On Tue, Jan 24, 2012 at 07:05:37PM +0100, Petr Savicky wrote:
> On Tue, Jan 24, 2012 at 05:19:42PM +0000, yan jiao wrote:
> > Dear All,
> >
> > I'm wondering if there is a R function could give me all the
> > combinations of the grouping/cluster result, given the number of the groups.
> > e.g.
> > 3 objects: x1 x2 x3, number of groups is 2
> > so the result will be
> > group1:x1,x2; group2: x3
> > group1: x1;group2: x2,x3
> > group1: x1,x3;group2: x2
[...]
> For more groups, a similar approach may be used.
> Eliminating equivalent groupings may be done,
> for example, by requiring that if an element is
> assigned to group i, then each of the groups
> 1, ..., i-1 is assigned to some earlier element.
>
> So, 1, 2, 3, 2 is OK, but 1, 3, 2, 2 is not.
For example, all groupings of 5 elements into 3 groups
may be computed as follows.
check.row <- function(x, k)
{
y <- unique(x)
length(y) == k && all(y == 1:k)
}
gr <- as.matrix(expand.grid(x1=1, x2=1:2, x3=1:3, x4=1:3, x5=1:3))
ok <- apply(gr, 1, check.row, k=3)
gr <- gr[ok, ]
gr
x1 x2 x3 x4 x5
[1,] 1 2 3 1 1
[2,] 1 2 3 2 1
[3,] 1 2 1 3 1
[4,] 1 1 2 3 1
[5,] 1 2 2 3 1
[6,] 1 2 3 3 1
[7,] 1 2 3 1 2
[8,] 1 2 3 2 2
[9,] 1 2 1 3 2
[10,] 1 1 2 3 2
[11,] 1 2 2 3 2
[12,] 1 2 3 3 2
[13,] 1 2 1 1 3
[14,] 1 1 2 1 3
[15,] 1 2 2 1 3
[16,] 1 2 3 1 3
[17,] 1 1 1 2 3
[18,] 1 2 1 2 3
[19,] 1 1 2 2 3
[20,] 1 2 2 2 3
[21,] 1 2 3 2 3
[22,] 1 2 1 3 3
[23,] 1 1 2 3 3
[24,] 1 2 2 3 3
[25,] 1 2 3 3 3
According to
http://en.wikipedia.org/wiki/Partition_of_a_set
The number of partitions of an n-element set into exactly k nonempty
parts is the Stirling number of the second kind S(n, k).
and according to the table at
http://en.wikipedia.org/wiki/Stirling_number_of_the_second_kind
S(5, 3) = 25, so the above table "gr" has the correct number of rows.
Hope this helps.
Petr Savicky.
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