Python Regex – Why re.findall Not Returning Full Match?

pythonpython-reregex

I have a file that includes a bunch of strings like "size=XXX;". I am trying Python's re module for the first time and am a bit mystified by the following behavior: if I use a pipe for 'or' in a regular expression, I only see that bit of the match returned. E.g.:

>>> myfile = open('testfile.txt', 'r').read()
>>> re.findall('size=50;', myfile)
['size=50;', 'size=50;', 'size=50;', 'size=50;']

>>> re.findall('size=51;', myfile)
['size=51;', 'size=51;', 'size=51;']

>>> re.findall('size=(50|51);', myfile)
['51', '51', '51', '50', '50', '50', '50']

>>> re.findall(r'size=(50|51);', myfile)
['51', '51', '51', '50', '50', '50', '50']

The "size=" part of the match is gone (Yet it is certainly used in the search, otherwise there would be more results). What am I doing wrong?

Best Answer

The problem you have is that if the regex that re.findall tries to match captures groups (i.e. the portions of the regex that are enclosed in parentheses), then it is the groups that are returned, rather than the matched string.

One way to solve this issue is to use non-capturing groups (prefixed with ?:).

>>> import re
>>> s = 'size=50;size=51;'
>>> re.findall('size=(?:50|51);', s)
['size=50;', 'size=51;']

If the regex that re.findall tries to match does not capture anything, it returns the whole of the matched string.

Although using character classes might be the simplest option in this particular case, non-capturing groups provide a more general solution.

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Python – re.findall() Not Working as Expected

The unwanted behaviour comes down to the way you formulate regualar expression:

rexp=re.compile("(aa|bb)+")

Parentheses (aa|bb) forms a group.

And if we look at the docs of findall we will see this:

Return all non-overlapping matches of pattern in string, as a list of strings. The string is scanned left-to-right, and matches are returned in the order found. If one or more groups are present in the pattern, return a list of groups; this will be a list of tuples if the pattern has more than one group. Empty matches are included in the result unless they touch the beginning of another match.**

As you formed a group, it mathced first aa, then bb, then aa again (because of + quantifier). So this group holds aa in the end. And findall returns this value in the list ['aa'] (as there is only one match aabbaa of the whole expression, the list contains only one element aa which is saved in the group).

From the code you gave, you seemed to want to do this:

>>> rexp=re.compile("(?:aa|bb)+")
>>> rexp.findall(sequence)
['aabbaa']

(?: ...) doesnt create any group, so findall returns the match of the whole expression.

In the end of your question you show the desired output. This is achieved by just looking for aa or bb. No quantifiers (+ or *) are needed. Just do it the way is in the Inbar Rose's answer:

>>> rexp=re.compile("aa|bb")
>>> rexp.findall(sequence)
['aa', 'bb', 'aa']

Python Regex – Fixing Unwanted Results from re.findall

>>> re.findall("(?:100|[0-9][0-9]|[0-9])%", "89%")
['89%']

When there are capture groups findall returns only the captured parts. Use ?: to prevent the parentheses from being a capture group.

Best Answer

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