[R] How to coerce data frame elements into factors?
Andrew Perrin
aperrin at socrates.berkeley.edu
Thu Mar 1 21:27:23 CET 2001
Thank you - that did it. See comments below.
On 1 Mar 2001, Peter Dalgaard BSA wrote:
> Andrew Perrin <aperrin at socrates.berkeley.edu> writes:
>
[snip]
> > > is.factor(logic.resources.f.df$interests)
> > [1] FALSE
>
> apply will give you a vector or array, so you're doing is.factor(NULL)
>
Got it - thanks for that.
> >
> >
> > I've also tried individually,
> >
> > > foo<-data.frame(c(factor(mediate), factor(morality)))
>
> > > is.factor(foo$mediate)
> > [1] FALSE
>
> What gives you the idea that that would have a component called
> "mediate"?
>
Well, the fact that my original data frame contained a component called
"mediate". And that when I did the apply() version above, the resulting
data frame did contain "mediate" as determined using colnames().
> > x<-1:0
> > y<-0:1
> > data.frame(c(factor(x),factor(y)))
> c.factor.x...factor.y..
> 1 2
> 2 1
> 3 1
> 4 2
>
> Try this:
>
> > d<-data.frame(a=0:1,b=1:0)
> > d<-as.data.frame(lapply(d,factor))
>
Worked! Thanks - here's what I did:
> logic.resources.f.df<-as.data.frame(lapply(logics.resources.df,factor))
> colnames(logic.resources.f.df)
[1] "mediate" "interests" "enlightened" "morality" "capacity"
[6] "comparison" "information" "doubt" "mistrust" "actors"
[11] "metalogic" "policy"
> is.factor(logic.resources.f.df$mediate)
[1] TRUE
> --
> O__ ---- Peter Dalgaard Blegdamsvej 3
> c/ /'_ --- Dept. of Biostatistics 2200 Cph. N
> (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918
> ~~~~~~~~~~ - (p.dalgaard at biostat.ku.dk) FAX: (+45) 35327907
>
----------------------------------------------------------------------
Andrew J Perrin - Ph.D. Candidate, UC Berkeley, Dept. of Sociology
Chapel Hill, North Carolina, USA - http://demog.berkeley.edu/~aperrin
aperrin at socrates.berkeley.edu - aperrin at igc.apc.org
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