Matlab formula optimization: Radial Basis Function - matlab

Question

z - matrix of doubles, size Nx2;
x - matrix of doubles, size Nx2;
sup = x(i, :);
phi(1, i) = {#(z) exp(-g * sum((z - sup(ones([size(z, 1) 1]),:)) .^ 2, 2))};
this is a Radial Basis Function (RBF) for logistic regression. Here is the formula:
I need your advice, can i optimize this formula? coz it calls millions times, and it takes a lot of time...

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
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,:)
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

Answer 2

Amro has mentioned some really good methods. But the bsxfun can be further exploited by reshaping one of the matrices.
>> type r.m
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 every row of X from every row of Z. This usually is straight forward 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 just one vector. Here I chose X, but Z can be used interchangeably with X.