If, as in this example, i is always 1, 2, ... and has an odd length
in each group then:
do.call(rbind, by(DF, DF$id, function(x) x[median(x$i), ]))
On Tue, May 26, 2009 at 8:13 AM, Zeljko Vrba <zvrba at ifi.uio.no>
wrote:> I have a large data-frame with measurements such as:
>
> id i v1 ?v2 ?v3
> 1 ?1 1.1 1.2 1.3
> 1 ?2 1.4 1.5 1.6
> 1 ?3 1.5 1.7 1.8
> 2 ?1 2.1 2.2 2.3
> 2 ?2 2.7 2.5 2.6
> 2 ?3 2.4 2.8 2.9
>
> For each unique value of 'id' (which in the real data-set is a
combination of
> three variables) I want to compute the median of v1 within each group
('i'
> distinguishes measurements within a group), and copy the value of the
remaining
> columns (v2 and v3). ?Thus, the desired result for this small example is
>
> id i v1 ?v2 ?v3
> 1 ?2 1.4 1.5 1.6
> 2 ?3 2.4 2.8 2.9
>
> I have written a (rather clumsy, in my opinion) function to perform this
task
> (see below). ?Is there a more "standard" way of achieving this?
>
> The function is:
> agg.column <- function(df, key, groups, FUN)
> {
> ?for(i in 1:length(groups))
> ? ?groups[[i]] <- as.factor(groups[[i]])
> ?groups <- split(df, interaction(groups, lex.order=TRUE))
> ?ret <- data.frame()
>
> ?for(g in groups) {
> ? ?key.fun <- FUN(g[[key]])
> ? ?row.idx <- match(key.fun, g[[key]])
> ? ?ret <- rbind(ret, g[row.idx,])
> ?}
> ?ret
> }
>
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