Matching multiple random optional regex expressions

Question

Is it possible to match more than one optional regex to a string, in any order (but retrievable in a specific order?)

For example,

s = '(pattern1)(pattern2)(pattern3)'

such that

match = re.search(s, 'pattern2 pattern1')
match = re.search(s, 'pattern1 pattern3 pattern2')
match = re.search(s, 'pattern3 pattern1')

and every other permutation match, and furthermore

match.groups()

always returns pattern1, pattern2, pattern3 in the same order, even if one or more are None

I know this sounds unlikely -- just wondering if and how it can be done.

are you looking for s = 'pattern1|pattern2|pattern3' where the pipe separate your options? — Meitham
– Meitham, Commented Feb 18, 2013 at 21:47

mderk · Accepted Answer · 2013-02-18 21:58:47Z

1

Do you mean

s = '(pattern1|pattern2|pattern3)'
match = sorted(re.findall(s, 'pattern1 pattern3 pattern2'))
match
>>> ['pattern1', 'pattern2', 'pattern3']

?

answered Feb 18, 2013 at 21:58

mderk

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Comments

Vorsprung · Accepted Answer · 2013-02-18 22:09:10Z

step one read the documentation for itertools and see what itertools magic will generate the kind of sets of matches you want. For example

>>> import itertools
>>> a=['aaa','bbb','ccc']
>>> for q in itertools.permutations(a):
...   print q
... 
('aaa', 'bbb', 'ccc')
('aaa', 'ccc', 'bbb')
('bbb', 'aaa', 'ccc')
('bbb', 'ccc', 'aaa')
('ccc', 'aaa', 'bbb')
('ccc', 'bbb', 'aaa')

To ensure that the matches are returned in a consistent manner tag each part of the regexp with ?P for example

>>> rl=[]
>>> bigr=""
>>> for q in itertools.permutations(a):
...   r=""
...   for ms in q:
...     r = r + "(?P<" + ms + ">" + ms + ")"
...   rl.append(r)
... 
>>> rl
['(?P<aaa>aaa)(?P<bbb>bbb)(?P<ccc>ccc)', '(?P<aaa>aaa)(?P<ccc>ccc)(?P<bbb>bbb)', '(?P<bbb>bbb)(?P<aaa>aaa)(?P<ccc>ccc)', '(?P<bbb>bbb)(?P<ccc>ccc)(?P<aaa>aaa)', '(?P<ccc>ccc)(?P<aaa>aaa)(?P<bbb>bbb)', '(?P<ccc>ccc)(?P<bbb>bbb)(?P<aaa>aaa)']

In the example above I've used the match strings as the id tags in the P part of the expression. You could generate a "name1" "name2" or similar instead

Finally, join up all the little regexps into one giant regexp

onegiantregexp = "|".join(rl)

And use something like the re module "groupdict" to get the results

Hope this helps

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