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.