Do you mean if the numbers in each row are ordered? They are not, but if it's needed, we can order them. The matrix only has 5000 rows.
On Mon, Nov 16, 2009 at 1:34 PM, David Winsemius <dwinsem...@comcast.net>wrote: > > 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 > <dwinsem...@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 < >>>> dwinsem...@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<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<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 > > [[alternative HTML version deleted]] ______________________________________________ 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.
