Number of replications of a term

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