Create new columns based on rows

Question

If I had a matrix like:

     [,1] [,2] 
[1,]    1    7
[2,]    2    8
[3,]    3    9
[4,]    4   10
[5,]    5   11
[6,]    6   12

Does anyone have an idea as to how I might create a new matrix from the above that looks like:

     [,1] [,2] [,3] [,4] [,5] [,6]
[1,]    1    7  3    9    5    11
[2,]    2    8  4    10   6    12

Answer 1

We create a grouping variable with ?gl and use the arguments n=nrow(m1), k=2 and length=nrow(m1). We split the matrix ('m1'), unlist, and create a new matrix with nrow=2.

 matrix(unlist(split(m1,as.numeric(gl(nrow(m1), 2, nrow(m1))))),nrow=2)
 #     [,1] [,2] [,3] [,4] [,5] [,6]
 #[1,]    1    7    3    9    5   11
 #[2,]    2    8    4   10    6   12

Or another option is converting to array by specifying the dimensions. Here I used c(2, 2, 3) as we can get a 2x2 matrix for the first two dimensions and the third is based on the nrow(m1)/2. Then, we can permute the dimensions of the array using aperm, concatenate (c) to form a vector and convert to matrix.

 matrix(c(aperm(array(t(m1), c(2, 2,3)),c(2,1,3))), nrow=2)
 #     [,1] [,2] [,3] [,4] [,5] [,6]
 #[1,]    1    7    3    9    5   11
 #[2,]    2    8    4   10    6   12

data

m1 <- structure(1:12, .Dim = c(6L, 2L))

Answer 2

Here' another option: First the matrix is transformed into one with two rows, then the odd and even numbered columns are rearranged:

m3 <- m2 <- matrix(c(m),nrow = 2) #take data from original matrix, convert it into a matrix with two rows and store a copy in m2 and m3
m3[,seq(1,ncol(m2),2)] <- m2[,1:(ncol(m2)/2)] #define the odd-numbered columns of m3
m3[,seq(2,ncol(m2),2)] <- m2[,(ncol(m2)/2+1):ncol(m2)] # same for the even-numbered columns
> m3
#     [,1] [,2] [,3] [,4] [,5] [,6]
#[1,]    1    7    3    9    5   11
#[2,]    2    8    4   10    6   12