[R] concatenating factor from list

Gabor Grothendieck ggrothendieck at gmail.com
Thu Mar 16 02:54:54 CET 2006


On 3/15/06, Sebastian Luque <spluque at gmail.com> wrote:
> Sebastian Luque <spluque at gmail.com> wrote:
>
> "Gabor Grothendieck" <ggrothendieck at gmail.com> wrote:
>
> >> Is this ok or is it what you are trying to avoid:
>
> >> factor(unlist(lapply(cutYield, as.character)))
>
> > Thank you Gabor.  The problem with that is what if some levels do not
> > appear in any member of cutYield?
>
> This addition to your code takes care of that, although it's a bit
> expensive:
>
> factor(unlist(lapply(cutYield, as.character)),
>       levels = unique(unlist(lapply(test, levels))))
>
>

In thinking about this a bit more here is another solution which
does not disassemble and reassemble the factors and so may
be more along the lines you were looking for:

do.call("rbind", lapply(cutYield, function(x) data.frame(x = x)))$x




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