Matlab formula optimization: Radial Basis Function

Question

• z - matrix of doubles, size Nx2;
• x - matrix of doubles, size Nx2;

sup = x(i, :);

phi(1, i) = {@(z) exp(-g * sum((z -

Answer 1

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

Instead of using REPMAT or indexing to repeat the vector x(i,:) to match the rows of z, consider using the efficient BSXFUN function:

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

The above obviously loops over every row of x

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You can go one step further, and use the PDIST2 to compute the euclidean distance between every 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 every value in the matrix: rbf(i,j) corresponds to the function value between z(i,:) and x(j,:)

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EDIT:

I timed the different 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 )

The results:

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