How to populate a matrix quickly without nested for loops - r

Question

I have a dataset with values 0, 1, and 2.
data <- matrix(c(1, 0, 0, 1, 2, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 2, 0, 1, 1, 1), nrow = 5, ncol = 4)
> data
[,1] [,2] [,3] [,4]
[1,] 1 1 0 2
[2,] 0 1 0 0
[3,] 0 0 0 1
[4,] 1 1 0 1
[5,] 2 0 0 1
I would like to a create matrix based on this data such that the value 0 is (0, 0), 1 is (1, 0), and 2 is (0, 1). Below is the code that I'm using:
data.exp <- matrix(NA, nrow = nrow(data)*2, ncol = ncol(data))
for(i in 1:nrow(data)){
for(j in 1:(ncol(data))){
if(data[i,j] == 1){
vec <- c(1, 0)
}else if(data[i, j] == 0){
vec <- c(0, 0)
}else{
vec <- c(0, 1)
}
data.exp[((i*2-1):(i*2)), j] <- vec
}
}
> data.exp
[,1] [,2] [,3] [,4]
[1,] 1 1 0 0
[2,] 0 0 0 1
[3,] 0 1 0 0
[4,] 0 0 0 0
[5,] 0 0 0 1
[6,] 0 0 0 0
[7,] 1 1 0 1
[8,] 0 0 0 0
[9,] 0 0 0 1
[10,] 1 0 0 0
Is there a faster way to generate the matrix, data.exp, without having to use a nested for loop in R? As the sample size increases, the nested for loop approach is not very efficient.

Answer 1

apply should be pretty fast for matrices. Create a list, v, with appropriate values and subset by considering 0, 1, or 2 in data as indices of v
v = list(c(0, 0), c(1, 0), c(0, 1))
apply(data, 2, function(i) do.call(cbind, v[i + 1]))
# [,1] [,2] [,3] [,4]
# [1,] 1 1 0 0
# [2,] 0 0 0 1
# [3,] 0 1 0 0
# [4,] 0 0 0 0
# [5,] 0 0 0 1
# [6,] 0 0 0 0
# [7,] 1 1 0 1
# [8,] 0 0 0 0
# [9,] 0 0 0 1
# [10,] 1 0 0 0

Answer 2

Here is an option without any loop
t(
matrix(
scan(text = toString(c("0, 0", "1, 0", "0, 1")[data + 1]), sep = ","),
byrow = TRUE,
nrow = ncol(data)
)
)
which gives
[,1] [,2] [,3] [,4]
[1,] 1 1 0 0
[2,] 0 0 0 1
[3,] 0 1 0 0
[4,] 0 0 0 0
[5,] 0 0 0 1
[6,] 0 0 0 0
[7,] 1 1 0 1
[8,] 0 0 0 0
[9,] 0 0 0 1
[10,] 1 0 0 0
A more concise option (thank #akrun's contribution)
> matrix(unlist(list(c(0, 0), c(1, 0), c(0, 1))[data + 1]), nrow = nrow(data) * 2)
[,1] [,2] [,3] [,4]
[1,] 1 1 0 0
[2,] 0 0 0 1
[3,] 0 1 0 0
[4,] 0 0 0 0
[5,] 0 0 0 1
[6,] 0 0 0 0
[7,] 1 1 0 1
[8,] 0 0 0 0
[9,] 0 0 0 1
[10,] 1 0 0 0

Answer 3

Start by making two matrices of the same dimensions as data, one which sets all of the 2s to 0 and the other which sets all of the 1s to 0 and all of the 2s to 1. Then interleave the two matrices row by row.
The first part if accomplished using ifelse; for the second part, flodel's answer to this question helps.
Putting it all together, you have
data <- matrix(c(1, 0, 0, 1, 2, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 2, 0, 1, 1, 1), nrow = 5, ncol = 4)
l<-list(ifelse(data < 2, data, 0),
ifelse(data > 1, 1, 0))
do.call(rbind, l)[order(sequence(sapply(l, nrow))), ]
# [,1] [,2] [,3] [,4]
# [1,] 1 1 0 0
# [2,] 0 0 0 1
# [3,] 0 1 0 0
# [4,] 0 0 0 0
# [5,] 0 0 0 1
# [6,] 0 0 0 0
# [7,] 1 1 0 1
# [8,] 0 0 0 0
# [9,] 0 0 0 1
# [10,] 1 0 0 0