Here are several solutions:
#1
# replace first three and last 3 characters with nothing
x <- c("23:10:34", "01:02:03")
gsub("^...|...$", "", x)
#2
# backref = -1 says extract the portion in parens
# it would have returned a list so we use simplify = c
library(gsubfn)
strapply(x, ":(..):", backref = -1, simplify = c)
#3
# convert to POSIXct and use POSIXct's format
library(chron)
format(as.POSIXct(as.chron(times(x))), "%M")
#4
# chron times are fractions of a day
# this gives numeric result whereas others give character
library(chron)
floor(24 * 60 * as.numeric(times(x))) %% 60
On Wed, Oct 15, 2008 at 5:54 PM, Redding, Matthew
<[EMAIL PROTECTED]> wrote:
> Hi All,
>
> Is there a means to extract the "10" from "23:10:34" in one pass using
> strsplit (or something else)?
> tst <- "23:10:34"
>
> For example my attempt
> strsplit(as.character(tst),"^[0-9]*:")
> gives
> [[1]]
> [1] "" "" "34"
>
> Obviously it is matching the first two instances of [0-9]. Note that
> there may be only one digit before the first ":".
>
> How do I anchor the match to the begginning or better still, just
> extract the number I want in one pass?
>
> I can see that I can add "begin" to the beginning of the string, and
> match that and do something similar at the end, getting rid of empty
> strings
> etc - but I think it would take about 3 passess - and the files are
> large. And besides that code would be unlovely.
>
> Kind regards,
>
>
> Matt Redding
> ********************************DISCLAIMER**************...{{dropped:15}}
>
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