Create a symmetric matrix from circular shifts of a vector

Question

I'm struggling with the creation of a symmetric matrix.

Let's say a vector v <- c(1,2,3)

I want to create a matrix like this:

matrix(ncol = 3, nrow = 3, c(1,2,3,2,3,1,3,1,2), byrow = FALSE)
     [,1] [,2] [,3]
[1,]    1    2    3
[2,]    2    3    1
[3,]    3    1    2

(This is just an reprex, I have many vectors with different lengths.)

Notice this is a symmetric matrix with diagonal c(1,3,2) (different from vector v) and the manual process to create the matrix would be like this:

Using the first row as base (vector v) the process is to fill the empty spaces with the remaining values on the left side.

Any help is appreciated. Thanks!

Answer 1

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"

Answer 2

Another option is to use Reduce and make c(v[-1], v[1]) accumulative.

do.call(rbind, Reduce(function(x, y) c(x[-1], x[1]), v[-1], v, accumulate = TRUE))
#     [,1] [,2] [,3]
#[1,]    1    2    3
#[2,]    2    3    1
#[3,]    3    1    2