4 ms·

While it is a neat trick as one liner, I would recommend against doing anything like this in any software that requires maintenance. The code is hard, or imposs

While it is a neat trick as one liner, I would recommend against doing anything like this in any software that requires maintenance. The code is hard, or impossible to follow, no comments. Brittle and only few people can understand what it really does. Better option would be 10 lines of Python or JavaScript with some comments.

lrvick 4y ago
I understood it instantly on first read. Probably depends on how much shell you write.
rascul 4y ago
Comments can be added. Understanding it requires learning the tools. Just like understanding python or javascript requires learning python or javascript. It's not impossible to follow.
kragen 4y ago
I thought it was trivial to understand, though the comment above it helps a lot, and it's maybe an unfair advantage that I'd done the same thing in pretty much the same way four years ago. It probably depends on your background; I wouldn't write it that way for people who didn't know shell, just like I wouldn't write this comment in English for people who speak only Spanish. I'm not convinced that it's easier to understand in Python (even though I simplified it a bit, in part because one piece of the Python 3 braindamage was moving string.maketrans to bytes): import re def main(words): for word in words: word = word.strip().upper() if re.compile(r'[A-FOI]{6}$').match(word): print('#' + word.replace('I', '1').replace('O', '0')) if __name__ == '__main__': main(open('/usr/share/dict/words')) I think the shell version is clearly better for interactive improvisation, though.
- js2 4y ago
  I prefer search with an explicit '^' in the pattern to using match. For a throw-away script I'd probably do this: import re is_hex_like = re.compile(r"^[a-foi]{6}$", re.I).search for word in filter(is_hex_like, open("/usr/share/dict/words")): hexword = word.upper().replace("O", "0").replace("I", "1").rstrip() print(f"#{hexword}")
  - kragen 4y ago
    That's nicer than my version! I'm curious why you prefer search(), though.
    - js2 4y ago
      1. I don't have to remember which implicitly anchors to the start of the string and which doesn't. 2. I prefer the explicitness of '^' (maybe that's just another way of stating (1). 3. I can use re.M to modify '^' to match at the start of each line on multiline strings, whereas match will still keep searching from the front. 4. The asymmetry of anchoring the front but not the end is weird. Python now has fullmatch, but ugh, just use the pattern for that if you need it. 5. Off the top of my head, I can't think of another language that has a regex function that implicitly anchors the front.
      - kragen 4y ago
        Hmm, I see. Interesting! I think of regexps as state machines, so I think of the implicit loop to find a starting position as extra complexity, which can give rise to for example performance problems, though it's true that in many languages you can't avoid it.
  - Too 4y ago
    findall and multiline mode makes it even easier, at the cost of loading whole file into memory though, for that reason your alternaive is probably better import re wordlist = open("/usr/share/dict/words").read() for word in re.findall(r"^[a-foi]{6}$", wordlist, re.IGNORECASE | re.MULTILINE): hexword = word.upper().replace("O", "0").replace("I", "1") print(f"#{hexword}")

lrvick 4y ago

I understood it instantly on first read. Probably depends on how much shell you write.

rascul 4y ago

Comments can be added. Understanding it requires learning the tools. Just like understanding python or javascript requires learning python or javascript. It's not impossible to follow.

kragen 4y ago

I thought it was trivial to understand, though the comment above it helps a lot, and it's maybe an unfair advantage that I'd done the same thing in pretty much the same way four years ago. It probably depends on your background; I wouldn't write it that way for people who didn't know shell, just like I wouldn't write this comment in English for people who speak only Spanish. I'm not convinced that it's easier to understand in Python (even though I simplified it a bit, in part because one piece of the Python 3 braindamage was moving string.maketrans to bytes): import re def main(words): for word in words: word = word.strip().upper() if re.compile(r'[A-FOI]{6}$').match(word): print('#' + word.replace('I', '1').replace('O', '0')) if __name__ == '__main__': main(open('/usr/share/dict/words')) I think the shell version is clearly better for interactive improvisation, though.

js2 4y ago
I prefer search with an explicit '^' in the pattern to using match. For a throw-away script I'd probably do this: import re is_hex_like = re.compile(r"^[a-foi]{6}$", re.I).search for word in filter(is_hex_like, open("/usr/share/dict/words")): hexword = word.upper().replace("O", "0").replace("I", "1").rstrip() print(f"#{hexword}")
- kragen 4y ago
  That's nicer than my version! I'm curious why you prefer search(), though.
  - js2 4y ago
    1. I don't have to remember which implicitly anchors to the start of the string and which doesn't. 2. I prefer the explicitness of '^' (maybe that's just another way of stating (1). 3. I can use re.M to modify '^' to match at the start of each line on multiline strings, whereas match will still keep searching from the front. 4. The asymmetry of anchoring the front but not the end is weird. Python now has fullmatch, but ugh, just use the pattern for that if you need it. 5. Off the top of my head, I can't think of another language that has a regex function that implicitly anchors the front.
    - kragen 4y ago
      Hmm, I see. Interesting! I think of regexps as state machines, so I think of the implicit loop to find a starting position as extra complexity, which can give rise to for example performance problems, though it's true that in many languages you can't avoid it.
- Too 4y ago
  findall and multiline mode makes it even easier, at the cost of loading whole file into memory though, for that reason your alternaive is probably better import re wordlist = open("/usr/share/dict/words").read() for word in re.findall(r"^[a-foi]{6}$", wordlist, re.IGNORECASE | re.MULTILINE): hexword = word.upper().replace("O", "0").replace("I", "1") print(f"#{hexword}")