mylist <- list(matrix(c(1, 3, -1, 0, 2, 1), nrow = 2, byrow = TRUE),
matrix(c(-2, 0, 10, 1, 2, 9, 2, 0, 0), nrow = 3, byrow = TRUE))
> mylist
[[1]]
[,1] [,2] [,3]
[1,] 1 3 -1
[2,] 0 2 1
[[2]]
[,1] [,2] [,3]
[1,] -2 0 10
[2,] 1 2 9
[3,] 2 0 0
I have a list of matrices called mylist
where the dimensions of the matrices can differ. For each matrix, I want to double the row values and insert it as a new row underneath. My desired output is as follows:
[[1]]
[,1] [,2] [,3]
[1,] 1 3 -1
[2,] 2 6 -2
[3,] 0 2 1
[4,] 0 4 2
[[2]]
[,1] [,2] [,3]
[1,] -2 0 10
[2,] -4 0 20
[3,] 1 2 9
[4,] 2 4 18
[5,] 2 0 0
[6,] 4 0 0
CodePudding user response:
In an lapply
rep
eat each row index found using seq_len
with nrow
and multiply repeatedly by 1:2
exploiting recycling.
lapply(mylist, \(x) x[rep(seq_len(nrow(x)), each=2), ]*1:2)
# [[1]]
# [,1] [,2] [,3]
# [1,] 1 3 -1
# [2,] 2 6 -2
# [3,] 0 2 1
# [4,] 0 4 2
#
# [[2]]
# [,1] [,2] [,3]
# [1,] -2 0 10
# [2,] -4 0 20
# [3,] 1 2 9
# [4,] 2 4 18
# [5,] 2 0 0
# [6,] 4 0 0
CodePudding user response:
You can use rbind
but you need to permute rows to get multiplied row beneath:
lapply(
mylist,
function(x){
rbind(x, x * 2)[as.vector(t(matrix(seq_len(nrow(x) * 2), ncol = 2))),]
}
)
[[1]]
[,1] [,2] [,3]
[1,] 1 3 -1
[2,] 2 6 -2
[3,] 0 2 1
[4,] 0 4 2
[[2]]
[,1] [,2] [,3]
[1,] -2 0 10
[2,] -4 0 20
[3,] 1 2 9
[4,] 2 4 18
[5,] 2 0 0
[6,] 4 0 0
CodePudding user response:
Another base R approach which also utilises the rbind
function with do.call
.
lapply(mylist, function(x)
do.call(rbind, lapply(1:nrow(x), function(i) rbind(x[i, ], x[i,]*2))))
[[1]]
[,1] [,2] [,3]
[1,] 1 3 -1
[2,] 2 6 -2
[3,] 0 2 1
[4,] 0 4 2
[[2]]
[,1] [,2] [,3]
[1,] -2 0 10
[2,] -4 0 20
[3,] 1 2 9
[4,] 2 4 18
[5,] 2 0 0
[6,] 4 0 0