categorization
I don't know if it's "the fastest" way, but you can get there with
as.character(factor(exData$Condition, levels=c("c20", "c10", "c9",
"c5"), labels=c("AA", "BB", "CC", "DD")))
-Ista
On Mon, Nov 9, 2009 at 2:06 PM, phoebe kong <sityeekong at gmail.com> wrote:
Hi All, I have a dataset with a column named "Condition", Sample ? Condition 1 ? ? ? ? ? ?c20 2 ? ? ? ? ? ?c20 3 ? ? ? ? ? ?c10 4 ? ? ? ? ? ?c10 5 ? ? ? ? ? ?c9 6 ? ? ? ? ? ?c9 7 ? ? ? ? ? ?c5 8 ? ? ? ? ? ?c5 9 ? ? ? ? ? ?c20 10 ? ? ? ? ?c10 Could you let me know the fastest way to change c20->"AA", c20->"BB", c9->"CC", c5->"DD" without using for loop? Thanks phoebe ? ? ? ?[[alternative HTML version deleted]]
______________________________________________ R-help at 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.
Ista Zahn Graduate student University of Rochester Department of Clinical and Social Psychology http://yourpsyche.org