I have the following matrix
2 4 1
6 32 1
4 2 1
5 3 2
4 2 2
I want to make the following two matrices based on 3rd column
first
2 4
6 32
4 2
second
5 3
4 2
Best I can come up with, but I get an error
x <- cbind(mat[,1], mat[,2]) if mat[,3]=1
y <- cbind(mat[,1], mat[,2]) if mat[,3]=2
If mat is your matrix:
mat <- matrix(1:15,ncol=3)
mat[,3] <- c(1,1,1,2,2)
> mat
[,1] [,2] [,3]
[1,] 1 6 1
[2,] 2 7 1
[3,] 3 8 1
[4,] 4 9 2
[5,] 5 10 2
Then you can use split:
> lapply( split( mat[,1:2], mat[,3] ), matrix, ncol=2)
$`1`
[,1] [,2]
[1,] 1 6
[2,] 2 7
[3,] 3 8
$`2`
[,1] [,2]
[1,] 4 9
[2,] 5 10
The lapply of matrix is necessary because split drops the attributes that make a vector a matrix, so you need to add them back in.
Yet another example:
#test data
mat <- matrix(1:15,ncol=3)
mat[,3] <- c(1,1,1,2,2)
#make a list storing a matrix for each id as components
result <- lapply(by(mat,mat[,3],identity),as.matrix)
Final product:
> result
$`1`
V1 V2 V3
1 1 6 1
2 2 7 1
3 3 8 1
$`2`
V1 V2 V3
4 4 9 2
5 5 10 2
If you have a matrix A, this will get the first two columns when the third column is 1:
A[A[,3] == 1,c(1,2)]
You can use this to obtain matrices for any value in the third column.
Explanation: A[,3] == 1 returns a vector of booleans, where the i-th position is TRUE if A[i,3] is 1. This vector of booleans can be used to index into a matrix to extract the rows we want.
Disclaimer: I have very little experience with R, this is the MATLAB-ish way to do it.
split.data.frame could be used also to split a matrix.
mat <- matrix(1:15,ncol=3)
mat[,3] <- c(1,1,1,2,2)
x <- split.data.frame(mat[,-3], mat[,3])
x
#$`1`
# [,1] [,2]
#[1,] 1 6
#[2,] 2 7
#[3,] 3 8
#
#$`2`
# [,1] [,2]
#[1,] 4 9
#[2,] 5 10
str(x)
#List of 2
# $ 1: num [1:3, 1:2] 1 2 3 6 7 8
# $ 2: num [1:2, 1:2] 4 5 9 10
Or split the index and and use it in lapply to subset.
lapply(split(seq_along(mat[,3]), mat[,3]), \(i) mat[i, -3, drop=FALSE])
#$`1`
# [,1] [,2]
#[1,] 1 6
#[2,] 2 7
#[3,] 3 8
#
#$`2`
# [,1] [,2]
#[1,] 4 9
#[2,] 5 10
This is a functional version of pedrosorio's idea:
getthird <- function(mat, idx) mat[mat[,3]==idx, 1:2]
sapply(unique(mat[,3]), getthird, mat=mat) #idx gets sent the unique values
#-----------
[[1]]
[,1] [,2]
[1,] 1 6
[2,] 2 7
[3,] 3 8
[[2]]
[,1] [,2]
[1,] 4 9
[2,] 5 10
We can use by or tapply
> by(seq_along(mat[, 3]), mat[, 3], function(k) mat[k, -3])
mat[, 3]: 1
[,1] [,2]
[1,] 1 6
[2,] 2 7
[3,] 3 8
------------------------------------------------------------
mat[, 3]: 2
[,1] [,2]
[1,] 4 9
[2,] 5 10
> tapply(seq_along(mat[, 3]), mat[, 3], function(k) mat[k, -3])
$`1`
[,1] [,2]
[1,] 1 6
[2,] 2 7
[3,] 3 8
$`2`
[,1] [,2]
[1,] 4 9
[2,] 5 10
