[R] combinatorics

[Christos Hatzis]

Hi Robin,

This approach first generates all combinations and then eliminates the
non-feasible ones.  It should work fine for smallish vectors but might not
scale well for larger vectors.  Hopefully it gives you what you need for
this problem.

xx <- c("A","A","B","B","C")
yy <- 1:length(xx)
zz <- expand.grid(yy,yy,yy,yy,yy)

ss <- zz[ apply(zz, 1, FUN=function(x) length(unique(x))) == length(xx), ] 
ss <- as.matrix(ss)

pp <- apply(ss, 1, FUN=function(x,v) paste(v[as.vector(x)], collapse=""),
v=xx) 
res <- unique(pp)

> res
 [1] "CBBAA" "BCBAA" "BBCAA" "CBABA" "BCABA" "CABBA" "ACBBA" "BACBA" "ABCBA"
"BBACA" "BABCA"
[12] "ABBCA" "CBAAB" "BCAAB" "CABAB" "ACBAB" "BACAB" "ABCAB" "CAABB" "ACABB"
"AACBB" "BAACB"
[23] "ABACB" "AABCB" "BBAAC" "BABAC" "ABBAC" "BAABC" "ABABC" "AABBC"
> length(res)
[1] 30 

-Christos

Christos Hatzis, Ph.D.
Nuvera Biosciences, Inc.
400 West Cummings Park
Suite 5350
Woburn, MA 01801
Tel: 781-938-3830
www.nuverabio.com
 

-----Original Message-----
From: r-help-bounces at stat.math.ethz.ch
[mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of Robin Hankin
Sent: Friday, October 13, 2006 10:19 AM
To: R-help at r-project.org
Subject: [R] combinatorics

Hi

How do I generate all ways of ordering  sets of indistinguishable items?

suppose I have two A's, two B's and a C.

Then I want

AABBC
AABCB
AACBC
ABABC
. . .snip...
BBAAC
. . .snip...
CBBAA

[there are 5!/(2!*2!) = 30 arrangements.  Note AABBC != BBAAC]

How do I do this?





--
Robin Hankin
Uncertainty Analyst
National Oceanography Centre, Southampton European Way, Southampton SO14
3ZH, UK
  tel  023-8059-7743

______________________________________________
R-help at stat.math.ethz.ch 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.