That's a good solution, but if you're really, really sure that the timestamps
are in the format you gave, it's quite a bit faster to use substr and paste,
because you don't have to do any searching in the string.
HTH
Rex
> x = rep("09:30:00.000.633",1000000)
> system.time(y<-paste(substr(x,1,12),substr(x,14,16),sep=""))
user system elapsed
0.87 0.00 0.88
> system.time(y<-sub("\\.(\\d+)$", "\\1", x))
user system elapsed
1.65 0.00 1.65
> system.time(y<-sub("\\.(\\d+)$", "\\1", x))
user system elapsed
1.65 0.00 1.66
> system.time(y<-paste(substr(x,1,12),substr(x,14,16),sep=""))
user system elapsed
0.88 0.00 0.89
>
-----Original Message-----
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Henrique Dallazuanna
Sent: Friday, March 18, 2011 8:32 AM
To: rivercode
Cc: r-help@r-project.org
Subject: Re: [R] Replace split with regex for speed ?
Try this:
sub("\\.(\\d+)$", "\\1", ts)
On Thu, Mar 17, 2011 at 11:01 PM, rivercode <aqua...@gmail.com> wrote:
>
> Have timestamp in format HH:MM:SS.MMM.UUU and need to remove the last "." so
> it is in format HH:MM:SS.MMMUUU.
>
> What is the fastest way to do this, since it has to be repeated on millions
> of rows. Should I use regex ?
>
> Currently doing it with a string split, which is slow:
>
> >head(ts)
> [1] 09:30:00.000.245 09:30:00.000.256 09:30:00.000.633 09:30:00.001.309
> 09:30:00.003.635 09:30:00.026.370
>
>
> ts = strsplit(ts, ".", fixed = TRUE)
> ts=lapply(ts, function(x) { paste(x[1], ".", x[2], x[3], sep="") } ) #
> Remove last . from timestamp, from HH:MM:SS.MMM.UUU to HH:MM:SS.MMMUUU
> ts = unlist(ts)
>
> Thanks,
> Chris
>
> --
> View this message in context:
> http://r.789695.n4.nabble.com/Replace-split-with-regex-for-speed-tp3386098p3386098.html
> Sent from the R help mailing list archive at Nabble.com.
>
> ______________________________________________
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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>
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40" S 49° 16' 22" O
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