[R] Splitting the string at the last sub-string
Tuszynski, Jaroslaw W.
JAROSLAW.W.TUSZYNSKI at saic.com
Thu Sep 15 17:00:17 CEST 2005
Thanks for suggestions. I suspect the "regexpr" version will be better than
my version, since I use it to find an string towards the end of a large (up
to ~30Mb) test/XML file.
Thanks again.
Jarek
====================================================\====
Jarek Tuszynski, PhD. o / \
Science Applications International Corporation <\__,|
(703) 676-4192 "> \
Jaroslaw.W.Tuszynski at saic.com ` \
-----Original Message-----
From: r-help-bounces at stat.math.ethz.ch
[mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of Prof Brian Ripley
Sent: Thursday, September 15, 2005 10:43 AM
To: Barry Rowlingson
Cc: r-help at stat.math.ethz.ch
Subject: Re: [R] Splitting the string at the last sub-string
On Thu, 15 Sep 2005, Barry Rowlingson wrote:
> Prof Brian Ripley wrote:
>
>>> substring(str, c(1, 26), c(25,length(str)))
>
> nchar(str) surely?
Yes, or anything larger: I actually tested 10000.
> regexps can be rather slow though. Here's two functions:
But that's not the way to do this repeatedly for the same pattern. (It is
normally compiling regexps that is slow, and regexpr is vectorized.) Not
that I would call 300us `slow'.
> byRipley =
> function(str,sub){
> lp=attr(regexpr(paste(".*",sub,sep=""),str),'match.length')
> return(substring(str, c(1, lp+1), c(lp,nchar(str)))) }
>
> byJarek =
> function(str,sub){
> y = unlist(strsplit(str,sub))
> return(cbind(paste(y[-length(y)], sub, sep="", collapse = ""),
> y[length(y)]))
> }
>
> and a quick test:
>
> > system.time(for(i in 1:100000){byJarek(str,sub)})
> [1] 15.55 0.10 16.06 0.00 0.00
>
> > system.time(for(i in 1:100000){byRipley(str,sub)})
> [1] 30.28 0.07 31.86 0.00 0.00
>
> Baz
>
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--
Brian D. Ripley, ripley at stats.ox.ac.uk
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UK Fax: +44 1865 272595
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