On Thu, Mar 17, 2011 at 02:54:49PM -0400, Jim Silverton wrote: > I have a matrix say: > > 23 1 > 12 12 > 0 0 > 0 1 > 0 1 > 0 2 > 23 2 > > I want to count of number of distinct rows and the number of disinct element > in the second column and put these counts in a column. SO at the end of the > day I should have: > > c(1, 1, 1, 2, 2, 1, 1) for the distinct rows and c(1, 1, 1, 2, 2, 2, 2) for > the counts of how many times the elements in the second column exists. Any > help is greatly appreciated.
Hi. I understand the first as follows. For each row compute the number of rows, which are equal to the given one. If this is correct, then the following can be used. a <- cbind( c(23, 12, 0, 0, 0, 0, 23), c(1, 12, 0, 1, 1, 2, 2)) u <- rep(1, times=nrow(a)) ave(u, a[, 1], a[, 2], FUN=sum) [1] 1 1 1 2 2 1 1 I am not sure, whether i understand the second correctly. Can you explain it in more detail? I would expect ave(u, a[, 2], FUN=sum) [1] 3 1 1 3 3 2 2 However, this is different from your expected output. Do you count only consecutive equal elements? Hope this helps. Petr Savicky. ______________________________________________ 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.
