How to speed up my code?

Question

George Bashkatov 2022-2-20

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1654510-how-to-speed-up-my-code

评论： George Bashkatov 2022-2-22

I have a code with a lot of cycles and matrix multiplication. Input data: vector M of matrices 2x2, n2 - length of vector M (number of matrices 2x2). k - current matrix. My matrix do cyclic multiplication. It goes from the k-th element to the first, after that to last and after that to the (k+1)th.

E.g. n2=6, k=2: M2*M1*M2*M3*M4*M5*M6*M5*M4*M3

I was trying to turn vector M to gpuArray, also I was trying to turn cycles, that started from "for i=1:n2" to parfor but it didn't help. So, it's the problem with number of cycles? Can you give me some advices how to solve these problems and how avoid problems like these in future?

for i=1:n2
     K(:,:,i)=M(:,:,i);
 end
 for i=1:n2 %return the original matrix for the new cycle
           J(:,:,i)=K(:,:,i);
 end
%cycle from k-th element to the beginning 
if k == 1
else
  for i=k-1:-1:1
    M(:,:,i)=M(:,:,i+1)*M(:,:,i);
    K(:,:,i)=M(:,:,i);
  end
%return the original matrix, change the first element
  for i=1:n2
    M(:,:,i)=J(:,:,i);
  end
    M(:,:,1)=K(:,:,1);
end
%loop from beginning to end
  for i=1:n2-1
    M(:,:,i+1)=M(:,:,i)*M(:,:,i+1);
    K(:,:,i+1)=M(:,:,i+1);
  end
%return the original matrix, change the "last" element
  for i=1:n2
    M(:,:,i)=J(:,:,i);
  end
M(:,:,n2)=K(:,:,n2);
if n2 == k
M(:,:,1)=K(:,:,n2-1);
else
%loop from end to (k+1)th element
  for i=n2:-1:k+2
    M(:,:,i-1)=M(:,:,i)*M(:,:,i-1);
  end
end

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Akira Agata 2022-2-20

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1654510-how-to-speed-up-my-code#answer_900740

在 MATLAB Online 中打开

Avoiding for-loop will enhance computational efficiency.

For example, the first for-loop:

for i=1:n2
  K(:,:,i)=M(:,:,i);
end

should be as follows:

K(:,:,1:n2)=M(:,:,1:n2);

3 个评论
显示 1更早的评论隐藏 1更早的评论

Walter Roberson 2022-2-22

@George Bashkatov

No, for two different reasons:

the * operator is not vectorizable. The closest you can get to that https://www.mathworks.com/help/matlab/ref/pagemtimes.html
With you looping like that, you are creating a cummulative matrix product, M(:,:,end), M(:,:,end)*M(:,:,end-1), M(:,:,end)*M(:,:,end-1)*M(:,:,end-2), and so on. So the results for any given iteration depend upon the results for the previous iterations. That is not something that can be vectorized... not even with pagemtimes .

If you were using the .* operator instead of the * operator then with a permute(), reshape(), flipud(), cumprod(), flipud(), reshape(), permute() you could get the answer without a loop. But with the * operator, you need a loop.

George Bashkatov 2022-2-22