[R] extracting a matched string using regexpr

Gabor Grothendieck ggrothendieck at gmail.com
Thu May 6 01:52:29 CEST 2010


I am using Vista.  Another thing to try is strapply using the tcl
engine (assuming you do have tcltk capabilities) and the R engine.  On
Vista R 2.11.0 patched I get the same result:

> capabilities()[["tcltk"]]
[1] TRUE
> strapply(test, "\\d{5}", c, engine = "tcl")[[1]]
[1] "88958"
> strapply(test, "\\d{5}", c, engine = "R")[[1]]
[1] "88958"

On Vista with R 2.9.2 I do get bad results:

> test<-"</tr><tr><th>88958</th><th>Abcdsef</th><th>67.8S</th><th>68.9\nW</th><th>26m</th>"
> sub(".*(\\d{5}).*", "\\1", test)
[1] "</tr><tr><th>88958</th><th>Abcdsef</th><th>67.8S</th><th>68.9\nW</th><th>26m</th>"
> sub(".*(\\d{5}).*", "\\1", test, extended = TRUE)
[1] "</tr><tr><th>88958</th><th>Abcdsef</th><th>67.8S</th><th>68.9\nW</th><th>26m</th>"
> R.version.string
[1] "R version 2.9.2 Patched (2009-09-08 r49647)"
> win.version()
[1] "Windows Vista (build 6002) Service Pack 2"


On Wed, May 5, 2010 at 6:20 PM, steven mosher <moshersteven at gmail.com> wrote:
> Hmm.
> I have R11 just downloaded fresh.
> I'll reload a new session..and revert. I will note that I've had trouble
> with \\d
> which is why I was using [0-9]
> MAC here.
>
> On Wed, May 5, 2010 at 3:00 PM, Gabor Grothendieck <ggrothendieck at gmail.com>
> wrote:
>>
>> That's not what I get:
>>
>> >
>> > test<-"</tr><tr><th>88958</th><th>Abcdsef</th><th>67.8S</th><th>68.9\nW</th><th>26m</th>"
>> > sub(".*(\\d{5}).*", "\\1", test)
>> [1] "88958"
>> > R.version.string
>> [1] "R version 2.10.1 (2009-12-14)"
>>
>> I also got the above in R 2.11.0 patched as well.
>>
>>
>> On Wed, May 5, 2010 at 5:55 PM, steven mosher <moshersteven at gmail.com>
>> wrote:
>> >  test
>> > [1]
>> >
>> > "</tr><tr><th>88958</th><th>Abcdsef</th><th>67.8S</th><th>68.9\nW</th><th>26m</th>"
>> >> sub(".*(\\d{5}).*", "\\1", test)
>> > [1] "</th>"
>> >> sub(".*([0-9]{5}).*","\\1",test)
>> > [1] "88958"
>> >>
>> >
>> > I think the "</" in  the source throws something off.
>> > as the group capture appears to not be working, except the bracket
>> > version
>> > it did.
>> >
>> > On Wed, May 5, 2010 at 2:35 PM, Gabor Grothendieck
>> > <ggrothendieck at gmail.com>
>> > wrote:
>> >>
>> >> Here are two ways to extract 5 digits.
>> >>
>> >> In the first one \\1 refers to the portion matched between the
>> >> parentheses in the regular expression.
>> >>
>> >> In the second one strapply is like apply where the object to be worked
>> >> on is the first argument (array for apply, string for strapply) the
>> >> second modifies it (which dimension for apply, regular expression for
>> >> strapply) and the last is a function which acts on each value
>> >> (typically each row or column for apply and each match for strapply).
>> >> In this case we use c as our function to just return all the results.
>> >> They are returned in a list with one component per string but here
>> >> test is just a single string so we get a list one long and we ask for
>> >> the contents of the first component using [[1]].
>> >>
>> >> # 1 - sub
>> >> sub(".*(\\d{5}).*", "\\1", test)
>> >>
>> >> # 2 - strapply - see http://gsubfn.googlecode.com
>> >> library(gsubfn)
>> >> strapply(test, "\\d{5}", c)[[1]]
>> >>
>> >>
>> >>
>> >> On Wed, May 5, 2010 at 5:13 PM, steven mosher <moshersteven at gmail.com>
>> >> wrote:
>> >> > Given a text like
>> >> >
>> >> > I want to be able to extract a matched regular expression from a
>> >> > piece
>> >> > of
>> >> > text.
>> >> >
>> >> > this apparently works, but is pretty ugly
>> >> > # some html
>> >> >
>> >> >
>> >> > test<-"</tr><tr><th>88958</th><th>Abcdsef</th><th>67.8S</th><th>68.9\nW</th><th>26m</th>"
>> >> > # a pattern to extract 5 digits
>> >> >> pattern<-"[0-9]{5}"
>> >> > # regexpr returns a start point[1] and an attribute "match.length"
>> >> > attr(,"match.length)
>> >> > # get the substring from the start point to the stop point.. where
>> >> > stop
>> >> > =
>> >> > start +length-1
>> >> >>
>> >> >
>> >> >
>> >> > answer<-substr(test,regexpr(pattern,test)[1],regexpr(pattern,test)[1]+attr(regexpr(pattern,test),"match.length")-1)
>> >> >> answer
>> >> > [1] "88958"
>> >> >
>> >> > I tried using sub(pattern, replacement, x )  with a regexp that
>> >> > captured
>> >> > the
>> >> > group. I'd found an example of this in the mails
>> >> > but it didnt seem to work..
>> >
>> >
>
>



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