使矩阵对称 [英] Make a matrix symmetric

问题描述

我有一个矩阵，根据理论应该是对称的，但是在我的数据中可能不会被视为对称的.我想通过使用两个比较的单元格中的最大值来使它对称.

I have a matrix that should be symmetric according to theory, but might not be observed as symmetric in my data. I would like to force this to be symmetric by using the maximum of the two compared cells.

test_matrix <- matrix(c(0,1,2,1,0,1,1.5,1,0), nrow = 3)
test_matrix
#>     [,1] [,2] [,3]
#>[1,]    0    1  1.5
#>[2,]    1    0  1.0
#>[3,]    2    1  0.0

使用双循环执行此操作很容易.

It's easy enough to do this with a double loop.

for(i in 1:3){
  for(j in 1:3){
    test_matrix[i, j] <- max(test_matrix[i, j], test_matrix[j, i]) 
   }
}
test_matrix
#>      [,1] [,2] [,3]
#> [1,]    0    1    2
#> [2,]    1    0    1
#> [3,]    2    1    0

但是我的矩阵大于$ 3x3 $，R的循环问题已得到充分证明.我也对使我的代码尽可能整洁感兴趣.实际上，我考虑过将其放在代码高尔夫上，但这是一个真正的问题，我认为其他人可能对此感兴趣.

But my matrix is larger than $3x3$, and R's problems with loops are well-documented. I'm also interested in making my code as clean as possible. In fact, I considered putting this on code golf, but this is a real problem that I think others might be interested in.

我见过这以及

I've seen this one as well as this one, but mine is different in that those op's seemed to actually have a symmetric matrix that just needed reordering, and I have a matrix that I need to change to be symmetric.

推荐答案

您可以使用pmax()，它返回一对向量的逐元素最大值.

You could use pmax(), which returns the element-wise maxima of a pair of vectors.

pmax(test_matrix, t(test_matrix))
#      [,1] [,2] [,3]
# [1,]    0    1    2
# [2,]    1    0    1
# [3,]    2    1    0

它将与一对矩阵一起使用，因为在这里，因为:(1)在R中，矩阵是具有(维)属性的正好"向量； (2)用于实现pmax()的代码足以将其第一个参数的属性重新附加到它返回的值上.

It'll work with a pair of matrices, as here, because: (1) in R, matrices are 'just' vectors with attached (dimension) attributes; and (2) the code used to implement pmax() is nice enough to reattach the attributes of it's first argument to the value that it returns.

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