Hello, I believe that the following solves it:
aggregate(SD[, 3:ncol(SD)], by=list(ID), mean) aggregate(SD[, 3:ncol(SD)], by=list(ID), mean, na.rm=TRUE) It's the second you want, it will compute the means for groups that aren't only NA and return NaN for groups with all values NA. Rui Barradas -- View this message in context: http://r.789695.n4.nabble.com/Conditional-Loop-For-Data-Frame-Columns-tp4276821p4280750.html Sent from the R help mailing list archive at Nabble.com. ______________________________________________ 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.
