On Mon, Jun 8, 2009 at 7:18 PM, Wacek
Kusnierczyk<waclaw.marcin.kusnierc...@idi.ntnu.no> wrote:
> Gabor Grothendieck wrote:
>> Try this. See ?regex for more.
>>
>>
>>> x <- 'This happened in the 21. century." (the dot behind 21 is'
>>> regexpr("(?![0-9]+)[.]", x, perl = TRUE)
>>>
>> [1] 24
>> attr(,"match.length")
>> [1] 1
>>
>
> yes, but
>
> gregexpr('(?![0-9]+)[.]', 'a. 1. a1.', perl=TRUE)
> # 2 5 9
Yes, it should be:
> gregexpr('(?<=[0-9])[.]', 'a. 1. a1.', perl=TRUE)
[[1]]
[1] 5 9
attr(,"match.length")
[1] 1 1
which displays the position of every dot that is preceded
immediately by a digit. Or just replace gregexpr with regexpr
if its intended that it match only one.
>
> which, i guess, is not what you want. if what you want is to match all
> and only dots that follow at least one digit preceded by a word
> boundary, then the following should do, as far as i can see:
>
> gregexpr('\\b[0-9]+\\K[.]', 'a. 1. a1.', perl=TRUE)
> # 5
>
> vQ
>
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