I'll submit a bug report.
On 25/01/2023 8:38 p.m., Andrew Simmons wrote:
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.dun...@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.dun...@gmail.com
<mailto:murdoch.dun...@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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