Hi Marc.
For question 1.
I know in Perl that regular expressions when captured can be saved if not
overwritten. \\1 is the capture variable in your R examples.
So the 2nd regular expression does not match but \\1 still has 1980
captured from the previous expression, hence the result.
Maybe if you restart R and try your 2nd expression first, \\1 will be empty
or no match result.
Just speculation :)
John
On 9 Aug 2018 08:58, "Marc Girondot via R-help" <r-help@r-project.org>
wrote:
> Hi everybody,
>
> I have some questions about the way that sub is working. I hope that
> someone has the answer:
>
> 1/ Why the second example does not return an empty string ? There is no
> match.
>
> subtext <- "-1980-"
> sub(".*(1980).*", "\\1", subtext) # return 1980
> sub(".*(1981).*", "\\1", subtext) # return -1980-
>
> 2/ Based on sub documentation, it replaces the first occurence of a
> pattern: why it does not return 1980 ?
>
> subtext <- " 1980 1981 "
> sub(".*(198[01]).*", "\\1", subtext) # return 1981
>
> 3/ I want extract year from text; I use:
>
> subtext <- "bla 1980 bla"
> sub(".*[ \\.\\(-]([12][01289][0-9][0-9])[ \\.\\)-].*", "\\1", subtext) #
> return 1980
> subtext <- "bla 2010 bla"
> sub(".*[ \\.\\(-]([12][01289][0-9][0-9])[ \\.\\)-].*", "\\1", subtext) #
> return 2010
>
> but
>
> subtext <- "bla 1010 bla"
> sub(".*[ \\.\\(-]([12][01289][0-9][0-9])[ \\.\\)-].*", "\\1", subtext) #
> return 1010
>
> I would like exclude the case 1010 and other like this.
>
> The solution would be:
>
> 18[0-9][0-9] or 19[0-9][0-9] or 200[0-9] or 201[0-9]
>
> Is there a solution to write such a pattern in grep ?
>
> Thanks a lot
>
> Marc
>
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