[R] Minimal match to regexp?

[Andrew Simmons]

It seems like a bug to me. Using perl = TRUE, I see the desired result:

```
x <- "\n```html\nblah blah \n```\n\n```r\nblah blah\n```\n"

pattern2 <- "\n([`]{3,})html\n.*?\n\\1\n"

cat(regmatches(x, regexpr(pattern2, x, perl = TRUE)))
```

If you change it to something like:

```
x <- c(
    "\n```html\nblah blah \n```\n\n```r\nblah blah\n```\n",
    "\n```html\nblah blah \n```\n"
)

pattern2 <- "\n([`]{3,})html\n.*?\n\\1\n"

print(regmatches(x, regexpr(pattern2, x)), width = 10)
```

you can see that it does find the match, so the combination of *? and
\\1 must be messing up regexpr(). They seem to work perfectly fine on
their own.

On Wed, Jan 25, 2023 at 7:57 PM Duncan Murdoch <murdoch.duncan using gmail.com> wrote:
>
> Thanks for pointing out my mistake.  I oversimplified the real problem.
>
> I'll try to post a version of it that comes closer:  Suppose I have a
> string like this:
>
> x <- "\n```html\nblah blah \n```\n\n```r\nblah blah\n```\n"
>
> If I cat() it, I see that it is really markdown source:
>
>    ```html
>    blah blah
>    ```
>
>    ```r
>    blah blah
>    ```
>
> I want to find the part that includes the html block, but not the r
> block.  So I want to match "```html", followed by a minimal number of
> characters, then "```".  Then this pattern works:
>
>    pattern <- "\n```html\n.*?\n```\n"
>
> and we get the right answer:
>
>    cat(regmatches(x, regexpr(pattern, x)))
>
>    ```html
>    blah blah
>    ```
>
> Okay, but this flavour of markdown says there can be more backticks, not
> just 3.  So the block might look like
>
>    ````html
>    blah blah
>    ````
>
> I need to have the same number of backticks in the opening and closing
> marker.  So I make the pattern more complicated, and it doesn't work:
>
>    pattern2 <- "\n([`]{3,})html\n.*?\n\\1\n"
>
> This matches all of x:
>
>    > pattern2 <- "\n([`]{3,})html\n.*?\n\\1\n"
>    > cat(regmatches(x, regexpr(pattern2, x)))
>
>    ```html
>    blah blah
>    ```
>
>    ```r
>    blah blah
>    ```
>
>
> Is that a bug, or am I making a silly mistake again?
>
> Duncan Murdoch
>
>
>
> On 25/01/2023 7:34 p.m., Andrew Simmons wrote:
> > 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 using gmail.com
> > <mailto:murdoch.duncan using 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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>