On 09/21/2010 05:02 AM, Gregory Ryslik wrote: > Hi, > > I think I've found away around that issue. The following works. If this method is inefficient and one has something faster, I'll appreciate it though! > > lapply(mylist, function(x) as.numeric(as.character(x)))
You could avoid making them factors in the first place.... Otherwise, the common trick is as.numeric(levels(x))[x] or, if you are sure that the levels are always 0/1, c(0,1)[x]. Of course if that is true you could just accept the 1/2 solution and subtract 1 at the end. (But beware that you might have one element as a factor with a single level "1", and that would get treated exactly the same as a factor with only "0"...) -- Peter Dalgaard Center for Statistics, Copenhagen Business School Phone: (+45)38153501 Email: pd....@cbs.dk Priv: pda...@gmail.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.
