compute all permutation with repetition of list R - r

Question

i have this list of vector out:
[[1]]
[1] (1,5)(3)(4)(6)
[[2]]
[1] (3,6)(1)(4)(5)
[[3]]
[1] (3,4)(1)(5)(6)
[[4]]
[1] (3,5)(1)(4)(6)
[[5]]
[1] (4,6)(1)(3)(5)
[[5]]
[1] (4,5)(1)(3)(6)
[[6]]
[1] (5,6)(1)(3)(4)
take one elemente for example out[[5]] (4,6)(1)(3)(5), i would generate all permutation of element with all possible order, for example:
{(4,6)(3)(1)(5)},{(3)(4,6)(5)(1)}...

Answer 1

I think this is what the OP is looking for:
out <- list(c("(1,5)","(3)","(4)","(6)"), c("(3,6)","(1)","(4)","(5)"),
c("(3,4)","(1)","(5)","(6)"),c("(3,5)","(1)","(4)","(6)"),
c("(4,6)","(1)","(3)","(5)"), c("(4,5)","(1)","(3)","(6)"),
c("(5,6)","(1)","(3)","(4)"))
library(RcppAlgos)
myPerms <- lapply(out, function(x) {
unlist(permuteGeneral(x, length(x), FUN = function(y) {
paste0(c("{", y, "}"), collapse = "")
}))
})
We use permuteGeneral from RcppAlgos (I am the author) as we can utilize the FUN argument to pass a custom function that will be applied to each permutation (i.e. paste0(c("{", y, "}"), collapse = "")).
Here is the output of the 5th element:
myPerms[[5]]
[1] "{(4,6)(1)(3)(5)}" "{(4,6)(1)(5)(3)}" "{(4,6)(3)(1)(5)}"
[4] "{(4,6)(3)(5)(1)}" "{(4,6)(5)(1)(3)}" "{(4,6)(5)(3)(1)}"
[7] "{(1)(4,6)(3)(5)}" "{(1)(4,6)(5)(3)}" "{(1)(3)(4,6)(5)}"
[10] "{(1)(3)(5)(4,6)}" "{(1)(5)(4,6)(3)}" "{(1)(5)(3)(4,6)}"
[13] "{(3)(4,6)(1)(5)}" "{(3)(4,6)(5)(1)}" "{(3)(1)(4,6)(5)}"
[16] "{(3)(1)(5)(4,6)}" "{(3)(5)(4,6)(1)}" "{(3)(5)(1)(4,6)}"
[19] "{(5)(4,6)(1)(3)}" "{(5)(4,6)(3)(1)}" "{(5)(1)(4,6)(3)}"
[22] "{(5)(1)(3)(4,6)}" "{(5)(3)(4,6)(1)}" "{(5)(3)(1)(4,6)}"
If you really want permutations with repetition, simply set repetition = TRUE in permuteGeneral. Of course, if you want a more useable output, we can drop the custom FUN altogether.
UPDATE
After learning more about how the OP obtained out above, we can better attack the problem. First we find out that the OP uses listParts from the library partitions. Looking at the source code we have:
listParts
function (x)
{
f <- function(pp) {
out <- split(seq_along(pp), pp)
class(out) <- c(class(out), "equivalence")
out
}
apply(setparts(x), 2, f)
}
<bytecode: 0x10d7b09f8>
<environment: namespace:partitions>
We can alter this to obtain all permutations:
permListParts <- function (x)
{
f <- function(pp) {
out <- split(seq_along(pp), pp)
myPerms <- perms(length(out))
apply(myPerms, 2, function(x) {
temp <- out[x]
class(temp) <- c(class(temp), "equivalence")
temp
})
}
apply(setparts(x), 2, f)
}
We note that we are going against the intent of listParts... quoting the documentation:
"Note that (12)(3)(4) is the same partition as, for example, (3)(4)(21) as the equivalence relation is the same."
Oh well... here is the output for permutations of length 3:
permListParts(3)
[[1]]
[[1]][[1]]
[1] (1,2,3)
[[2]]
[[2]][[1]]
[1] (1,3)(2)
[[2]][[2]]
[1] (2)(1,3)
[[3]]
[[3]][[1]]
[1] (1,2)(3)
[[3]][[2]]
[1] (3)(1,2)
[[4]]
[[4]][[1]]
[1] (2,3)(1)
[[4]][[2]]
[1] (1)(2,3)
[[5]]
[[5]][[1]]
[1] (1)(2)(3)
[[5]][[2]]
[1] (1)(3)(2)
[[5]][[3]]
[1] (2)(1)(3)
[[5]][[4]]
[1] (2)(3)(1)
[[5]][[5]]
[1] (3)(1)(2)
[[5]][[6]]
[1] (3)(2)(1)

Answer 2

You could try this:
test <- list(list(1,5),3,4,6)
combn(test,2)
> combn(test,2)
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] List,2 List,2 List,2 3 3 4
[2,] 3 4 6 4 6 6
This would give you all the possible combinations of 2 elements from your list. If you want to do that for each element of your main list you can then use a for or lapply on it.
try:
lapply(out, function(x) combn(x,2))

Answer 3

How about gtools packages
out <- list(
c(1.5,3,4,6),
c(3.6,1,4,5),
c(3.4,1,5,6),
c(3.5,1,4,6),
c(4.6,1,3,5),
c(4.5,1,3,6),
c(5.6,1,3,4))
require(gtools)
allPerm <- function(x){
return(permutations(length(x), length(x), x))
}