matlab:矢量化4D矩阵和

时间:2015-03-09 09:08:00

标签: performance matlab matrix vectorization nested-loops

我需要在MATLAB中执行以下计算:

其中w和v是具有N个元素的向量,A是四维矩阵(N ^ 4个元素)。这可以通过以下迂腐代码来实现:

N=10;
A=rand(N,N,N,N);
v=rand(N,1);
w=zeros(N,1);

for pp=1:N
  for ll=1:N
    for mm=1:N
      for nn=1:N
        w(pp)=w(pp)+A(pp,ll,mm,nn)*v(ll)*v(mm)*conj(v(nn));
      end
    end
  end
end

这是非常慢的。有没有办法在MATLAB中对这种求和进行矢量化?

2 个答案:

答案 0 :(得分:6)

方法#1

很少有reshapematrix multiplication -

A1 = reshape(A,N^3,N)*conj(v)
A2 = reshape(A1,N^2,N)*v
w = reshape(A2,N,N)*v

方法#2

使用一个bsxfunreshapematrix-multiplication -

A1 = reshape(A,N^3,N)*conj(v)
vm = bsxfun(@times,v,v.')
w = reshape(A1,N,N^2)*vm(:)

基准

本节比较了本文中列出的两种方法的运行时,Shai's post中首次测试的方法以及问题中列出的原始方法。

基准代码

N=100;
A=rand(N,N,N,N);
v=rand(N,1);

disp('----------------------------------- With Original Approach')
tic
%// .... Code from the original post   ...//
toc

disp('----------------------------------- With Shai Approach #1')
tic
s4 = sum( bsxfun( @times, A, permute( conj(v), [4 3 2 1] ) ), 4 ); 
s3 = sum( bsxfun( @times, s4, permute( v, [3 2 1] ) ), 3 );
w2 = s3*v; 
toc

disp('----------------------------------- With Divakar Approach #1')
tic
A1 = reshape(A,N^3,N)*conj(v);
A2 = reshape(A1,N^2,N)*v;
w3 = reshape(A2,N,N)*v;
toc

disp('----------------------------------- With Divakar Approach #2')
tic
A1 = reshape(A,N^3,N)*conj(v);
vm = bsxfun(@times,v,v.');
w4 = reshape(A1,N,N^2)*vm(:);
toc

运行时结果

----------------------------------- With Original Approach
Elapsed time is 4.604767 seconds.
----------------------------------- With Shai Approach #1
Elapsed time is 0.334667 seconds.
----------------------------------- With Divakar Approach #1
Elapsed time is 0.071905 seconds.
----------------------------------- With Divakar Approach #2
Elapsed time is 0.058877 seconds.

<强>结论

这篇文章中的第二种方法似乎是提供了比原始方法更快的 80x

答案 1 :(得分:4)

您可以尝试使用

假设vN - 1列向量(否则,应该稍微修改排列)。

% sum over n (4th dim)
s4 = sum( bsxfun( @times, A, permute( conj(v), [4 3 2 1] ) ), 4 ); 

现在,中期结果仅为N - 按 - N - 按 - N

% sum over m (3rd dim)
s3 = sum( bsxfun( @times, s4, permute( v, [3 2 1] ) ), 3 )

继续上一笔费用

% sum over l (2nd dim)
w = s3*v; 

<子> 考虑一下,您是否考虑在其多维版本中使用dot?我没有测试它,但它应该工作(可能是一些小的修正)。

s4 = dot( A, permute( conj(v), [4 3 2 1] ), 4 );
s3 = dot( s4, permute( v, [3 2 1] ), 3 );
w = s3*v;