Immaterial, yes, but it is always good to test :) and your solution *is*
faster and it is even faster if you can assume byte strings:
> strings = sprintf('f:/foo/bar//%s.tif', replicate(1000,
paste(sample(letters, 10), collapse='')))
> library(rbenchmark)
> benchmark(columns=c('test', 'elapsed'), replications=1000, order=NULL,
'one-pass, perl'=sub('.*//(.*)[.]tif$', '\\1', strings, perl=TRUE),
'two-pass, perl'=sub('.tif$', '', basename(strings), perl=TRUE),
'one-pass, no perl'=sub('.*//(.*)[.]tif$', '\\1', strings, perl=FALSE),
'two-pass, no perl'=sub('.tif$', '', basename(strings), perl=FALSE),
'fixed'=sub(".tif", "", basename(strings), fixed=TRUE),
'fixed, bytes'=sub(".tif", "", basename(strings), fixed=TRUE,
useBytes=TRUE))
test elapsed
1 one-pass, perl 2.946
2 two-pass, perl 3.858
3 one-pass, no perl 15.884
4 two-pass, no perl 3.788
5 fixed 2.264
6 fixed, bytes 1.813
Allan
Gabor Grothendieck wrote:
Although speed is really immaterial here this is likely
to be faster than all shown so far:
sub(".tif", "", basename(metr_list), fixed = TRUE)
It does not allow file names with .tif in the middle
of them since it will delete the first occurrence rather
than the last but such a situation is highly unlikely.
On Tue, May 26, 2009 at 4:24 PM, Wacek Kusnierczyk
<waclaw.marcin.kusnierc...@idi.ntnu.no> wrote:
Monica Pisica wrote:
Hi everybody,
Thank you for the suggestions and especially the explanation Waclaw provided
for his code. Maybe one day i will be able to wrap my head around this.
Thanks again,
you're welcome. note that if efficiency is an issue, you'd better have
perl=TRUE there:
output = sub('.*//(.*)[.]tif$', '\\1', input, perl=TRUE)
with perl=TRUE, the one-pass solution is somewhat faster than the
two-pass solution of gabor's -- which, however, is probably easier to
understand; with perl=FALSE (the default), the performance drops:
strings = sprintf(
'f:/foo/bar//%s.tif',
replicate(1000, paste(sample(letters, 10), collapse='')))
library(rbenchmark)
benchmark(columns=c('test', 'elapsed'), replications=1000, order=NULL,
'one-pass, perl'=sub('.*//(.*)[.]tif$', '\\1', strings, perl=TRUE),
'two-pass, perl'=sub('.tif$', '', basename(strings), perl=TRUE),
'one-pass, no perl'=sub('.*//(.*)[.]tif$', '\\1', strings,
perl=FALSE),
'two-pass, no perl'=sub('.tif$', '', basename(strings), perl=FALSE))
# 1 one-pass, perl 3.391
# 2 two-pass, perl 4.944
# 3 one-pass, no perl 18.836
# 4 two-pass, no perl 5.191
vQ
Monica
----------------------------------------
Date: Tue, 26 May 2009 15:46:21 +0200
From: waclaw.marcin.kusnierc...@idi.ntnu.no
To: pisican...@hotmail.com
CC: r-help@r-project.org
Subject: Re: [R] split strings
Monica Pisica wrote:
Hi everybody,
I have a vector of characters and i would like to extract certain parts. My
vector is named metr_list:
[1] "F:/Naval_Live_Oaks/2005/data//BE.tif"
[2] "F:/Naval_Live_Oaks/2005/data//CH.tif"
[3] "F:/Naval_Live_Oaks/2005/data//CRR.tif"
[4] "F:/Naval_Live_Oaks/2005/data//HOME.tif"
And i would like to extract BE, CH, CRR, and HOME in a different vector named
"names.id"
one way that seems reasonable is to use sub:
output = sub('.*//(.*)[.]tif$', '\\1', input)
which says 'from each string remember the substring between the
rigthmost two slashes and a .tif extension, exclusive, and replace the
whole thing with the captured part'. if the pattern does not match, you
get the original input:
sub('.*//(.*)[.]tif$', '\\1', 'f:/foo/bar//buz.tif')
# buz
vQ
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