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