.6875 0.0625;0.0625 0.8125;0.1875 0.6875;0.3125 0.3125;
0.4375 0.9375;0.5625 0.5625;0.9375 0.1875;0.8125 0.4375];
v=16;
tic
d(:,:)=squareform(pdist(design(1:v,:))).^2;
toc
This code needs about 0.009 seconds to run. I need something faster, because this calculation is in a for loop. Thank you in advance.
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Walter Roberson
am 21 Jul. 2016
You can use bsxfun of @minus to do the subtractions, then .^2 to square the components, sum() along the appropriate dimension, and .^(1/2) the result. This should be marginally faster than calling pdist, but probably not much faster.
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