Re: [R] generating unordered combinations

[Dan Halligan]

That's brilliant - thanks.

On 17 Sep 2009, at 23:36, William Dunlap wrote:

There is a 1-1 correspondance between your n-sets
consisting of m possible element types (0 through m-1
in your example) and the number of n-subsets of a (n+m-1)-set.
E.g., your example had m=3 and n=3 and subtracting
1:3 from each column of combn(3+3-1,3) gives your result:

t(combn(3+3-1, 3)-(1:3))

     [,1] [,2] [,3]
[1,]    0    0    0
[2,]    0    0    1
[3,]    0    0    2
[4,]    0    1    1
[5,]    0    1    2
[6,]    0    2    2
[7,]    1    1    1
[8,]    1    1    2
[9,]    1    2    2
[10,]    2    2    2

Bill Dunlap
TIBCO Software Inc - Spotfire Division
wdunlap tibco.com

-----Original Message-----
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Dan Halligan
Sent: Thursday, September 17, 2009 1:31 PM
To: r-help@r-project.org
Subject: [R] generating unordered combinations

Hi,

I am trying to generate all unordered combinations of a set of

numbers / characters, and I can only find a (very) clumsy way of doing

this using expand.grid.  For example, all unordered combinations of
the numbers 0, 1, 2 are:
0, 0, 0
0, 0, 1
0, 0, 2
0, 1, 1
0, 1, 2
0, 2, 2
1, 1, 1
1, 1, 2
1, 2, 2
2, 2, 2

(I have not included, for example, 1, 0, 0, since it is equivalent to
0, 0, 1).

I have found a way to generate this data.frame using expand.grid as
follows:

g <- expand.grid(c(0,1,2), c(0,1,2), c(0,1,2))
for(i in 1:nrow(g)) {
        g[i,] <- sort(as.character(g[i,]))
}
o <- order(g$Var1, g$Var2, g$Var3)
unique(g[o,]).

This is obviously quite clumsy and hard to generalise to a greater
number of characters, so I'm keen to find any other solutions.  Can
anyone suggest a better (more general, quicker) method?

Cheers

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and provide commented, minimal, self-contained, reproducible code.