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在张量内矢量化矩阵乘法

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我在渲染部分代码时遇到了一些麻烦 . 我有一个（n，n，m）张量，我想将每个切片乘以m乘以秒（n乘n）矩阵（非元素） .

这是for循环的样子：

Tensor=zeros(2,2,3);
Matrix = [1,2; 3,4];

for j=1:n
    Matrices_Multiplied = Tensor(:,:,j)*Matrix;
    Recursive_Matrix=Recursive_Matrix + Tensor(:,:,j)/trace(Matrices_Multiplied);
end

如何以矢量化方式对张量内的各个矩阵执行矩阵乘法？是否有像tensor-dot这样的内置函数可以处理这个还是更聪明？

matlab matrix vectorization matrix-multiplication

1 回答

  • 1

    Bsxfunning并使用efficient matrix-multiplication，我们可以 -

    % Calculate trace values using matrix-multiplication
T = reshape(Matrix.',1,[])*reshape(Tensor,[],size(Tensor,3));

% Use broadcasting to perform elementwise division across all slices
out = sum(bsxfun(@rdivide,Tensor,reshape(T,1,1,[])),3);

    同样，可以用一个矩阵乘法替换最后一步，以进一步提高性能 . 因此，所有矩阵乘法专用解决方案将是 -

    [m,n,r] = size(Tensor);
out = reshape(reshape(Tensor,[],size(Tensor,3))*(1./T.'),m,n)

    Runtime test

    基准代码 -

    % Input arrays
n = 100; m = 100;
Tensor=rand(n,n,m);
Matrix=rand(n,n);
num_iter = 100; % Number of iterations to be run for

tic
disp('------------ Loopy woopy doops : ')
for iter = 1:num_iter
    Recursive_Matrix = zeros(n,n);
    for j=1:n
        Matrices_Multiplied = Tensor(:,:,j)*Matrix;
        Recursive_Matrix=Recursive_Matrix+Tensor(:,:,j)/trace(Matrices_Multiplied);
    end
end
toc, clear iter  Recursive_Matrix  Matrices_Multiplied

tic
disp('------------- Bsxfun matrix-mul not so dull : ')
for iter = 1:num_iter
    T = reshape(Matrix.',1,[])*reshape(Tensor,[],size(Tensor,3));
    out = sum(bsxfun(@rdivide,Tensor,reshape(T,1,1,[])),3);
end
toc, clear T out

tic
disp('-------------- All matrix-mul having a ball : ')
for iter = 1:num_iter
    T = reshape(Matrix.',1,[])*reshape(Tensor,[],size(Tensor,3));
    [m,n,r] = size(Tensor);
    out = reshape(reshape(Tensor,[],size(Tensor,3))*(1./T.'),m,n);
end
toc

    计时 -

    ------------ Loopy woopy doops : 
Elapsed time is 3.339464 seconds.
------------- Bsxfun matrix-mul not so dull : 
Elapsed time is 1.354137 seconds.
-------------- All matrix-mul having a ball : 
Elapsed time is 0.373712 seconds.
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