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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