[R] pairs
David Winsemius
dwinsemius at comcast.net
Mon Nov 16 02:19:48 CET 2009
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
>
> ______________________________________________
> 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.
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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