# [R] Number of replications of a term

Ray Brownrigg ray at mcs.vuw.ac.nz
Tue Jan 24 22:32:32 CET 2006

```There's an even faster one, which nobody seems to have mentioned yet:

rep(l <- rle(ids)\$lengths, l)

Timing on my 2.8GHz NetBSD system shows:

> length(ids)
[1] 45150
> # Gabor:
> system.time(for (i in 1:100) ave(as.numeric(factor(ids)), ids, FUN =
length))
[1] 3.45 0.06 3.54 0.00 0.00
> # Barry (and others I think):
> system.time(for (i in 1:100) table(ids)[ids])
[1] 2.13 0.05 2.20 0.00 0.00
> Me:
> system.time(for (i in 1:100) rep(l <- rle(ids)\$lengths, l))
[1] 1.60 0.00 1.62 0.00 0.00

Of course the difference between 21 milliseconds and 16 milliseconds is
not great, unless you are doing this a lot.

Ray Brownrigg

> From: Gabor Grothendieck <ggrothendieck at gmail.com>
>
> Nice.  I timed it and its much faster than mine too.
>
> On 1/24/06, Barry Rowlingson <B.Rowlingson at lancaster.ac.uk> wrote:
> > Laetitia Marisa wrote:
> > > Hello,
> > >
> > > Is there a simple and fast function that returns a vector of the number
> > > of replications for each object of a vector ?
> > > For example :
> > > I have a vector of IDs :
> > > ids <- c( "ID1", "ID2", "ID2", "ID3", "ID3","ID3", "ID5")
> > >
> > >  I want the function returns the following vector where each term is the
> > > number of replicates for the given id :
> > > c( 1, 2, 2, 3,3,3,1 )
> >
> > One-liner:
> >
> >  > table(ids)[ids]
> > ids
> > ID1 ID2 ID2 ID3 ID3 ID3 ID5
> >   1   2   2   3   3   3   1
> >
> >  'table(ids)' computes the counts, then the subscripting [ids] looks it
> > all up.
> >
> >  Now try it on your 40,000-long vector!
> >
> > Barry

```