On Mon, 15 Mar 2004 [EMAIL PROTECTED] wrote: > Full_Name: George Leigh > Version: 1.8.1 > OS: Windows 2000 > Submission from: (NULL) (203.25.1.208) > > > The following example gives the correct answer when the first argument of tapply > is a numeric vector, but an incorrect answer when it is a factor. If the > function used by tapply is "length", the type and contents of the first argument > should make no difference, provided it has the same length as the second > argument.
Not so: > split(x, y) $"1" [1] 1 > split(y, y) $"1" [1] <NA> 1 Levels: 1 Note that as there is only one level, NA must be 1 in y, whereas it does not have to be in x. So the answer for a factor in your problem is definitely correct, if fortuitous. R does the same as S in this example. If there were more than one level in y, the issue is less clearcut. Probably y[[k]] <- x[f == k] in split.default should be x[f %in% k] Note too z <- x; class(x) <- "foo" > split(z, y) $"1" [1] NA 1 > x = c(NA, 1) > > y = factor(x) > > tapply(x, y, length) > 1 > 1 > > tapply(y, y, length) > 1 > 2 -- 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 272860 (secr) Oxford OX1 3TG, UK Fax: +44 1865 272595 ______________________________________________ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-devel
