Assuming that the number of columns is 4, then consider this approach:

> prs <-scan()
1: 2 5 1 6
5: 1 7 8 2
9: 3 7 6 2
13: 9 8 5 7
17:
Read 16 items
prmtx <- matrix(prs, 4,4, byrow=T)

#Now make copus of x.y and y.x

pair.str <- sapply(1:nrow(prmtx), function(z) c(apply(combn(prmtx[z,], 2), 2,function(x) paste(x[1],x[2], sep=".")) , apply(combn(prmtx[z,], 2), 2,function(x) paste(x[2],x[1], sep="."))) )
tpair <-table(pair.str)

# This then gives you a duplicated list
> tpair[tpair>1]
pair.str
1.2 2.1 2.6 2.7 6.2 7.2 7.8 8.7
  2   2   2   2   2   2   2   2

# So only take the first half of the pairs:
> head(tpair[tpair>1], sum(tpair>1)/2)

pair.str
1.2 2.1 2.6 2.7
  2   2   2   2

--
David.


On Nov 15, 2009, at 8:06 PM, David Winsemius wrote:

I could of course be wrong but have you yet specified the number of columns for this pairing exercise?

On Nov 15, 2009, at 5:26 PM, cindy Guo wrote:

Hi, All,

I have an n by m matrix with each entry between 1 and 15000. I want to know the frequency of each pair in 1:15000 that occur together in rows. So for
example, if the matrix is
2 5 1 6
1 7 8 2
3 7 6 2
9 8 5 7
Pair (2,6) (un-ordered) occurs together in rows 1 and 3. I want to return the value 2 for this pair as well as that for all pairs. Is there a fast way
to do this avoiding loops? Loops take too long.

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

David Winsemius, MD
Heritage Laboratories
West Hartford, CT

______________________________________________
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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