Creating a symmetric matrix in R

问题

I have a matrix in R that is supposed to be symmetric, however, due to machine precision the matrix is never symmetric (the values differ by around 10^-16). Since I know the matrix is symmetric I have been doing this so far to get around the problem:

s.diag = diag(s)
s[lower.tri(s,diag=T)] = 0
s = s + t(s) + diag(s.diag,S)

Is there a better one line command for this?

回答1:

Is the workaround really necessary if the values only differ by that much?

Someone pointed out that my previous answer was wrong. I like some of the other ones better, but since I can't delete this one (accepted by a user who left), here's yet another solution using the micEcon package:

symMatrix(s[upper.tri(s, TRUE)], nrow=nrow(s), byrow=TRUE)

回答2:

s<-matrix(1:25,5)
s[lower.tri(s)] = t(s)[lower.tri(s)]

回答3:

You can force the matrix to be symmetric using forceSymmetric function in Matrix package in R:

library(Matrix)
x<-Matrix(rnorm(9), 3)
> x
3 x 3 Matrix of class "dgeMatrix"
           [,1]       [,2]       [,3]
[1,] -1.3484514 -0.4460452 -0.2828216
[2,]  0.7076883 -1.0411563  0.4324291
[3,] -0.4108909 -0.3292247 -0.3076071

A <- forceSymmetric(x)
> A
3 x 3 Matrix of class "dsyMatrix"
           [,1]       [,2]       [,3]
[1,] -1.3484514 -0.4460452 -0.2828216
[2,] -0.4460452 -1.0411563  0.4324291
[3,] -0.2828216  0.4324291 -0.3076071

回答4:

 s<-matrix(1:25,5)
 pmean <- function(x,y) (x+y)/2
 s[] <- pmean(s, matrix(s, nrow(s), byrow=TRUE))
 s
#-------
     [,1] [,2] [,3] [,4] [,5]
[1,]    1    4    7   10   13
[2,]    4    7   10   13   16
[3,]    7   10   13   16   19
[4,]   10   13   16   19   22
[5,]   13   16   19   22   25

来源：https://stackoverflow.com/questions/18165320/creating-a-symmetric-matrix-in-r

标签

symmetric