Chris Bergstresser <chris <at> subtlety.com> writes: : : Date: Tue, 01 Mar 2005 16:45:36 -0600 : From: Chris Bergstresser <[EMAIL PROTECTED]> : To: <r-help@stat.math.ethz.ch> : Subject: [R] How to convert a factor to a numeric? : : : Hi all -- : : I've got two columns, both of which correspond to three factor : levels (e.g., column1 is "a", "b", or "c"; column2 is "x", "y", or "z"). : I'd like to generate a third column, consisting on whether the two : factors are correctly aligned for a given case (in this example, "a" : corresponds to "x", "b" to "y", and "c" to "z"). For example: : : a x TRUE : a y FALSE : b y TRUE : c z TRUE : b x FALSE : : Several questions: : : The easiest way seemed to me to be comparing the numeric values : across columns, but the encodings are (a=1, b=2, c=3) and (x=1, y=3, : z=2). Is there a way to change the underlying value representing each : factor, so I could just run an equality on them?
If f1 and f2 are the two factors: as.numeric(f1) == as.numeric(factor(as.character(f2))) : Is there a simple way to check for correspondence without recoding : the factors? I am not sure I would recommend this but it could be done like this: as.numeric(f1) == ifelse(f2=="x", 1, 5-as.numeric(f2)) : In the help for factor(), it says "In particular, 'as.numeric' : applied to a factor is meaningless, and may happen by implicit coercion. : To "revert" a factor 'f' to its original numeric values, : 'as.numeric(levels(f))[f]' is recommended and slightly more efficient : than 'as.numeric(as.character(f))'." However, I get the following : results. What's going on? I suspect they were thinking of the case where the levels themselves are of class numeric as in factor(c(10,10,11,11,12,12)) since it does not seem to be correct otherwise. ______________________________________________ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
