I am puzzled by the advice to use is.na(x) <- TRUE instead of x <- NA.
?NA says Function `is.na<-' may provide a safer way to set missingness. It behaves differently for factors, for example. However, "MAY provide" is a bit scary, and it doesn't say WHAT the difference in behaviour is. I must say that "is.na(x) <- ..." is rather repugnant, because it doesn't work. What do I mean? Well, as the designers of SETL who many years ago coined the term "sinister function call" to talk about f(...)<-..., pointed out, if you do f(x) <- y then afterwards you expect f(x) == y to be true. So let's try it: > x <- c(1,NA,3) > is.na(x) <- c(FALSE,FALSE,TRUE) > x [1] 1 NA NA > is.na(x) [1] FALSE TRUE TRUE vvvvv So I _assigned_ c(FALSE,FALSE,TRUE) to is.na(x), but I _got_ c(FALSE,TRUE, TRUE)> instead. ^^^^^ That is not how a well behaved sinister function call should work, and it's enough to scare someone off is.na()<- forever. The obvious way to set elements of a variable to missing is ... <- NA. Wouldn't it be better if that just plain worked? Can someone give an example of is.na()<- and <-NA working differently with a factor? I just tried it: > x <- factor(c(3,1,4,1,5,9)) > y <- x > is.na(x) <- x==1 > y[y==1] <- NA > x [1] 3 <NA> 4 <NA> 5 9 Levels: 1 3 4 5 9 > y [1] 3 <NA> 4 <NA> 5 9 Levels: 1 3 4 5 9 Both approaches seem to have given the same answer. What did I miss? ______________________________________________ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help