Creating a Symmetric Matrix in R

Creating a symmetric matrix in R

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)

How do I make a symmetric matrix using R?

You can try this -

mat[upper.tri(mat)] <- mat[upper.tri(mat)] + mat[lower.tri(mat)]
mat[lower.tri(mat)] <- mat[upper.tri(mat)]
mat

#     [,1] [,2] [,3]
#[1,]    1    6   10
#[2,]    6    5   14
#[3,]   10   14    9

data

mat <- structure(c(1L, 4L, 7L, 2L, 5L, 8L, 3L, 6L, 9L), .Dim = c(3L, 3L))
mat

#     [,1] [,2] [,3]
#[1,]    1    2    3
#[2,]    4    5    6
#[3,]    7    8    9

Most Efficient way to create a symmetric matrix

Just for completion I would like to also show this technique. The addition of the transpose will not work if the lower part of the matrix (under the diagonal) has values filled in, as it will add them to the upper part of the matrix.

Using the Matrix package we can create a sparse Matrix, which in case of creating the symmetric of a big matrix would require much less memory and even speed it up.

In order to create a symmetric sparse matrix from matrix e we would do:

library(Matrix)
rowscols <- which(upper.tri(e), arr.ind=TRUE)
sparseMatrix(i=rowscols[,1],    #rows to fill in
             j=rowscols[,2],    #cols to fill in
             x=e[upper.tri(e)], #values to use (i.e. the upper values of e)
             symmetric=TRUE,    #make it symmetric
             dims=c(nrow(e),nrow(e))) #dimensions

Output:

5 x 5 sparse Matrix of class "dsCMatrix"

[1,] .  2  3  4  5
[2,] 2  .  6  8 10
[3,] 3  6  . 12 15
[4,] 4  8 12  . 20
[5,] 5 10 15 20  .

Microbenchmark:

Let's make a function to make a symmetric Matrix out of a matrix (copies the upper part of matrix to the lower by default):

symmetrise <- function(mat){
  rowscols <- which(upper.tri(mat), arr.ind=TRUE)
  sparseMatrix(i=rowscols[,1], 
               j=rowscols[,2], 
               x=mat[upper.tri(mat)], 
               symmetric=TRUE, 
               dims=c(nrow(mat),ncol(mat)) )  
}

And test:

> microbenchmark(
e + t(e),
symmetrise(e),
e[lower.tri(e)] <- t(e)[lower.tri(e)],
times=1000
)
Unit: microseconds
                                  expr      min       lq      mean   median        uq       max neval cld
                              e + t(e)   75.946   99.038  117.1984  110.841  134.9590   246.825  1000 a  
                         symmetrise(e) 5530.212 6246.569 6950.7681 6921.873 7034.2525 48662.989  1000   c
 e[lower.tri(e)] <- t(e)[lower.tri(e)]  261.193  322.771  430.4479  349.968  395.3815 36873.894  1000  b 

As you can see symmetrise is actually much slower than e + t(e) or df[lower.tri(df)] <- t(df)[lower.tri(df)] but at least you have a function that automatically symmetrises a matrix (takes the upper part and creates the lower by default) and in case you have a big matrix where memory is an issue, this might come in handy.

P.S. Wherever you see a . in the Matrix this represents a zero. By using a different system a sparse Matrix is a kind of 'compressed' object making it more memory efficient.

Create a matrix symmetric from long table in r

You can get the data in wide format first.

library(dplyr)
library(tidyr)

mat <- df %>%
  pivot_wider(names_from = Var2, values_from = pre, values_fill = 0) %>%
  column_to_rownames('Var1') %>%
  as.matrix()

mat
#  a b  c  d e
#a 1 0  0  0 0
#b 2 1  0  0 0
#c 3 6  1  0 0
#d 4 7  9  1 0
#e 5 8 10 11 1

Since you have a symmetric matrix you can copy the lower triangular matrix to upper triangle.

mat[upper.tri(mat)] <- t(mat)[upper.tri(mat)]
mat

#  a b  c  d  e
#a 1 2  3  4  5
#b 2 1  6  7  8
#c 3 6  1  9 10
#d 4 7  9  1 11
#e 5 8 10 11  1

data

df <- data.frame (Var1  = c("a", "b", "c","d","e","b","c","d","e","c","d","e","d","e","e"),
                  Var2 = c("a","a","a","a","a","b","b","b","b","c","c","c","d","d","e"),
                  pre = c(1,2,3,4,5,1,6,7,8,1,9,10,1,11,1) )

R: make symmetric matrix from lower diagonal

You need to make sure that the elements you're copying are ordered correctly:

m <- matrix(NA,4,4)
m[lower.tri(m,diag=TRUE)] <- 1:10
     [,1] [,2] [,3] [,4]
[1,]    1   NA   NA   NA
[2,]    2    5   NA   NA
[3,]    3    6    8   NA
[4,]    4    7    9   10

makeSymm <- function(m) {
   m[upper.tri(m)] <- t(m)[upper.tri(m)]
   return(m)
}
makeSymm(m)

Or you can use the built-in

Matrix::forceSymmetric(m,uplo="L")

Create a symmetric matrix from circular shifts of a vector

Let me answer my own question in order to close it properly, using the incredible simple and easy solution from Henrik's comment:

matrix(v, nrow = 3, ncol = 4, byrow = TRUE)[ , 1:3]

Maybe the byrow = TRUE matches the three steps of the illustration best conceptually, but the output is the same with:

matrix(v, nrow = 4, ncol = 3)[1:3, ]
#      [,1] [,2] [,3]
# [1,]    1    2    3
# [2,]    2    3    1
# [3,]    3    1    2

Because there may be "many vectors with different lengths", it could be convenient to make a simple function and apply it to the vectors stored in a list:

cycle = function(x){
  len = length(x)
  matrix(x, nrow = len + 1, ncol = len)[1:len , ]
}

l = list(v1 = 1:3, v2 = letters[1:4])    

lapply(l, cycle)    
# $v1
#      [,1] [,2] [,3]
# [1,]    1    2    3
# [2,]    2    3    1
# [3,]    3    1    2
# 
# $v2
#     [,1] [,2] [,3] [,4]
# [1,] "a"  "b"  "c"  "d" 
# [2,] "b"  "c"  "d"  "a" 
# [3,] "c"  "d"  "a"  "b" 
# [4,] "d"  "a"  "b"  "c"

How to create a symmetric matrix in R counting how often two columns have the same values?

In order to create a co-occurrence matrix from your data, you first need to convert your NAs into 0s, then do a cross-product of your data without the first ID column:

x = data.frame(ID = c(1:4), sp1 = c(NA,0,1,1), sp2 = c(1,0,NA,1), sp3 = c(1,1,0,1))
x[is.na(x)] = 0
crossprod(t(x[-1]))

     [,1] [,2] [,3] [,4]
[1,]    2    1    0    2
[2,]    1    1    0    1
[3,]    0    0    1    1
[4,]    2    1    1    3

Constructing symmetric matrix on R

m <- matrix(0.5, ncol = 1000, nrow = 1000)
diag(m) <- 1

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