Christoph,
This is more complicated than your analysis.
1) apply takes a matrix as an argument, not a data frame, and so first
coerced 'dat' to a character matrix.
2) unlist is working quite correctly. The issue is array(), which
contains as.vector(data). Thus although the result could be a factor
matrix, as.vector is coercing it to a character matrix. It might be
desirable to return a factor matrix, but we are not going to do that in
feature freeze (if ever) and I really don't think it would be what you
wanted.
Perhaps the help page should contain an explicit statement that the result
will be coerced to a basic vector type by as.vector().
On Mon, 25 Sep 2006, Christoph Buser wrote:
> Dear R-core
>
> There is a different output for the apply function due to the
> change of unlist as mentioned in the R news.
>
> Newly, applying as.factor() (or factor()) in
>
> str(dat <- data.frame(x = 1:10, f1 = gl(2,5,labels = c("A",
"B"))))
> (d1 <- apply(dat,2,as.factor))
>
> newly returns a character matrix while in R-2.3.1 the same
> command resulted in an integer matrix that was consistent (up to
> the ordering of the factor levels) with data.matrix().
That's coincidence -- try x=11:20.
> The change is caused by the change of unlist() that, used for a
> list of factors, newly returns a single factor instead of an
> integer. I am happy with this change, but:
>
> Is it desirable to change apply so that it does not return a
> character matrix in the example above or include a warning for
> such a case?
>
> Thank you very much for an answer.
>
> Regards,
>
> Christoph Buser
>
> --------------------------------------------------------------
> Christoph Buser <buser at stat.math.ethz.ch>
> Seminar fuer Statistik, LEO C13
> ETH Zurich 8092 Zurich SWITZERLAND
> phone: x-41-44-632-4673 fax: 632-1228
> http://stat.ethz.ch/~buser/
>
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>
--
Brian D. Ripley, ripley at stats.ox.ac.uk
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