[R] Number of replications of a term
tlumley at u.washington.edu
Wed Jan 25 01:44:38 CET 2006
On Wed, 25 Jan 2006, Ray Brownrigg wrote:
> There's an even faster one, which nobody seems to have mentioned yet:
> rep(l <- rle(ids)$lengths, l)
I considered this but it wasn't clear to me from the initial post that
each ID occupied a contiguous section of the vector.
Also, lazy evaluation makes code like this
rep(l <- rle(ids)$lengths, l)
a bit worrying. It relies on rep() using the first argument before it uses
the second one. In this case, clearly, it works, but it is not a style I
would encourage and it's easy to construct functions where it fails.
> Timing on my 2.8GHz NetBSD system shows:
>  45150
>> # Gabor:
>> system.time(for (i in 1:100) ave(as.numeric(factor(ids)), ids, FUN =
>  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])
>  2.13 0.05 2.20 0.00 0.00
>> system.time(for (i in 1:100) rep(l <- rle(ids)$lengths, l))
>  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:
>>>> 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 )
>>> > table(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!
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Thomas Lumley Assoc. Professor, Biostatistics
tlumley at u.washington.edu University of Washington, Seattle
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