re.findall not returning full match?

Question

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?

Answer 1

When a regular expression contains parentheses, they capture their contents to groups, changing the behaviour of findall() to only return those groups. Here's the relevant section from the docs:

(...)

Matches whatever regular expression is inside the parentheses, and indicates the start and end of a group; the contents of a group can be retrieved after a match has been performed, and can be matched later in the string with the \number special sequence, described below. To match the literals '(' or ')', use \( or \), or enclose them inside a character class: [(] [)].

To avoid this behaviour, you can use a non-capturing group:

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

Again, from the docs:

(?:...)

A non-capturing version of regular parentheses. Matches whatever regular expression is inside the parentheses, but the substring matched by the group cannot be retrieved after performing a match or referenced later in the pattern.

Answer 2

In some cases, the non-capturing group is not appropriate, for example with regex which detects repeated words (example from python docs)

r'(\b\w+)\s+\1'

In this situation to get whole match one can use

[groups[0] for groups in re.findall(r'((\b\w+)\s+\2)', text)]

Note that \1 has changed to \2.

Answer 3

'size=(50|51);' means you are looking for size=50 or size=51 but only matching the 50 or 51 part (note the parentheses), therefore it does not return the sign=.

If you want the sign= returned, you can do:

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

Answer 4

I think what you want is using [] instead of (). [] indicates a set of characters while () indicates a group match. Try something like this:

re.findall('size=5[01];', myfile)

Answer 5

Here is a clean solution: https://www.ocpsoft.org/tutorials/regular-expressions/or-in-regex/ if the website dies here is the example (try on regex101.com):

regex: ^I like (dogs|penguins), but not (lions|tigers).$ try with: I like dogs, but not lions. I like dogs, but not tigers. I like penguins, but not lions. I like penguins, but not tigers.

Match 1 Full match 2-29 I like dogs, but not lions. Group 1. 9-13 dogs Group 2. 23-28 lions ...

but with regex: ^I like (?:dogs|penguins), but not (?:lions|tigers).$ Match 1 Full match 2-29 I like dogs, but not lions. Match 2 Full match 30-58 I like dogs, but not tigers. ...

Answer 6

As others mentioned, the "problem" with re.findall is that it returns a list of strings/tuples-of-strings depending on the use of capture groups. If you don't want to change the capture groups you're using (not to use character groups [] or non-capturing groups (?:)), you can use finditer instead of findall. This gives an iterator of Match objects, instead of just strings. So now you can fetch the full match, even when using capture groups:

import re

s = 'size=50;size=51;'
for m in re.finditer('size=(50|51);', s):
    print(m.group())

Will give:

size=50;
size=51;

And if you need a list, similar to findall, you can use a list-comprehension:

>>> [m.group() for m in re.finditer('size=(50|51);', s)]
['size=50;', 'size=51;']