5 ms·

It’s definitely not doing what the prompt asked for. https://regex101.com/r/ZNQa9X/1 https://regex101.com/r/ZNQa9X/1 The generated regex is the same as (

It’s definitely not doing what the prompt asked for. https://regex101.com/r/ZNQa9X/1 https://regex101.com/r/ZNQa9X/1 The generated regex is the same as (\bdog\b)|(\bcat\b) https://regex101.com/r/vTtEU4/1 https://regex101.com/r/vTtEU4/1 I’m currently trying to figure out how to match a word starting with dog without using \bdog.* because .* would proceed to eat the rest of the line. So I was thinking I could say \bdog[^\b]* But that doesn’t work, it also ends up eating the rest of the line as well.

shagie 4y ago
Use \S which is the opposite set of \s which avoids eating word boundaries too. \b(dog\S*)|(\S*cat)\b You could also use a \B instead of a \S though there are different meanings there.
- codetrotter 4y ago
  It almost does the trick https://regex101.com/r/sbpy8s/1 https://regex101.com/r/sbpy8s/1 But this matches for example dog.cat as one single word. But I would like that it matches separately dog and cat in this case. Likewise, I’d want for example dogapple-bananacat to be matched as two separate words dogapple and bananacat After a bit more reading online I thought that maybe the following regex would do what I want: \b(dog\p{L}*)|(\p{L}*cat)\b https://regex101.com/r/1NT5Ie/1 https://regex101.com/r/1NT5Ie/1 But that does not match dog42 as a word. What I want is a way to include everything after dog that is not \b And likewise everything preceding cat that is not \b Edit: I think I’ve found it after reading https://stackoverflow.com/questions/4541573/what-are-non-word-boundary-in-regex-b-compared-to-word-boundary https://stackoverflow.com/questions/4541573/what-are-non-wor... (\bdog\w*)|(\w*cat\b) Seems to behave exactly like I want. https://regex101.com/r/f3uJUE/1 https://regex101.com/r/f3uJUE/1
  - btown 4y ago
    After all, who needs wget when you have \wcat!
  - deleted 4y ago
    [deleted]
  - KronisLV 4y ago
    Out of curiosity: if humans have trouble coming up with anything non-trivial, like regexes, why should something that has been trained on the output of humans do much better? To me it feels like if 90% of $TASK content out there would be bad and people would struggle with it, then the AI-genrated $TASK output would be similarly flawed, be it regarding a programming language or something else. As a silly example, consider how much bad legacy PHP code is out there and what the answers to some PHP questions could become because of that. But it's still possible to get answers to simplistic problems reasonably fast, or at least get workable examples to then test and iterate upon, which can easily save some time.
mminer237 4y ago
\bdog\w\*
- codetrotter 4y ago
  Yup. See my response to the other sibling comment. In particular: (\bdog\w*)|(\w*cat\b) Seems to behave exactly like I want. https://regex101.com/r/f3uJUE/1 https://regex101.com/r/f3uJUE/1
paulclinger 4y ago
Agree; the ChatGPT answer is not correct, as the assignment is to match a word that starts with `dog` and ends with `cat`. You can make .* non-greedy by adding ? at the end, but it's not needed in this case, as the engine should backtrack. Something like this should work: /\bdog[\w_-]*cat\b/ (assuming _ and - should be allowed inside words). You can also specify word-separators ([^ ] instead of [\w_-]) if that's easier to read.

shagie 4y ago

Use \S which is the opposite set of \s which avoids eating word boundaries too. \b(dog\S*)|(\S*cat)\b You could also use a \B instead of a \S though there are different meanings there.

codetrotter 4y ago
It almost does the trick https://regex101.com/r/sbpy8s/1 https://regex101.com/r/sbpy8s/1 But this matches for example dog.cat as one single word. But I would like that it matches separately dog and cat in this case. Likewise, I’d want for example dogapple-bananacat to be matched as two separate words dogapple and bananacat After a bit more reading online I thought that maybe the following regex would do what I want: \b(dog\p{L}*)|(\p{L}*cat)\b https://regex101.com/r/1NT5Ie/1 https://regex101.com/r/1NT5Ie/1 But that does not match dog42 as a word. What I want is a way to include everything after dog that is not \b And likewise everything preceding cat that is not \b Edit: I think I’ve found it after reading https://stackoverflow.com/questions/4541573/what-are-non-word-boundary-in-regex-b-compared-to-word-boundary https://stackoverflow.com/questions/4541573/what-are-non-wor... (\bdog\w*)|(\w*cat\b) Seems to behave exactly like I want. https://regex101.com/r/f3uJUE/1 https://regex101.com/r/f3uJUE/1
- btown 4y ago
  After all, who needs wget when you have \wcat!
- deleted 4y ago
  [deleted]
- KronisLV 4y ago
  Out of curiosity: if humans have trouble coming up with anything non-trivial, like regexes, why should something that has been trained on the output of humans do much better? To me it feels like if 90% of $TASK content out there would be bad and people would struggle with it, then the AI-genrated $TASK output would be similarly flawed, be it regarding a programming language or something else. As a silly example, consider how much bad legacy PHP code is out there and what the answers to some PHP questions could become because of that. But it's still possible to get answers to simplistic problems reasonably fast, or at least get workable examples to then test and iterate upon, which can easily save some time.

mminer237 4y ago

\bdog\w\*

codetrotter 4y ago
Yup. See my response to the other sibling comment. In particular: (\bdog\w*)|(\w*cat\b) Seems to behave exactly like I want. https://regex101.com/r/f3uJUE/1 https://regex101.com/r/f3uJUE/1

paulclinger 4y ago

Agree; the ChatGPT answer is not correct, as the assignment is to match a word that starts with `dog` and ends with `cat`. You can make .* non-greedy by adding ? at the end, but it's not needed in this case, as the engine should backtrack. Something like this should work: /\bdog[\w_-]*cat\b/ (assuming _ and - should be allowed inside words). You can also specify word-separators ([^ ] instead of [\w_-]) if that's easier to read.