On Wed, 22 Feb 2006, Robin Hankin wrote: > Hi Dmitris, and list > > On 22 Feb 2006, at 09:24, Dimitris Rizopoulos wrote: > >> another approach is: >> >> names(which(table(a) == 1)) >> >> but I don't know if you find this more elegant :)
Since the names of the table are the levels of the factor, I would use levels(a)[table(a) %in% 1] Well, almost. If you have NA or NaN as a factor level then all these solutions need to be more complicated. > well, thank you for this (which() is good here!) but this is still > "inelegant" IMHO > because it uses the names() of a table. > > If I had > > > a <- as.factor(c(1,1,1,2,3,4,4,4,4,5)) > > names(which(table(a)==1)) > [1] "2" "3" "5" > > > > this gives a character vector. > > I could coerce using as.integer() here, but this seems so....inelegant. But a has a character vector of levels, and there is nothing there to tell R that you wanted integers and not decimal-digit character strings. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UK Fax: +44 1865 272595 ______________________________________________ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
