What about: lapply(levels(d$fac), function(x)head(d[d$fac == x,], 1))
Thanks for the reproducible example. If you put set.seed(123) before the call to sample, then everyone who tries it will get the same data frame d. Sarah On Fri, Dec 13, 2013 at 4:15 PM, Gang Chen <gangch...@gmail.com> wrote: > Suppose I have a dataframe defined as > > L3 <- LETTERS[1:3] > (d <- data.frame(cbind(x = 1, y = 1:10), fac = sample(L3, 10, replace > = TRUE))) > > x y fac > 1 1 1 C > 2 1 2 A > 3 1 3 B > 4 1 4 C > 5 1 5 B > 6 1 6 B > 7 1 7 A > 8 1 8 A > 9 1 9 B > 10 1 10 A > > I want to extract those rows that are the first occurrences for each level > of factor 'fac', which are basically the first three rows above. How can I > achieve that? The real dataframe is more complicated than the example > above, and I can't simply list all the levels of factor 'fac' by > exhaustibly listing all the levels like the following > > d[d$fac=='A' | d$fac=='B' | d$fac=='C', ] > > Thanks, > Gang -- Sarah Goslee http://www.functionaldiversity.org ______________________________________________ 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.
