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
Cel biometrie, methodologie en kwaliteitszorg / Section biometrics,
methodology and quality assurance
Gaverstraat 4
9500 Geraardsbergen
Belgium
tel. + 32 54/436 185
[EMAIL PROTECTED]
www.inbo.be
To call in the statistician after the experiment is done may be no more
than asking him to perform a post-mortem examination: he may be able to
say what the experiment died of.
~ Sir Ronald Aylmer Fisher
The plural of anecdote is not data.
~ Roger Brinner
The combination of some data and an aching desire for an answer does not
ensure that a reasonable answer can be extracted from a given body of
data.
~ John Tukey
-----Oorspronkelijk bericht-----
Van: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
Namens Lucien Lemmens
Verzonden: zondag 31 augustus 2008 15:58
Aan: [EMAIL PROTECTED]
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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