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 ______________________________________________ R-help@r-project.org <mailto:R-help@r-project.org> mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help <https://stat.ethz.ch/mailman/listinfo/r-help> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html <http://www.R-project.org/posting-guide.html> and provide commented, minimal, self-contained, reproducible code.
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