两个数组的相应2d切片的乘法和数组切片的反演

Question

我有两个相同维度A的数组B和1000 x 3 x 20 x 20。我想生成维度C的第三个数组3 x 3 x 20 x 20，它将是A和B的相应切片的矩阵乘法的结果，即C(:,:,i,j) = A(:,:,i,j)'*B(:,:,i,j)。然后，我需要通过反转相应的C矩阵（即D）将数组3 x 3转换为新数组D(:,:,i,j) = inv(C(:,:,i,j))。同样，很清楚如何使用循环来完成此操作。有没有办法绕过400个项目？

编辑：用于比较不同解决方案的效果的基准代码将是 -

%// 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
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%// 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));
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%// 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.

Answer 1

<强>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);

multinv的功能代码可用here，声称效率非常高。

对于第一部分，您也可以试试这个 -

C = squeeze(sum(bsxfun(@times,permute(A,[2 1 5 3 4]),permute(B,[5 1 2 3 4])),2));

这个似乎正在重新安排这些元素，而不是像＃34;破坏性地＆＃34;正如上面代码中提到的那个，但缺点是需要squeeze可能会减慢它的速度。我会留给你，也鼓励你选择更好的基准。

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为什么bsxfun + GPU？

我增加了循环限制，因为这可能是循环代码和矢量化代码之间的真正测试。所以，这是第1部分的修改代码 -

%// 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);
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我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！