Quoting Marc Girondot via R-help <[email protected]>:

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-

This is as documented in ?sub:
   "Elements of character vectors x which are not
    substituted will be returned unchanged"

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

Because the pattern matches the whole string,
not just the year:

    regexpr(".*(198[01]).*", subtext)
    ## [1] 1
    ## attr(,"match.length")
    ## [1] 11
    ## attr(,"useBytes")
    ## [1] TRUE

From this match, the RE engine will give you the last backreference-match,
which is "1981". If you want to _extract_ the first year, use a non-greedy RE instead:

    sub(".*?(198[01]).*", "\\1", subtext)
    ## [1] "1980"

I say _extract_ because you may _replace_ the pattern, as expected:

    sub("198[01]", "YYYY", subtext)
    ## [1] " YYYY 1981 "

That is because the pattern does not match the whole string.
Perhaps this example makes it clearer:

    test <- "1 2 3 4 5"
    sub("([0-9])", "\\1\\1", test)
    ## [1] "11 2 3 4 5"
    sub(".*([0-9]).*", "\\1\\1", test)
    ## [1] "55"
    sub(".*?([0-9]).*", "\\1\\1", test)
    ## [1] "11"



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 ?

You answered this yourself, I think.


Thanks a lot

Marc



--
Enrico Schumann
Lucerne, Switzerland
http://enricoschumann.net

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