3 ms·

How would you do this? if (match := re.match(r1, s)): o = match.group(1) elif (match := re.match(r2, s)): o = match.group(1)

PurpleRamen 8y ago
match = re.match(r1, s) if match: o = match.group(1) else: match = re.match(r2, s) if match: o = match.group(1) or a bit shorter: match = re.match(r1, s) if not match: match = re.match(r2, s) if match: o = match.group(1) You could also just loop: for pattern in (r1, r2, ...): match = re.match(pattern, s) if match: o = match.group(1) break else: do_failure_handling() But this goes a bit beyond the original question.
- cup-of-tea 8y ago
  I made a mistake in the question, it should have been: if (match := re.match(r1, s)): o = match.group(1) # plus some code here elif (match := re.match(r2, s)): o = match.group(2) # plus some other code here In this case only your first solution works, I think. Leaving aside that having those deeply nested ifs is incredibly ugly, I find it hard to accept that something which completely changes the possible structure of the code is just "syntactic sugar".
  - PurpleRamen 8y ago
    It's the syntactic sugars job to beautify the ugly. And your example is also a bit unfair, because you also use another syntactic sugar, the elif, to make it more beautyful. It only falls back to the ugly solution because I can't sweeten sugar for a similar dish. But the overall question is: when is the sugar just syntactical, and at what point does it become a complete new taste?
  - Demiurge 8y ago
    Why is it incredibly ugly when it actually conveys the explicit logic of that which is happening? Can the logic be simplified? Are there any unnecessary characters? I don't see that. This is why I like Python, it's easy to read, and it is easy to refactor and see how you can improve the logic.
est 8y ago
well, that looks easy. m = re.match(r1, s) or re.match(r2, s) o = m.group(1) if m else None
- cup-of-tea 8y ago
  I made a mistake in my question. The arguments to group shouldn't have been the same. What I want to get at is when the two branches are completely different code.
  - est 8y ago
    I think it's a case where PEP 572 tries to fix bad library design. re.match shouldn't return None at all. I often write helper functions like: matcher = lambda r, s: getattr(re.match(r, s), 'group', lambda i: '') o = matcher(r1, s)(1) or matcher(r2, s)(3) here matcher have a fixed, static return type, string.

PurpleRamen 8y ago

match = re.match(r1, s) if match: o = match.group(1) else: match = re.match(r2, s) if match: o = match.group(1) or a bit shorter: match = re.match(r1, s) if not match: match = re.match(r2, s) if match: o = match.group(1) You could also just loop: for pattern in (r1, r2, ...): match = re.match(pattern, s) if match: o = match.group(1) break else: do_failure_handling() But this goes a bit beyond the original question.

cup-of-tea 8y ago
I made a mistake in the question, it should have been: if (match := re.match(r1, s)): o = match.group(1) # plus some code here elif (match := re.match(r2, s)): o = match.group(2) # plus some other code here In this case only your first solution works, I think. Leaving aside that having those deeply nested ifs is incredibly ugly, I find it hard to accept that something which completely changes the possible structure of the code is just "syntactic sugar".
- PurpleRamen 8y ago
  It's the syntactic sugars job to beautify the ugly. And your example is also a bit unfair, because you also use another syntactic sugar, the elif, to make it more beautyful. It only falls back to the ugly solution because I can't sweeten sugar for a similar dish. But the overall question is: when is the sugar just syntactical, and at what point does it become a complete new taste?
- Demiurge 8y ago
  Why is it incredibly ugly when it actually conveys the explicit logic of that which is happening? Can the logic be simplified? Are there any unnecessary characters? I don't see that. This is why I like Python, it's easy to read, and it is easy to refactor and see how you can improve the logic.

est 8y ago

well, that looks easy. m = re.match(r1, s) or re.match(r2, s) o = m.group(1) if m else None

cup-of-tea 8y ago
I made a mistake in my question. The arguments to group shouldn't have been the same. What I want to get at is when the two branches are completely different code.
- est 8y ago
  I think it's a case where PEP 572 tries to fix bad library design. re.match shouldn't return None at all. I often write helper functions like: matcher = lambda r, s: getattr(re.match(r, s), 'group', lambda i: '') o = matcher(r1, s)(1) or matcher(r2, s)(3) here matcher have a fixed, static return type, string.