Dear all, is there an easy way to get all possible combinations (?) with replacement. If n=6, k=3, i want something like 0 0 6 0 5 1 0 4 2 0 3 3 0 2 4 . . . 5 0 1 5 1 0 6 0 0 I tried to look at combn() but I could not get this done with it. Thank you in advance: Daniel
probably expand.grid(), e.g., expand.grid(rep(list(0:6), 3)) I hope it helps. Best, Dimitris On 4/21/2011 9:28 PM, Kehl D?niel wrote:> Dear all, > > is there an easy way to get all possible combinations (?) with replacement. > If n=6, k=3, i want something like > > 0 0 6 > 0 5 1 > 0 4 2 > 0 3 3 > 0 2 4 > . > . > . > 5 0 1 > 5 1 0 > 6 0 0 > > I tried to look at combn() but I could not get this done with it. > > Thank you in advance: > Daniel > > ______________________________________________ > 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. >-- Dimitris Rizopoulos Assistant Professor Department of Biostatistics Erasmus University Medical Center Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands Tel: +31/(0)10/7043478 Fax: +31/(0)10/7043014 Web: http://www.erasmusmc.nl/biostatistiek/
Thank you. I only need those where the rowsum = n. I could choose those with code, but I dont think it is efficient that way. daniel 2011-04-21 12:33 keltez?ssel, Dimitris Rizopoulos ?rta:> > expand.grid(rep(list(0:6), 3))
On Thu, Apr 21, 2011 at 12:52:34PM -0700, Kehl D?niel wrote:> Thank you. > I only need those where the rowsum = n. > I could choose those with code, but I dont think it is efficient that way.Efficiency of using expand.grid() may be improved, if expand.grid() is used only to k-1 columns, then the k-th column is computed and the rows with a negative value in it are discarded. n <- 6 k <- 3 a <- as.matrix(expand.grid(rep(list(0:n), k - 1))) a <- cbind(a, n - rowSums(a)) colnames(a) <- NULL a <- a[0 <= a[, k], ] nrow(a) == choose(n + k - 1, k - 1) [1] TRUE In this way, we select choose(n + k - 1, k - 1) among n^(k - 1) rows and not among n^k. Hope this helps. Petr Savicky.
Does the following do what you want? It should
generate all the (unordered) NPart-partitions of Sum
by mapping the output of combn(Sum+NParts-1,NParts-1).
f <- function (Sum, NParts)
{
cm <- combn(Sum + NParts - 1, NParts - 1)
cm <- rbind(cm, Sum + NParts)
if (NParts > 1) {
r <- 2:NParts
cm[r, ] <- cm[r, ] - cm[r - 1, ]
}
t(cm - 1)
}
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -----Original Message-----
> From: r-help-bounces at r-project.org
> [mailto:r-help-bounces at r-project.org] On Behalf Of Kehl D?niel
> Sent: Thursday, April 21, 2011 12:29 PM
> To: r-help at r-project.org
> Subject: [R] all combinations with replacement
>
> Dear all,
>
> is there an easy way to get all possible combinations (?)
> with replacement.
> If n=6, k=3, i want something like
>
> 0 0 6
> 0 5 1
> 0 4 2
> 0 3 3
> 0 2 4
> .
> .
> .
> 5 0 1
> 5 1 0
> 6 0 0
>
> I tried to look at combn() but I could not get this done with it.
>
> Thank you in advance:
> Daniel
>
> ______________________________________________
> 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.
>