Minimal match to regexp?
grep(value = TRUE) just returns the strings which match the pattern. You
have to use regexpr() or gregexpr() if you want to know where the matches
are:
```
x <- "abaca"
# extract only the first match with regexpr()
m <- regexpr("a.*?a", x)
regmatches(x, m)
# or
# extract every match with gregexpr()
m <- gregexpr("a.*?a", x)
regmatches(x, m)
```
You could also use sub() to remove the rest of the string:
`sub("^.*(a.*?a).*$", "\\1", x)`
keeping only the match within the parenthesis.
On Wed, Jan 25, 2023, 19:19 Duncan Murdoch <murdoch.duncan at gmail.com> wrote:
The docs for ?regexp say this: "By default repetition is greedy, so the
maximal possible number of repeats is used. This can be changed to
?minimal? by appending ? to the quantifier. (There are further
quantifiers that allow approximate matching: see the TRE documentation.)"
I want the minimal match, but I don't seem to be getting it. For example,
x <- "abaca"
grep("a.*?a", x, value = TRUE)
#> [1] "abaca"
Shouldn't I have gotten "aba", which is the first match to "a.*a"? If
not, what would be the regexp that would give me the first match to
"a.*a", without greedy expansion of the .*?
Duncan Murdoch
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