矢量和嵌套矩阵乘法

Question

这里是原代码：

K = zeros(N*N) 
for a=1:N 
    for i=1:I 
     for j=1:J 
      M = kron(X(:,:,a).',Y(:,:,a,i,j)); 

      %A function that essentially adds M to K. 
     end 
    end 
end 

的目标是向量化的kroniker乘法电话。我的直觉是将X和Y看作矩阵的容器（供参考，被馈送给kron的X和Y的切片是7x7阶的矩形矩阵）。在此容器方案下，X显示一维容器，Y显示为三维容器。我的下一个猜测是将Y重塑为二维容器或更好的一维容器，然后再进行X和Y的元素乘法运算。问题是：如何以保留M的轨迹的方式进行重构？ matlab甚至可以在这个容器构思中处理这个想法，还是需要进一步重构容器以进一步暴露内部矩阵元素？

Answer 1

方法＃1：矩阵乘法与6D置换

% Get sizes 
[m1,m2,~] = size(X); 
[n1,n2,N,n4,n5] = size(Y); 

% Lose the third dim from X and Y with matrix-multiplication 
parte1 = reshape(permute(Y,[1,2,4,5,3]),[],N)*reshape(X,[],N).'; 

% Rearrange the leftover dims to bring kron format 
parte2 = reshape(parte1,[n1,n2,I,J,m1,m2]); 

% Lose dims correspinding to last two dims coming in from Y corresponding 
% to the iterative summation as suggested in the question 
out = reshape(permute(sum(sum(parte2,3),4),[1,6,2,5,3,4]),m1*n1,m2*n2) 

方法2：简单7D置换

% 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])); 

% Lose the two dims with summation reduction for final output 
out = sum(reshape(mults,m1*n1,m2*n2,[]),3); 

---

验证

下面是运行原始和提出的方法的设置 -

% Setup inputs 
X = rand(10,10,10); 
Y = rand(10,10,10,10,10); 

% Original approach 
[n1,n2,N,I,J] = size(Y); 
K = zeros(100); 
for a=1:N 
    for i=1:I 
     for j=1:J 
      M = kron(X(:,:,a).',Y(:,:,a,i,j)); 
      K = K + M; 
     end 
    end 
end 

% Approach #1 
[m1,m2,~] = size(X); 
[n1,n2,N,n4,n5] = size(Y); 
mults = bsxfun(@times,permute(X,[4,2,5,1,3]),permute(Y,[1,6,2,7,3,4,5])); 
out1 = sum(reshape(mults,m1*n1,m2*n2,[]),3); 

% Approach #2 
[m1,m2,~] = size(X); 
[n1,n2,N,n4,n5] = size(Y); 
parte1 = reshape(permute(Y,[1,2,4,5,3]),[],N)*reshape(X,[],N).'; 
parte2 = reshape(parte1,[n1,n2,I,J,m1,m2]); 
out2 = reshape(permute(sum(sum(parte2,3),4),[1,6,2,5,3,4]),m1*n1,m2*n2); 

运行后，我们看到了最大。与提出的方法相对于原始方法的绝对偏差 -

>> error_app1 = max(abs(K(:)-out1(:))) 
error_app1 = 
    1.1369e-12 
>> error_app2 = max(abs(K(:)-out2(:))) 
error_app2 = 
    1.1937e-12 

值看起来不错！

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标杆

使用相同的大数据集的时序这三种方法作为用于验证，我们得到这样的事情 -

----------------------------- With Loop 
Elapsed time is 1.541443 seconds. 
----------------------------- With BSXFUN 
Elapsed time is 1.283935 seconds. 
----------------------------- With MATRIX-MULTIPLICATION 
Elapsed time is 0.164312 seconds. 

好像矩阵乘法做得相当不错为这些尺寸的数据集！