Matlab Formula Optimization: Radial Base Function

Question

• z is the doubling matrix, size Nx2;
• x is the doubling matrix, size Nx2;

sup = x(i, :);

phi(1, i) = {@(z) exp(-g * sum((z - sup(ones([size(z, 1) 1]),:)) .^ 2, 2))};

This is the radial base function (RBF) for logistic regression. Here is the formula:

I need your advice, can I optimize this formula? because he calls millions of times and it takes a lot of time ...

Answer 1

It seems that in the recent changes you introduced some syntax errors, but I think I understood what you were trying to do (from the first version).

Instead of REPMAT or indexing, to repeat the vector x(i,:) to match the z strings, consider using an effective BSXFUN :

 rbf(:,i) = exp( -g .* sum(bsxfun(@minus,z,x(i,:)).^2,2) ); 

The above obviously intersects each row x

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You can go one step further and use PDIST2 to calculate the Euclidean distance between each pair of rows in z and x :

 %# some random data X = rand(10,2); Z = rand(10,2); g = 0.5; %# one-line solution rbf = exp(-g .* pdist2(Z,X,'euclidean').^2); 

Now each value in the matrix: rbf(i,j) corresponds to the value of the function between z(i,:) and x(j,:)

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EDIT:

I have timed various methods, here is the code I used:

 %# some random data N = 5000; X = rand(N,2); Z = rand(N,2); g = 0.5; %# PDIST2 tic rbf1 = exp(-g .* pdist2(Z,X,'euclidean').^2); toc %# BSXFUN+loop tic rbf2 = zeros(N,N); for j=1:N rbf2(:,j) = exp( -g .* sum(bsxfun(@minus,Z,X(j,:)).^2,2) ); end toc %# REPMAT+loop tic rbf3 = zeros(N,N); for j=1:N rbf3(:,j) = exp( -g .* sum((Z-repmat(X(j,:),[N 1])).^2,2) ); end toc %# check if results are equal all( abs(rbf1(:)-rbf2(:)) < 1e-15 ) all( abs(rbf2(:)-rbf3(:)) < 1e-15 ) 

Results:

 Elapsed time is 2.108313 seconds. # PDIST2 Elapsed time is 1.975865 seconds. # BSXFUN Elapsed time is 2.706201 seconds. # REPMAT 

Answer 2

Amro mentioned some really good methods. But bsxfun can be additionally used by changing one of the matrices.

 >> type rm N = 5000; X = rand(N,2); Z = rand(N,2); g = 0.5; %BSXFUN+loop tic rbf2 = zeros(N,N); for j=1:N rbf2(:,j) = exp( -g .* sum(bsxfun(@minus,Z,X(j,:)).^2,2) ); end toc tic diffs = bsxfun(@minus, reshape(X', [1, 2, N]), Z); dist = reshape(sum(diffs.^2, 2), [N, N]); rbf3 = exp(-g .* dist); toc >> r Elapsed time is 2.235527 seconds. Elapsed time is 0.877833 seconds. >> r Elapsed time is 2.253943 seconds. Elapsed time is 1.047295 seconds. >> r Elapsed time is 2.234132 seconds. Elapsed time is 0.856302 seconds. >> max(abs(rbf2(:) - rbf3(:))) ans = 0 

You want to subtract each row X from each row Z. This is usually straightforward when one of them is a vector and the other is a matrix. But if both of them are matrices, we can do this by making sure that each matrix in the volume contains only one vector. I chose X here, but Z can be used interchangeably with X.