Matlab中的张量收缩 [英] Tensor contraction in Matlab
问题描述
可能重复:
MATLAB:如何对两个矩阵进行矢量乘积? a>
Possible Duplicate:
MATLAB: How to vector-multiply two arrays of matrices?
在Matlab中是否可以收缩高维张量?
Is there a way to contract higher-dimensional tensors in Matlab?
例如,假设我有两个具有以下大小的3维数组:
For example, suppose I have two 3-dimensional arrays, with these sizes:
size(A) == [M,N,P]
size(B) == [N,Q,P]
我想分别在第二个和第一个索引上收缩A
和B
.换句话说,我想将A
视为大小为[M,N]
的矩阵的数组,并将B
视为等长长度的[N,Q]
矩阵的数组;我想将这些数组逐个元素地相乘以得到大小为[M,Q,P]
的东西.
I want to contract A
and B
on the second and first indices, respectively. In other words, I want to consider A
to be an array of matrices of size [M,N]
and B
to be equal length array of [N,Q]
matrices; I want to multiply these arrays element-by-element (matrix-by-matrix) to get something of size [M,Q,P]
.
我可以通过for循环来做到这一点:
I can do this via a for-loop:
assert(size(A,2) == size(B,1));
assert(size(A,3) == size(B,3));
M = size(A,1);
P = size(A,3);
Q = size(B,2);
C = zeros(M, Q, P);
for ii = 1:size(A,3)
C(:,:,ii) = A(:,:,ii) * B(:,:,ii);
end
有没有一种方法可以避免for循环? (也许可以处理任意维数的数组?)
Is there a way to do this that avoids the for-loop? (And perhaps works with arrays of an arbitrary number of dimensions?)
推荐答案
Here is a solution (similar to what was done here) that computes the result in a single matrix-multiplication operation, although it involves heavy manipulation of the matrices to put them into desired shape. I then compare it to the simple for-loop computation (which I admit is a lot more readable)
%# 3D matrices
A = rand(4,2,3);
B = rand(2,5,3);
[m n p] = size(A);
[n q p] = size(B);
%# single matrix-multiplication operation (computes more products than needed)
AA = reshape(permute(A,[2 1 3]), [n m*p])'; %'# cat(1,A(:,:,1),...,A(:,:,p))
BB = reshape(B, [n q*p]); %# cat(2,B(:,:,1),...,B(:,:,p))
CC = AA * BB;
[mp qp] = size(CC);
%# only keep "blocks" on the diagonal
yy = repmat(1:qp, [m 1]);
xx = bsxfun(@plus, repmat(1:m,[1 q])', 0:m:mp-1); %'
idx = sub2ind(size(CC), xx(:), yy(:));
CC = reshape(CC(idx), [m q p]);
%# compare against FOR-LOOP solution
C = zeros(m,q,p);
for i=1:p
C(:,:,i) = A(:,:,i) * B(:,:,i);
end
isequal(C,CC)
请注意,上面执行的乘法比所需的更多,但有时 任何人谁添加,会(从执行时间开始)减损". .遗憾的是事实并非如此,因为这里的FOR循环要快得多:)
Note that the above is performing more multiplications than needed, but sometimes "Anyone who adds, detracts (from execution time)". Sadly this is not the case, as the FOR-loop is much faster here :)
我的观点是证明向量化并不容易,并且基于循环的解决方案并不总是不好的...
My point was to show that vectorization is not easy, and that loop-based solutions are not always bad...
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