Collapsing data frame by selecing one row per group

Question

I'm trying to collapse a data frame by removing all but one row from each group of rows with identical values in a particular column. In other words, the first row from each group.

For example, I'd like to convert this

> d = data.frame(x=c(1,1,2,4),y=c(10,11,12,13),z=c(20,19,18,17))
> d
  x  y  z
1 1 10 20
2 1 11 19
3 2 12 18
4 4 13 17

Into this:

    x  y  z
1   1 11 19
2   2 12 18
3   4 13 17

I'm using aggregate to do this currently, but the performance is unacceptable with more data:

> d.ordered = d[order(-d$y),]
> aggregate(d.ordered,by=list(key=d.ordered$x),FUN=function(x){x[1]})

I've tried split/unsplit with the same function argument as here, but unsplit complains about duplicate row numbers.

Is rle a possibility? Is there an R idiom to convert rle's length vector into the indices of the rows that start each run, which I can then use to pluck those rows out of the data frame?

Answer 1

Maybe duplicated() can help:

R> d[ !duplicated(d$x), ]
  x  y  z
1 1 10 20
3 2 12 18
4 4 13 17
R> 

Edit Shucks, never mind. This picks the first in each block of repetitions, you wanted the last. So here is another attempt using plyr:

R> ddply(d, "x", function(z) tail(z,1))
  x  y  z
1 1 11 19
2 2 12 18
3 4 13 17
R> 

Here plyr does the hard work of finding unique subsets, looping over them and applying the supplied function -- which simply returns the last set of observations in a block z using tail(z, 1).

Answer 2

Just to add a little to what Dirk provided... duplicated has a fromLast argument that you can use to select the last row:

d[ !duplicated(d$x,fromLast=TRUE), ]