Excellent! I wonder why I haven't seen aggregate before.
Thanks!
ben
On Tue, Feb 28, 2012 at 4:51 PM, ilai <ke...@math.montana.edu> wrote:
> aggregate(val~lvls+nm,data=x,FUN='median')
>
>
>
> On Tue, Feb 28, 2012 at 4:43 PM, Ben quant <ccqu...@gmail.com> wrote:
> > Hello,
> >
> > I can get the median for each factor, but I'd like another column to go
> > with each factor. The nm column is a long name for the lvls column. So
> > unique work except for the order can get messed up.
> >
> > Example:
> > x =
> >
> data.frame(val=1:10,lvls=c('cat2',rep("cat1",4),rep("cat2",4),'cat1'),nm=c('longname2',rep("longname1",4),rep("longname2",4),'longname1'))
> > x
> > val lvls nm
> > 1 1 cat2 longname2
> > 2 2 cat1 longname1
> > 3 3 cat1 longname1
> > 4 4 cat1 longname1
> > 5 5 cat1 longname1
> > 6 6 cat2 longname2
> > 7 7 cat2 longname2
> > 8 8 cat2 longname2
> > 9 9 cat2 longname2
> > 10 10 cat1 longname1
> >
> > unique doesn't work in data.frame:
> > mdn = do.call(rbind,lapply(split(x[,1], x[,2]), median))
> > data.frame(mdn,ln=as.character(unique(x[,3])))
> > mdn ln
> > cat1 4 longname2
> > cat2 7 longname1
> >
> > I want:
> > mdn ln
> > cat1 4 longname1
> > cat2 7 longname2
> >
> > Thank you very much!
> >
> > PS - looking for simple'ish solutions. I know I can do it with loops and
> > merges, but is there an option I am not using here?
> >
> > Ben
> >
> > [[alternative HTML version deleted]]
> >
> > ______________________________________________
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> > PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
>
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