Splitting the string at the last sub-string

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

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)
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