Getting all splits of numeric sequence in R - r

Question

I'm trying to get all the possible splits of a sequence [1:n] in R. E.g.:
getSplits(0,3)
Should return all possible splits of the sequence 123, in other words (in a list of vectors):
[1] 1
[2] 1 2
[3] 1 2 3
[4] 1 3
[5] 2
[6] 2 3
[7] 3
Now I've created a function which does get to these vectors recursively, but having trouble combining them into one as above. My function is:
getSplits <- function(currentDigit, lastDigit, split) {
splits=list();
for (nextDigit in currentDigit: lastDigit)
{
currentSplit <- c(split, c(nextDigit));
print(currentSplit);
if(nextDigit < lastDigit) {
possibleSplits = c(list(currentSplit), getSplits(nextDigit+1, lastDigit, currentSplit));
}else{
possibleSplits = currentSplit;
}
splits <- c(splits, list(possibleSplits));
}
return(splits);
}
Where printing each currentSplit results in all the right vectors I need, but somehow the final returnt list (splits) nests them into deeper levels of lists, returning:
[1] 1
[[1]][[2]]
[[1]][[2]][[1]]
[1] 1 2
[[1]][[2]][[2]]
[1] 1 2 3
[[1]][[3]]
[1] 1 3
[[2]]
[[2]][[1]]
[1] 2
[[2]][[2]]
[1] 2 3
[[3]]
[1] 3
For the corresponding function call getSplits(1, 3, c()).
If anyone could help me out on getting this to work the way I described above, it'd be much appreciated!

Answer 1

character vector output
Try combn:
k <- 3
s <- unlist(lapply(1:k, combn, x = k, toString))
s
## [1] "1" "2" "3" "1, 2" "1, 3" "2, 3" "1, 2, 3"
data frame output
If you would prefer that the output be in the form of a data frame:
read.table(text = s, header = FALSE, sep = ",", fill = TRUE, col.names = 1:k)
giving:
X1 X2 X3
1 1 NA NA
2 2 NA NA
3 3 NA NA
4 1 2 NA
5 1 3 NA
6 2 3 NA
7 1 2 3
list output
or a list:
lapply(s, function(x) scan(textConnection(x), quiet = TRUE, sep = ","))
giving:
[[1]]
[1] 1
[[2]]
[1] 2
[[3]]
[1] 3
[[4]]
[1] 1 2
[[5]]
[1] 1 3
[[6]]
[1] 2 3
[[7]]
[1] 1 2 3
Update: Have incorporated improvement mentioned in comments as well as one further simplification and also added data frame and list output.

Answer 2

Here is another approach:
f <- function(nums) sapply(1:length(nums), function(x) t(combn(nums, m = x)))
f(1:3)
This yields
[[1]]
[,1]
[1,] 1
[2,] 2
[3,] 3
[[2]]
[,1] [,2]
[1,] 1 2
[2,] 1 3
[3,] 2 3
[[3]]
[,1] [,2] [,3]
[1,] 1 2 3

Answer 3

The OP is looking for the Power set of c(1,2,3). There are several packages that will quickly get you this in one line. Using the package rje, we have:
library(rje)
powerSet(c(1,2,3))
[[1]]
numeric(0)
[[2]]
[1] 1
[[3]]
[1] 2
[[4]]
[1] 1 2
[[5]]
[1] 3
[[6]]
[1] 1 3
[[7]]
[1] 2 3
[[8]]
[1] 1 2 3
... and with iterpc:
library(iterpc)
getall(iterpc(c(2,1,1,1), 3, labels = 0:3))
[,1] [,2] [,3]
[1,] 0 0 1
[2,] 0 0 2
[3,] 0 0 3
[4,] 0 1 2
[5,] 0 1 3
[6,] 0 2 3
[7,] 1 2 3
More generally,
n <- 3
getall(iterpc(c(n-1,rep(1, n)), n, labels = 0:n)) ## same as above