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Minimal match to regexp?

7 messages · Andrew Simmons, Jeff Newmiller, Duncan Murdoch

Duncan Murdoch
# 
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
Andrew Simmons
# 
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:

            

      
      
      
    

        

      
      
    

        

  
  


  


      

    

    

      
      

        


  

        

      Duncan Murdoch
    

    

      
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On 25/01/2023 7:19 p.m., Duncan Murdoch wrote:

            

      
      
      
    

        
Sorry, that was a dumb question.  Of course grep returned the whole 
thing.  I should be using regexpr() or some related function to extract 
the match.

Duncan Murdoch

  
  


  


      

    

    

      
      

        


  

        

      Jeff Newmiller
    

    

      
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Perhaps

sub( "^.*(a.*?a).*$", "\\1", x )

        
On January 25, 2023 4:19:01 PM PST, Duncan Murdoch <murdoch.duncan at gmail.com> wrote:

            

      
      
      
    

        

      
      
    

        

  
    

      
      
    

  


  


      

    

    

      
      

        


  

        

      Duncan Murdoch
    

    

      
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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:

            

      
      
      
    

        

      
      
    

  
  


  


      

    

    

      
      

        


  

        

      Andrew Simmons
    

    

      
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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 at gmail.com> wrote:

            

      
      
      
    

        

      
      
    

            

      
      
      
    

  
  


  


      

    

    

      
      

        


  

        

      Duncan Murdoch
    

    

      
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I'll submit a bug report.

        
On 25/01/2023 8:38 p.m., Andrew Simmons wrote: