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",......
[[alternative HTML version deleted]]
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"> ? > > > [[alternative HTML version deleted]] > > > > ______________________________________________ > R-help <at> r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. >
A more generic solution is
allComb <- expand.grid(rep(list(letters[1:5]), 7))
whichComb <- sapply(seq_len(ncol(allComb) - 1), x = allComb, function(i,
x){
whichCombination <- sapply(seq(i + 1, ncol(x)), y = x, function(j,
y){
as.numeric(y[, i]) <= as.numeric(y[, j])
})
apply(whichCombination, 1, all)
})
allComb[apply(whichComb, 1, all), ]
HTH,
Thierry
------------------------------------------------------------------------
----
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature
and Forest
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say what the experiment died of.
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~ Roger Brinner
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ensure that a reasonable answer can be extracted from a given body of
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-----Oorspronkelijk bericht-----
Van: r-help-bounces op r-project.org [mailto:r-help-bounces op r-project.org]
Namens Lucien Lemmens
Verzonden: zondag 31 augustus 2008 15:58
Aan: r-help op stat.math.ethz.ch
Onderwerp: Re: [R] give all combinations
Another solution requiring also a bit of programming is:
l<-letters[1:3]
c2<-c()
for(i in 1:3){c2<-c(c2,paste(letters[i],letters[i:3],sep=""))}
c2
[1] "aa" "ab" "ac" "bb" "bc"
"cc"
n<-length(c2)
c3<-c();for(i in
1:n){c3<-c(c3,paste(c2[i],letters[ceiling(i/2):3],sep=""))}
c3
[1] "aaa" "aab" "aac" "aba"
"abb" "abc" "acb" "acc" "bbb"
"bbc" "bcc"
"ccc"
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Hi Yuan, Lucien, List.
try this:
f <-
function (...)
{
args <- list(...)
if(length(args)==0){
return(NULL)
}
if (length(args) == 1) {
return(args[[1]])
}
if (length(args) > 2) {
jj <- do.call("Recall", c(args[-1]))
return(do.call("Recall", c(list(args[[1]]), list(jj) )))
}
a <- args[[1]]
b <- args[[2]]
if (is.null(b)) {
return(a)
}
jj <- outer(a,b,paste)
return(jj[!lower.tri(jj)])
}
[the difficult bit (IMO) is to make f() work with any number of
arguments; thus
f(a,a) and f(a,a,a,a,a,b,b,a,b) and whatever should also work and this
is why
the Recall bit is needed].
Comments anyone?
HTH
rksh
Lucien Lemmens wrote:>
>
> Another solution requiring also a bit of programming is:
>
> l<-letters[1:3]
> c2<-c()
> for(i in
1:3){c2<-c(c2,paste(letters[i],letters[i:3],sep=""))}
> c2
> [1] "aa" "ab" "ac" "bb"
"bc" "cc"
> n<-length(c2)
> c3<-c();for(i in
1:n){c3<-c(c3,paste(c2[i],letters[ceiling(i/2):3],sep=""))}
> c3
> [1] "aaa" "aab" "aac" "aba"
"abb" "abc" "acb" "acc" "bbb"
"bbc" "bcc" "ccc"
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
I seem to be missing something here:
given a set X:{a,b,c,whatever...}
the mathematical definition of 'permutation' is the set of all possible
sequences of the elements of X.
The definition of 'combination' is all elements of 'permutation'
which
cannot be re-ordered to become a different element.
example: X:{a,b,c}
perm(X) = ab, ab, bc, ba, ca, cb
comb(X) = ab, ac, bc
So maybe a better question for this mailing list is: Are there
functions available in any R package which produce perm() and comb()
(perhaps as well as other standard combinatoric functions) ?
Carl