我有两个数组，A并且B维数相同1000 x 3 x 20 x 20。我想生成一个第三阵列C尺寸的3 x 3 x 20 x 20，这将是对应的切片的矩阵乘法的结果A和B，即C(:,:,i,j) = A(:,:,i,j)'*B(:,:,i,j)。然后，我需要通过反转相应的矩阵（即）将数组C转换为新数组。同样，很清楚如何使用循环执行此操作。有没有一种方法可以使物品仿制循环？D3 x 3D(:,:,i,j) = inv(C(:,:,i,j))400
编辑：比较不同解决方案性能的基准代码将是-
%// Inputs
n1 = 50;
n2 = 200;
A = rand(n1,3,n2,n2);
B = rand(n1,3,n2,n2);
%// A. CPU loopy code
tic
C = zeros(3,3,n2,n2);
for ii = 1:n2
for jj = 1:n2
C(:,:,ii,jj) = A(:,:,ii,jj)'*B(:,:,ii,jj); %//'
end
end
toc
%// B. Vectorized code (using squeeze)
tic
C1 = squeeze(sum(bsxfun(@times,permute(A,[2 1 5 3 4]),permute(B,[5 1 2 3 4])),2));
toc
%// C. Vectorized code (avoiding squeeze)
tic
C2 = sum(bsxfun(@times,permute(A,[2 5 3 4 1]),permute(B,[5 2 3 4 1])),5);
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%// D. GPU vectorized code
tic
A = gpuArray(A);
B = gpuArray(B);
C3 = sum(bsxfun(@times,permute(A,[2 5 3 4 1]),permute(B,[5 2 3 4 1])),5);
C3 = gather(C3);
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运行时结果-
Elapsed time is 0.287511 seconds.
Elapsed time is 0.250663 seconds.
Elapsed time is 0.337628 seconds.
Elapsed time is 1.259207 seconds.
代码
%// Part - 1
C = sum(bsxfun(@times,permute(A,[2 5 3 4 1]),permute(B,[5 2 3 4 1])),5);
%// Part - 2: Use MATLAB file-exchange tool multinv
D = multinv(C);
对于第一部分，您还可以尝试以下操作-
C = squeeze(sum(bsxfun(@times,permute(A,[2 1 5 3 4]),permute(B,[5 1 2 3 4])),2));
这似乎是在重新排列元素，而不是上面代码中提到的那样“破坏性”，但是缺点是需要这样做squeeze可能会使它变慢一点。我会把它留给您，也鼓励您进行基准测试并选择更好的一个。
bsxfun+ GPU？我增加了循环限制，因为这可能是对循环代码和矢量化代码之间的真实测试。因此，这是第1部分的修改后的代码-
%// Inputs
n1 = 50;
n2 = 200;
A = rand(n1,3,n2,n2);
B = rand(n1,3,n2,n2);
%// A. CPU loopy code
tic
C = zeros(3,3,n2,n2);
for ii = 1:n2
for jj = 1:n2
C(:,:,ii,jj) = A(:,:,ii,jj)'*B(:,:,ii,jj); %//'
end
end
toc
%// B. GPU vectorized code
tic
A = gpuArray(A);
B = gpuArray(B);
C1 = sum(bsxfun(@times,permute(A,[2 5 3 4 1]),permute(B,[5 2 3 4 1])),5);
C1 = gather(C1);
toc
我的系统的运行时结果是-
Elapsed time is 0.310056 seconds.
Elapsed time is 0.172499 seconds.
所以你看！
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