Reputation: 529
I'm not even sure how to give this a better, and not obviously duplicate type, title but I think this is a different question about expand.grid.
I have a list of variables for which I need a data.frame or list of every possible combination to feed into a bit of ordinal regression.
The list:
> indVars <- as.list(c("T.P","T.M","T.S","E"))
Desired output:
> out List of (?)
: "T.P"
: "T.M"
: "T.S"
: "E"
: "T.P" "T.M"
: "T.P" "T.S"
: "T.P" "E"
.
.
.
: "T.P" "T.M" "T.S" "E"
Attempted:
expand.grid(indVars)
gives a single row
> expand.grid(indVars)
Var1 Var2 Var3 Var4
1 T.P T.M T.S E
expand.grid(indVars,indVars)
gives 16 rows of all two variable combinations but doesn't do 3 or four AND where indVars[i]==indVars[i]
(so you get rows like
> expand.grid(indVars,indVars)[1,]
Var1 Var2
1 T.P T.P
Logic says expand.grid(indVars,indVars,indVars,indVars)
to give all up to combinations (256 of them) but again you end up with rows with multiple instances of the same indVar. For example:
> expand.grid(indVars,indVars,indVars,indVars)[241,]
Var1 Var2 Var3 Var4
241 T.P T.P E E
Request: Can someone point out how to expand this list of 4 variables into every combination of 1,2,3 and 4 of them with no duplicates?
Upvotes: 1
Views: 924
Reputation: 132651
Possibly this is an XY problem and there is a better aproach to do the ordinal regression.
I suspect that order doesn't matter. Use combn
:
res <- lapply(seq_along(indVars), combn, x = indVars, simplify = FALSE)
unlist(res, FALSE)
# [[1]]
# [1] "T.P"
#
# [[2]]
# [1] "T.M"
#
# [[3]]
# [1] "T.S"
#
# [[4]]
# [1] "E"
#
# [[5]]
# [1] "T.P" "T.M"
#
# [[6]]
# [1] "T.P" "T.S"
#
# [[7]]
# [1] "T.P" "E"
#
# [[8]]
# [1] "T.M" "T.S"
#
# [[9]]
# [1] "T.M" "E"
#
# [[10]]
# [1] "T.S" "E"
#
# [[11]]
# [1] "T.P" "T.M" "T.S"
#
# [[12]]
# [1] "T.P" "T.M" "E"
#
# [[13]]
# [1] "T.P" "T.S" "E"
#
# [[14]]
# [1] "T.M" "T.S" "E"
#
# [[15]]
# [1] "T.P" "T.M" "T.S" "E"
Upvotes: 3