
On Nov 16, 2009, at 2:32 PM, cindy Guo wrote:

> I forgot to say that there are no ties in each row. So any number  
> can occur only once in each row. Also as I mentioned earlier,  
> actually I only need the top 50 most frequent pairs, is there a more  
> efficient way to do it? Because I have 15000 numbers, output of all  
> the pairs would be too long.

?order


>
> Thank you,
>
> Cindy
>
> On Mon, Nov 16, 2009 at 7:02 AM, David Winsemius <dwinsemius@comcast.net 
> > wrote:
> I stuck in another "7" in one of the lines with a 2 and reasoned  
> that we could deal with the desire for non-ordered "pair counting"  
> by pasting min(x,y) to max(x,y);
>
> > dput(prmtx)
> structure(c(2, 1, 3, 9, 5, 7, 7, 8, 1, 7, 6, 5, 6, 2, 2, 7), .Dim =  
> c(4L,
> 4L))
> > prmtx
>     [,1] [,2] [,3] [,4]
> [1,]    2    5    1    6
> [2,]    1    7    7    2
> [3,]    3    7    6    2
> [4,]    9    8    5    7
>
> > pair.str <- sapply(1:nrow(prmtx), function(z)   
> apply(combn(prmtx[z,], 2), 2,function(x) paste(min(x[2],x[1]),  
> max(x[2],x[1]), sep=".")))
>
> The logic:
> sapply(1:nrow(prmtx), ... just loops over the rows of the matrix.
> combn(prmtx[z,], 2)  ... returns a two row matrix of combination in  
> a single row.
> apply(combn(prmtx[z,], 2), 2 ... since combn( , 2)  returns a matrix  
> that has two _rows_ I needed to loop over the columns.
> paste(min(x[2],x[1]), max(x[2],x[1]), sep=".") ... stick the minimum  
> of a pair in front of the max and separates them with a period to  
> prevent two+ digits from being non-unique
>
> Then using table() and logical tests in an index for the desired  
> multiple pairs:
>
>
> > tpair <-table(pair.str)
> > tpair
> pair.str
> 1.2 1.5 1.6 1.7 2.3 2.5 2.6 2.7 3.6 3.7 5.6 5.7 5.8 5.9 6.7 7.7 7.8  
> 7.9 8.9
>  2   1   1   2   1   1   2   3   1   1   1   1   1   1   1   1   1    
> 1   1
>
> > tpair[tpair>1]
> pair.str
> 1.2 1.7 2.6 2.7
>  2   2   2   3
>
> -- 
> David.
>
>
> On Nov 16, 2009, at 7:02 AM, David Winsemius wrote:
>
> I'm not convinced it's right. In fact, I'm pretty sure the last step  
> taking only the first half of the list is wrong. I also do not know  
> if you have considered how you want to count situations like:
>
> 3 2 7 4 5 7 ...
> 7 3 8 6 1 2 9 2 ......
>
> How many "pairs" of 2-7/7-2 would that represent?
>
> -- 
> David
> On Nov 15, 2009, at 11:06 PM, cindy Guo wrote:
>
> Hi, David,
>
> The matrix has 20 columns.
> Thank you very much for your help. I think it's right, but it seems  
> I need some time to figure it out. I am a green hand. There are so  
> many functions here I never used before. :)
>
> Cindy
>
> On Sun, Nov 15, 2009 at 5:19 PM, David Winsemius <dwinsemius@comcast.net 
> > wrote:
> 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@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
>
>
>
> David Winsemius, MD
> Heritage Laboratories
> West Hartford, CT
>
> ______________________________________________
> R-help@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
>
>

David Winsemius, MD
Heritage Laboratories
West Hartford, CT


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