Thanks Marc. It never occurred to me that I would need a ""stringsAsFactors" expression in a data.frame. I could have sworn I never did before when mocking up some data but clearly I was wrong or there has been a change in R v. 3.4.1 which seems unlikely.
On Friday, July 7, 2017, 10:37:29 AM EDT, Marc Schwartz <marc_schwa...@me.com> wrote: > On Jul 7, 2017, at 6:03 AM, John Kane via R-help <r-help@r-project.org> wrote: > > This is not serious problem but I just wonder if someone can explain what is > happening. > The same command within a dataframe is giving me a factor and as a plain > vector is giving me a character. It's probably something simple that I have > read and forgotten but I thought I'd ask. > Thanks > > #================================================ > dat1 <- data.frame(aa = letters[1:10]) > str(dat1) > data.frame': 10 obs. of 1 variable: > $ aa....letters.1.10.: Factor w/ 10 levels "a","b","c","d",..: 1 2 3 4 5 6 7 > 8 9 10#============================================================= > bb = letters[1:10] > str(bb) > chr [1:10] "a" "b" "c" "d" "e" "f" "g" "h" "i" "j" > #============================================================== > See the 'stringsAsFactors' argument in ?data.frame. dat1 <- data.frame(aa = letters[1:10], stringsAsFactors = FALSE) > str(dat1) 'data.frame': 10 obs. of 1 variable: $ aa: chr "a" "b" "c" "d" ... Regards, Marc Schwartz [[alternative HTML version deleted]] ______________________________________________ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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.