Make a list of equal length vectors

Question

In a question here on SO (LINK) a poster asked a question and I gave an answer that works but there's a part that bugs me, creating a list from a vector to pass as a list of indices. So les say I have this vector:

n <- 1:10
#> n
# [1]  1  2  3  4  5  6  7  8  9 10

Let's say I want to break it up into a list of vectors and each vector is of length 3. What's the best (shortest amount of code & or fastest) way to accomplish this? We want to toss out item 10 since it there's a remainder of 1 (10 %% 3) from 10/3 (length(n) - 10 %% 3).

This is the desired outcome

list(1:3, 4:6, 7:9)

This will give us the indices of those that can't make a group of three:

(length(n) + 1 - 10 %% 3):length(n)

EDIT

Here's an interesting approach posted by Wojciech Sobala on the other thread this is linked to (I asked them to answer here and if they do I'll remove this edit)

n <- 100
l <- 3
n2 <- n - (n %% l)
split(1:n2, rep(1:n2, each=l, length=n2))

As a function:

indices <- function(n, l){
    if(n > l) stop("n needs to be smaller than or equal to l")
    n2 <- n - (n %% l)
    cat("numbers", (n + 1 - n %% l):n, "did not make an index of length", l)
    split(1:n2, rep(1:n2, each=l, length=n2))
}

Answer 1

Not sure if this does the job?

x = function(x, n){ 
    if(n > x) stop("n needs to be smaller than or equal to x")
    output = matrix(1:(x-x%%n), ncol=(x-x%%n)/n, byrow=FALSE)
    output
}

Edit: changed the output to a list

x = function(x, n){ 
    if(n > x) stop("n needs to be smaller than or equal to x")
    output = matrix(1:(x-x%%n), ncol=(x-x%%n)/n, byrow=TRUE)
    split(output, 1:nrow(output))
}

Example:
x(10, 3)
$`1`
[1] 1 2 3

$`2`
[1] 4 5 6

$`3`
[1] 7 8 9

Answer 2

It's not the shortest, but here's a little recursive version:

wrap <- function(n,x,lx,y) {
    if (lx < n) return (y)
    z <- x[-(1:n)]
    wrap(n, z, length(z), c(y, list(x[1:n])))
}

wrapit <- function(x,n) {
    wrap(n,x,length(x),list())
}

> wrapit(1:10,3)
[[1]]
[1] 1 2 3

[[2]]
[1] 4 5 6

[[3]]
[1] 7 8 9

Answer 3

xx <- 1:10
xxr <- rle(0:(length(1:10)-1) %/% 3)  # creates an rle object
fac3 <- rep( xxr$values[xxr$lengths == 3], each=3)  #selects the one of length 3
                                     # and recreates the shortened grouping vector
tapply(xx[ 1:length(fac3)],          # a shortened original vector
                       fac3, list)   # split into little lists
$`0`                                # Hope you don't mind having names on your list
[1] 1 2 3

$`1`
[1] 4 5 6

$`2`
[1] 7 8 9