Performance – How to vectorize many ranked first external products?

% U and V are m-by-s matrices
A = zeros(s^2, m);% preallocate
for k = 1:m
 Ak = U(k,:)' * V(k,:);
 Ak = (Ak + Ak')/2;
 A(:, k) = Ak(:);
end

s = 5; m = 100000;
U = rand(m, s);
V = rand(m, s);< br />
% Iterate over large dimension
tic
B = zeros(s^2, m);
for k = 1:m
 Ak = U( k,:)' * V(k,:);
 Ak = (Ak + Ak')/2;
 B(:, k) = Ak(:);
end< br />toc

% Iterate over small dimension
tic
A = zeros(s, s, m);
for i = 1:s
 A(i,i,:) = U(:, i) .* V(:, i);
 for j = i+1:s
 A(i,j,:) = (U(:,i).*V(:,j) + U(:, j).*V(:, i))/2;
 A(j,i,:) = A(i ,j,:);
 end
end
A = reshape(A, [s^2, m]);
toc

% bsxfun -based solution
tic
A = bsxfun( @times, per mute( U, [1 3 2] ), permute( V, [1 2 3]) );
A = .5 * (A + permute( A, [1 3 2]) );
B = reshape( A, [m, s^2] )';
toc

Elapsed time is 0.547053 seconds.
Elapsed time is 0.042639 seconds.
Elapsed time is 0.039296 seconds.

% the outer product
A = bsxfun( @times, permute( U, [1 3 2] ), permute( V, [1 2 3]) );
% symmetrization
A = .5 * (A + permute( A, [1 3 2] ) );
% to vector (per product)
B = reshape( A, [ms^2] )';

Elapsed time is 0.217695 seconds.

Elapsed time is 0.037538 seconds.

 Elapsed time is 0.021507 seconds.

Can the following loop be vectorized? Each iteration in the loop forms an outer product, then symmetric and store the result as a column in a matrix. Expect m to be large (e.g., 1e4) and s to be small (e.g., 10).

% U and V are m-by-s matrices
A = zeros(s^2, m);% preallocate
for k = 1:m
 Ak = U(k,:)' * V(k,:);
 Ak = (Ak + Ak')/2;
 A(:, k) = Ak(:);
end

Edit

The following is a comparison of 3 different methods: iterative large dimension m, iterative small dimension s and the solution based on bsxfun (accepted and fastest answer) .

s = 5; m = 100000;
U = rand(m, s);
V = rand(m, s);< br />
% Iterate over large dimension
tic
B = zeros(s^2, m);
for k = 1:m
 Ak = U( k,:)' * V(k,:);
 Ak = (Ak + Ak')/2;
 B(:, k) = Ak(:);
end< br />toc

% Iterate over small dimension
tic
A = zeros(s, s, m);
for i = 1:s
 A(i,i,:) = U(:, i) .* V(:, i);
 for j = i+1:s
 A(i,j,:) = (U(:,i).*V(:,j) + U(:, j).*V(:, i))/2;
 A(j,i,:) = A(i ,j,:);
 end
end
A = reshape(A, [s^2, m]);
toc

% bsxfun -based solution
tic
A = bsxfun( @times, permute( U , [1 3 2] ), permute( V, [1 2 3]) );
A = .5 * (A + permute( A, [1 3 2]) );
B = reshape( A, [m, s^2] )';
toc

This is a time comparison:

Elapsed time is 0.547053 seconds.
Elapsed time is 0.042639 seconds.
Elapsed time is 0.039296 seconds.

Use bsxfun (this is how it is done a lot Fun):

% the outer product
A = bsxfun( @times, permute( U, [1 3 2] ), permute( V, [1 2 3]) );
% symmetrization
A = .5 * (A + permute( A, [1 3 2]) );
% to vector (per product)< br />B = reshape( A, [ms^2] )';

Benchmark test results (my machine):

>Original method (iteration big dim):< /p>

Elapsed time is 0.217695 seconds.

>”New” method (with smaller iterations):

Elapsed time is 0.037538 seconds.

>Interesting bsxfun:

Elapsed time is 0.021507 seconds.

As you As you can see, bsxfun needs ~2 / 3-1/2 of the fastest cycle…

Isn’t bsxfun very interesting?