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Is there a way to determine how many capture groups there are in a given regular expression?
I would like to be able to do the follwing:
def groups(regexp, s): """ Returns the first result of re.findall, or an empty default >>> groups(r'(\d)(\d)(\d)', '123') ('1', '2', '3') >>> groups(r'(\d)(\d)(\d)', 'abc') ('', '', '') """ import re m = re.search(regexp, s) if m: return m.groups() return ('',) * num_of_groups(regexp) This allows me to do stuff like:
first, last, phone = groups(r'(\w+) (\w+) ([\d\-]+)', 'John Doe 555-3456') However, I don't know how to implement num_of_groups. (Currently I just work around it.)
EDIT: Following the advice from rslite, I replaced re.findall with re.search.
sre_parse seems like the most robust and comprehensive solution, but requires tree traversal and appears to be a bit heavy.
MizardX's regular expression seems to cover all bases, so I'm going to go with that.
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Capturing groups are a handy feature of regular expression matching that allows us to query the Match object to find out the part of the string that matched against a particular part of the regular expression. Anything you have in parentheses () will be a capture group.
Capturing groups are a way to treat multiple characters as a single unit. They are created by placing the characters to be grouped inside a set of parentheses. For example, the regular expression (dog) creates a single group containing the letters "d" "o" and "g" .
groups() method. This method returns a tuple containing all the subgroups of the match, from 1 up to however many groups are in the pattern. The default argument is used for groups that did not participate in the match; it defaults to None. In later versions (from 1.5.
def num_groups(regex): return re.compile(regex).groups If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
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