[R] column selection for aggregate()
Petr PIKAL
petr.pikal at precheza.cz
Tue Jan 19 10:07:41 CET 2010
Hi
If I really wanted aggregate all numerics by all non numerics this is how
I would do it
my.numerics <- which(sapply(zeta, is.numeric))
my.factor <- which(sapply(zeta, is.factor))
aggregate(zeta[, my.numerics], zeta[, my.factor], mean)
Regards
Petr
r-help-bounces at r-project.org napsal dne 18.01.2010 16:33:17:
> I didn't understand from the help what really does the function rowMeans
> but it looks like it doesn't take into account the categorical variables
> (I want to calculate the means when the values of all categorical
> variables are the same, second part of aggregate). Moreover, ssfa_num
> contains only numeric variables, meaning that the categories will not be
> associated with it.
> I'm kind of confused with this approach.
> You think it would work for me?
>
> Thanks
> Ivan
>
> b k a écrit :
> >
> >
> > On Mon, Jan 18, 2010 at 10:17 AM, Ivan Calandra
> > <ivan.calandra at uni-hamburg.de <mailto:ivan.calandra at uni-hamburg.de>>
> > wrote:
> >
> > Thanks for your answer, but it doesn't work...
> >
> > Here is what I get:
> > > ssfamean <- aggregate(ssfa[[10:24]],ssfa[c("SPECSHOR", "BONE",
> > "TO_POS", "FACETTE", "SHEARFAC", "ENA_BA")],mean)
> > Error in .subset2(x, i, exact = exact) :
> > recursive indexing failed at level 2
> >
> >
> > Wouldn't you be better off with rowMeans() ? Split your dataframe into
> > numeric matrix:
> >
> > ssfa_num <- ssfa[10:24]
> >
> > ssfameans <- rowMeans(ssfa_num)
> >
> >
> > Also col_index <- match("Asfc", ssfa) doesn't really work since
> > col_index is composed of 1227 NAs...
> >
> >
> >
> >
> >
> > Yes, it should be:
> >
> > col_index <- match("Asfc", names(ssfa))
> >
> > Ben
>
> [[alternative HTML version deleted]]
>
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