Reputation: 1604
What is the fastest way to construct the matrix above in R? Somehow I feel there has to be a better way than the below.
M <- t(matrix(c(1,1,1,-1,-1,-1),nrow=3))
M <- rbind(M, matrix(rep(0,9), nrow=3))
M <- cbind(M, matrix(rep(0,5*3), ncol=3))
M <- cbind(M,rbind(matrix(rep(0,2*3),ncol=3),diag(3)))
Upvotes: 1
Views: 67
Reputation: 887901
We could use bdiag
to do this in a single line
library(Matrix)
as.matrix(bdiag(cbind(rbind(rep(1, 3), rep(-1, 3)), 0, 0), diag(3)))
-output
# [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
#[1,] 1 1 1 0 0 0 0 0
#[2,] -1 -1 -1 0 0 0 0 0
#[3,] 0 0 0 0 0 1 0 0
#[4,] 0 0 0 0 0 0 1 0
#[5,] 0 0 0 0 0 0 0 1
Upvotes: 0
Reputation: 174526
If you want a one-liner you could do:
t(`[<-`(`[<-`(`[<-`(matrix(0, 9, 5), 1:3, 1), 10:12, -1), 7:9, 3:5, diag(3)))
#> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
#> [1,] 1 1 1 0 0 0 0 0 0
#> [2,] -1 -1 -1 0 0 0 0 0 0
#> [3,] 0 0 0 0 0 0 1 0 0
#> [4,] 0 0 0 0 0 0 0 1 0
#> [5,] 0 0 0 0 0 0 0 0 1
Or if you want some real code golf,
`[<-`(`[<-`(matrix(0,5,9),c(1+0:2*5,0:2*6+33),1),2+0:2*5,-1)
Upvotes: 2
Reputation: 1441
A matrix is just a vector with a dim
attribute. matrix(c(rep(c(1,-1,0,0,0),3), rep(0,17), 1, rep(c(rep(0,5), 1), 2)), ncol = 9)
Upvotes: 0
Reputation: 102710
What about this?
M <- matrix(0,nrow = 5,ncol = 9)
M[1,1:3] <- 1
M[2,1:3] <- -1
diag(M[3:5,7:9]) <- 1
Upvotes: 4