[R] How to convert a factor to a numeric?

[Gabor Grothendieck]

Chris Bergstresser <chris <at> subtlety.com> writes:

: 
: Date:   Tue, 01 Mar 2005 16:45:36 -0600 
: From:   Chris Bergstresser <chris at subtlety.com>
: To:   <r-help at 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.