重新发布更多的细节,这大大改变了我第一个问题的范围。这是原始代码:
K = zeros(N*N)
for a=1:N
for i=1:I
for j=1:J
M = kron(X(:,:,a).',Y(:,:,a,i,j));
pr = real(trace(E*M));
K = K+H(i,j,a)*M/pr;
end
end
end其中E是布尔掩码,H是包含N个IxJ直方图的三维矩阵。K是输出。
目标是将kroniker乘法调用向量化。我的直觉是把X和Y看作矩阵的容器(作为参考,给kron的X和Y切片是7x7阶的方阵)。在这个容器方案下,X出现一个一维容器,Y作为一个三维容器.我的下一个猜测是将Y重塑成一个二维容器,或者更好的是一个一维容器,然后对X和Y进行按元素方向的乘法。问题是:如何以一种保持M的痕迹的方式进行整形,matlab甚至可以在这个容器的概念中处理这个想法,或者容器需要进一步的整形才能进一步暴露内部矩阵元素?
发布于 2016-08-03 20:20:32
用7D permute进行矩阵乘法
% Get sizes
[m1,m2,~] = size(X);
[n1,n2,N,n4,n5] = size(Y);
% Perform kron format elementwise multiplication betwen the first two dims
% of X and Y, keeping the third dim aligned and "pushing out" leftover dims
% from Y to the back
mults = bsxfun(@times,permute(X,[4,2,5,1,3]),permute(Y,[1,6,2,7,3,4,5]));
mults3D = reshape(mults,m1*n1,m2*n2,[]);
Emults3D = reshape(E*reshape(mults3D,size(mults3D,1),[]),size(mults3D));
% Trace summations by using linear indices of diagonal on 3D slices in Emults3D
MN = m1*n1;
idx = 1:MN+1:MN^2;
idx2D = bsxfun(@plus,idx(:),MN^2*(0:size(Emults3D,3)-1));
pr_sums = sum(Emults3D(idx2D),1);
% Perform "M/pr" equivalent elementwise divisions and then use
% matrix-multiplication to reduce the iterative summations
Mp = bsxfun(@rdivide,mults3D,reshape(pr_sums,1,1,[]));
out = reshape(Mp,[],size(Mp,3))*reshape(permute(H,[3,1,2]),[],1);
out = reshape(out,m1*n1,m2*n2);标杆
输入的设置是这样的-
% Size parameter
n = 5;
% Setup inputs
X = rand(n,n,n);
Y = rand(n,n,n,n,n);
E = rand(n*n,n*n)>0.5;
H = rand(n,n,n);
num_iter = 500; % Number of iterations to run the approaches for运行时的结果是-
----------------------------- With Loop
Elapsed time is 8.806286 seconds.
----------------------------- With Vectorization
Elapsed time is 1.471877 seconds.将大小参数n设置为10,运行时为-
----------------------------- With Loop
Elapsed time is 5.068872 seconds.
----------------------------- With Vectorization
Elapsed time is 4.399783 seconds.https://stackoverflow.com/questions/38751675
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