Question

我有两个似乎密切相关的matlab问题。

我想找到一种最有效的方法（没有循环？）将（A×A）矩阵与3d矩阵的每个矩阵（A x A x N）相乘。此外，我想了解每个产品的痕迹。 http://en.wikipedia.org/wiki/Matrix_multiplication#Frobenius_product

这是内部frobenius产品。关于我在下面的糟糕代码，我正在使用更高效的二级定义。

我想将向量（N x 1）的每个元素与其3d矩阵（A x A x N）的“对应”矩阵相乘。

function Y_returned = problem_1(X_matrix, weight_matrix)

% X_matrix is the randn(50, 50, 2000) matrix
% weight_matrix is the randn(50, 50) matrix

[~, ~, number_of_matries] = size(X_matrix);
Y_returned = zeros(number_of_matries, 1);
for i = 1:number_of_matries
%     Y_returned(i) = trace(X_matrix(:,:,i) * weight_matrix');
    temp1 = X_matrix(:,:,i)';
    temp2 = weight_matrix';
    Y_returned(i) =  temp1(:)' * temp2(:);
end
end


function output = problem_2(vector, matrix)

% matrix is the randn(50, 50, 2000) matrix
% vector is the randn(2000, 1) vector

[n1, n2, number_of_matries] = size(matrix);
output = zeros(n1, n2, number_of_matries);
for i = 1:number_of_matries
    output(:, :, i) = vector(i) .* matrix(:, :, i);
end
output = sum(output, 3);

end

Answer 1

我认为你的意思是元素乘法：

使用bsxfun：

A = 10;
N = 4;
mat1 = randn(A,A);
mat2 = randn(A,A,N);
result = bsxfun(@times, mat1, mat2);

使用bsxfun与permute对齐维度：

A = 10;
N = 4;
vec1 = rand(N,1);
mat2 = randn(A,A,N);
result = bsxfun(@times, permute(vec1,[2 3 1]), mat2);

Matlab - 将矩阵与3d矩阵的每个矩阵相乘

1 个答案: