I had a similar problem recently where I used the following code:
v <- factor(letters[1:3])
all.cases <- expand.grid(v, v)
all.cases[as.numeric(all.cases[, 2]) >= as.numeric(all.cases[, 1]), ]
I'm not sure how to extend it cleanly to an arbitrary number of
vectors, though.
Baptiste
On 31 Aug 2008, at 11:18, Peter Dalgaard wrote:
Lucien Lemmens wrote:
Yuan Jian <jayuan2008 <at> yahoo.com> writes:
Hello,
is there a simple way to give all combinations for a given vector:
v<-c("a","b","c")
combination(v,v) becomes
"aa","ab","ac","bb","bc","cc'
combination(v,v,v) becomes
"aaa","aab","aac","abb",......
vv<-c(outer(v,v,paste))
vv
[1] "a a" "b a" "c a" "a b" "b b" "c b" "a c" "b c" "c c"
vvv<-c(outer(vv,v,paste)
vvv
[1] "a a a" "b a a" "c a a" "a b a" "b b a" "c b a" "a c a" "b c a"
"c c a" "a a b" "b a b" "c a b" "a b b" "b b b" "c b b"
[16] "a c b" "b c b" "c c b" "a a c" "b a c" "c a c" "a b c" "b b
c" "c b c" "a c c" "b c c" "c c c"
Notice that "ca" was not in the request for combination(v,v), so
it's a bit trickier. I was thinking ?combn, but that doesn't count
the duplicated cases:
> combn(v,2)
[,1] [,2] [,3]
[1,] "a" "a" "b"
[2,] "b" "c" "c"
--
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Baptiste Auguié
School of Physics
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