On 13/02/2009, at 9:06 AM, markle...@verizon.net wrote:
Hi Jason: below seems to work. you have to take the transpose because the apply returns the rows transposed. i'm also not sure how to make the NAs be the last ones but maybe someone can show us how to do that.
Pretty easy: na.at.end <- function(x){ i <- is.na(x) c(x[!i],rep(NA,sum(i))) }
mat <- matrix(c(2,7,2,7,9,10,10,6,8,6,1,9,7,2,0),byrow=TRUE,nrow=3) print(mat) t(apply(mat,1, function(.row) { .row[duplicated(.row)] <- NA .row }))
Then just change to: t(apply(mat,1, function(.row) { .row[duplicated(.row)] <- NA na.at.end(.row) })) cheers, Rolf ###################################################################### Attention:\ This e-mail message is privileged and confid...{{dropped:9}} ______________________________________________ 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.