I would like to rearrange the matrix b in the following way:
> b <- matrix(1:24, 6, 4)
> b
[,1] [,2] [,3] [,4]
[1,] 1 7 13 19
[2,] 2 8 14 20
[3,] 3 9 15 21
[4,] 4 10 16 22
[5,] 5 11 17 23
[6,] 6 12 18 24
> cbind(b[1:2, ], b[3:4, ], b[5:6, ])
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
[1,] 1 7 13 19 3 9 15 21 5 11 17 23
[2,] 2 8 14 20 4 10 16 22 6 12 18 24
The number of rows (2 here) is always going to be the same but the matrix b is going to be very large, on the dimension of about 1M x 100. Is there some fast way I can do this without a for loop? Thanks.
First, split b into chunks of two rows using lapply and then cbind the subgroups together using do.call.
do.call(cbind, lapply(seq(1, NROW(b), 2), function(i) b[i:(i+1),]))
# [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
#[1,] 1 7 13 19 3 9 15 21 5 11 17 23
#[2,] 2 8 14 20 4 10 16 22 6 12 18 24
Another way is to separate elements of odd and even rows into two vectors and rbind them
rbind(do.call(c, data.frame(t(b[(1:NROW(b) %% 2) == 1,]))),
do.call(c, data.frame(t(b[(1:NROW(b) %% 2) == 0,]))))
# X11 X12 X13 X14 X21 X22 X23 X24 X31 X32 X33 X34
#[1,] 1 7 13 19 3 9 15 21 5 11 17 23
#[2,] 2 8 14 20 4 10 16 22 6 12 18 24
Yet another way which is, for the most part, similar to the second approach
do.call(rbind, lapply(split(data.frame(b), 1:2), function(x) do.call(c, data.frame(t(x)))))
# X11 X12 X13 X14 X31 X32 X33 X34 X51 X52 X53 X54
#1 1 7 13 19 3 9 15 21 5 11 17 23
#2 2 8 14 20 4 10 16 22 6 12 18 24