sorting without order

Hi Marc,

continuing on Prof. Dalgaard's proposal, you could use:

ix <- unlist(split(seq(along=v), v), use.names=FALSE)

but even with this, `sort()' seems faster if you are interseted only 
in grouping:

v <- sample(1:25000, 50000, TRUE)
######
system.time(ix <- do.call("c",split(seq(along=v),v)), gcFirst=TRUE)
[1] 0.13 0.00 0.13   NA   NA

system.time(ix <- unlist(split(seq(along=v), v), use.names=FALSE), 
gcFirst=TRUE)
[1] 0.06 0.00 0.07   NA   NA

system.time(x <- sort(v), gcFirst=TRUE)
[1] 0.01 0.00 0.02   NA   NA


I hope it helps.

Best,
Dimitris

----
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven

Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/16/336899
Fax: +32/16/337015
Web: http://www.med.kuleuven.ac.be/biostat
     http://www.student.kuleuven.ac.be/~m0390867/dimitris.htm



----- Original Message ----- 
From: "Marc Mamin" <M.Mamin at intershop.de>
To: <r-help at stat.math.ethz.ch>
Sent: Tuesday, November 23, 2004 10:58 AM
Subject: [R] sorting without order

Hello,


In order to increase the performance of a script I'd like to sort 
very large vectors containing repeated integer values.
I'm not interesting in having the values sorted, but only grouped.
I also need the equivalent of index.return from the standard "sort" 
function:

 f(c(10,1,10,100,1,10))

 =>

 grouped: c(10,10,10,1,1,100)
 ix:   c(1,3,6,2,5,4)


is there a way to achieve this which would be faster than the 
standard sort function?

Thanks for any hints,

Marc Mamin

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