性能 – 如何矢量化许多排名第一的外部产品？

下面的循环可以矢量化吗？循环中的每次迭代形成一个外部产品,然后对称并将结果存储为矩阵中的列.预计m很大(例如,1e4)并且s很小(例如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

编辑

以下是3种不同方法的比较：迭代大维度m,迭代小维度s和基于bsxfun的解决方案(接受且最快的答案).

s = 5; m = 100000;
U = rand(m, s);
V = rand(m, s);

% 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
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

这是一个时间比较：

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

使用bsxfun(这是如何完成它的很多乐趣)：

% 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, [m s^2] )';

基准测试结果(我的机器)：

>原始方法(迭代大昏暗)：

Elapsed time is 0.217695 seconds.

>“新”方法(迭代较小的暗淡)：

Elapsed time is 0.037538 seconds.

>有趣的bsxfun：

Elapsed time is 0.021507 seconds.

正如你所看到的,bsxfun需要~2 / 3 – 最快循环的1/2时间……

bsxfun不是很有趣吗？