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下面的循环可以矢量化吗？循环中的每次迭代都会形成一个外积，然后将结果对称化并存储为矩阵中的一列。预计m大（例如，1e4）和s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bsxfun基于解决方案（公认且最快的答案）。

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

下面是时间对比：

Elapsed time is 0.547053 seconds.
Elapsed time is 0.042639 seconds.
Elapsed time is 0.039296 seconds.
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1 回答 1

1

使用bsxfun（这就是它的乐趣所在）：

% 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）：

  1. 原始方法（迭代大暗淡）：

     Elapsed time is 0.217695 seconds.

  2. “新”方法（迭代较小的暗淡）：

     Elapsed time is 0.037538 seconds.

  3. 乐趣bsxfun：

     Elapsed time is 0.021507 seconds.

如您所见bsxfun，大约需要最快循环时间的 2/3 - 1/2...

不是很有趣吗？

于 2013-07-10T06:15:01.303 回答