Matlab：我如何执行行操作而不用蛮力for循环？ [英] Matlab: how can I perform row operations without brute-force for loop?

问题描述

  N1 = size（X，1）; 
 N2 = size（Xtrain，1）; 
 Dist =零（N1，N2）; （i，j）= D-sum（X（i，：）== Xtrain（j）（1）：
 
对于i = 1：N1 
对于j = 1： ，:)）; 
 end 
 end 
  

（X和Xtrain是稀疏的逻辑矩阵） >

它工作正常，并通过测试，但我相信这不是非常优化和写得很好的解决方案。

我可以使用一些构建的Matlab函数来改进这个函数吗？对于Matlab来说，我绝对是新手，所以我不知道是否真的有机会让它更好。 解决方案

你想学习矢量化，这里有一些代码来比较不同的实现这种成对的距离。

首先我们建立两个二进制矩阵作为输入（其中每行是一个实例）：

  m = 5; 
 n = 4; 
 p = 3; 
 A = double（rand（m，p）> 0.5）; 
 B = double（rand（n，p）> 0.5）; 
  

1。在每对实例上进行双循环

$ $ $ $ $ $ $ $ D $ =零（m，n）; （i，j）= B（j，:)）/ p（i，j）= ;
end
end

2。 PDIST2

  D1 = pdist2（A，B，'hamming'）; 
  

3。在每个实例上针对所有其他实例的单循环

  D2 =零（m，n）; 
 for i = 1：n 
 D2（：，i）= sum（bsxfun（@ne，A，B（i，:)），2）./ p; 
 end 
  

4。使用网格索引进行矢量化，全部针对所有

  D3 =零（m，n）; 
 [x，y] = ndgrid（1：m，1：n）; 
 D3（:) = sum（A（x（:)，:)〜= B（y（:)，:)，2）./ p; 
  

5。向量化在第三维，所有对所有

  D4 = sum（bsxfun（@ne，A，reshape（B。'， 1 pn]）），2）./ p; 
 D4 =置换（D4，[1 3 2]）; 
  

最后我们比较所有方法是否相等

  assert（isequal（D0，D1，D2，D3，D4））
  

I need to do function that works like this :

N1 = size(X,1);
N2 = size(Xtrain,1);
Dist = zeros(N1,N2);

    for i=1:N1
        for j=1:N2
            Dist(i,j)=D-sum(X(i,:)==Xtrain(j,:));
        end
    end

(X and Xtrain are sparse logical matrixes)

It works fine and passes the tests, but I believe it's not very optimal and well-written solution.

How can I improve that function using some built Matlab functions? I'm absolutely new to Matlab, so I don't know if there really is an opportunity to make it better somehow.

解决方案

You wanted to learn about vectorization, here some code to study comparing different implementations of this pair-wise distance.

First we build two binary matrices as input (where each row is an instance):

m = 5;
n = 4;
p = 3;
A = double(rand(m,p) > 0.5);
B = double(rand(n,p) > 0.5);

1. double-loop over each pair of instances

D0 = zeros(m,n);
for i=1:m
    for j=1:n
        D0(i,j) = sum(A(i,:) ~= B(j,:)) / p;
    end
end

2. PDIST2

D1 = pdist2(A, B, 'hamming');

3. single-loop over each instance against all other instances

D2 = zeros(m,n);
for i=1:n
    D2(:,i) = sum(bsxfun(@ne, A, B(i,:)), 2) ./ p;
end

4. vectorized with grid indexing, all against all

D3 = zeros(m,n);
[x,y] = ndgrid(1:m,1:n);
D3(:) = sum(A(x(:),:) ~= B(y(:),:), 2) ./ p;

5. vectorized in third dimension, all against all

D4 = sum(bsxfun(@ne, A, reshape(B.',[1 p n])), 2) ./ p;
D4 = permute(D4, [1 3 2]);

Finally we compare all methods are equal

assert(isequal(D0,D1,D2,D3,D4))

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