how can i make this code to run faster using vectorization?

aslan shaghal

2018 9 14

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2 次查看（30 天）

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I have an adjacency matrix(A) of 0 and 1 elements. for each of the elements in the matrix that is equal to 1, I have to replace it with 0. then exponentiate the matrix until that element will become a none zero element. so I have written this code here, the problem is it's very slow and I have no idea how to improve that. I would appreciate it if anyone could help me. A=[0 1 1 1 0; 1 0 1 0 1; 1 1 0 1 1; 1 0 1 0 0; 0 1 1 0 0]

 clc;
clear;
adjacency=load ('sample');
A=adjacency.A;
n=size(A,1);
An=A;
b=zeros(n,n);
for i=1:n
    for j=i+1:n      %because the matrix is symmetric I do this to get rid of extra calculation
      if A(i,j)==1
          A(i,j)=0;
          An(i,j)=0;
        while An(i,j)==0
            An=An*A;
        end
            b(i,j)=An(i,j);
            A(i,j)=1;      % this line is for changing the A to the original matrix
            An=A;          % this line is for changing the An to the original matrix
      else
          b(i,j)=0;
      end
      b(j,i)=b(i,j);
    end
end

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Adam Danz 2018-9-14

编辑：Adam Danz 2018-9-14

在 MATLAB Online 中打开

The load() function is probably what's slowing your code down, not the loops. I replaced the first 4 lines of your code with

A=[0 1 1 1 0; 1 0 1 0 1; 1 1 0 1 1;1 0 1 0 0; 0 1 1 0 0];

and ran it 10,000 times which only took 0.063 seconds (using tic/toc). The mean run time was 0.0000063 seconds with a standard deviation of 0.000017. That's pretty fast. (using matlab 2018b)

aslan shaghal 2018-9-14

编辑：aslan shaghal 2018-9-14

在 MATLAB Online 中打开

I should mention that the actual A is much larger than this and it could be up to 2000 by 2000; after profiling the code, the part that slows down the code is:

An=An*A

I try to replace this part with