On Nov 16, 2009, at 2:41 PM, cindy Guo wrote: > 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. >
No, I mean type ?order at the R command line and read the help page. > 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 >> 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 > > 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.
