Why does re.findall return a list of tuples when my pattern only contains one group?

Question

Say I have a string s containing letters and two delimiters 1 and 2. I want to split the string in the following way:

• if a substring t falls between 1 and 2, return t
• otherwise, return each character

So if s = 'ab1cd2efg1hij2k', the expected output is ['a', 'b', 'cd', 'e', 'f', 'g', 'hij', 'k'].

I tried to use regular expressions:

import re
s = 'ab1cd2efg1hij2k'
re.findall( r'(1([a-z]+)2|[a-z])', s )

[('a', ''),
 ('b', ''),
 ('1cd2', 'cd'),
 ('e', ''),
 ('f', ''),
 ('g', ''),
 ('1hij2', 'hij'),
 ('k', '')]

From there i can do [ x[x[-1]!=''] for x in re.findall( r'(1([a-z]+)2|[a-z])', s ) ] to get my answer, but I still don't understand the output. The documentation says that findall returns a list of tuples if the pattern has more than one group. However, my pattern only contains one group. Any explanation is welcome.

Solution 1

You pattern has two groups, the bigger group:

(1([a-z]+)2|[a-z])

and the second smaller group which is a subset of your first group:

([a-z]+)

Here is a solution that gives you the expected result although mind you, it is really ugly and there is probably a better way. I just can't figure it out:

import re
s = 'ab1cd2efg1hij2k'
a = re.findall( r'((?:1)([a-z]+)(?:2)|([a-z]))', s )
a = [tuple(j for j in i if j)[-1] for i in a]

>>> print a
['a', 'b', 'cd', 'e', 'f', 'g', 'hij', 'k']

Solution 2

I am 5 years too late to the party, but I think I might have found an elegant solution to the re.findall() ugly tuple-ridden output with multiple capture groups.

In general, if you end up with an output which looks something like that:

[('pattern_1', '', ''), ('', 'pattern_2', ''), ('pattern_1', '', ''), ('', '', 'pattern_3')]

Then you can bring it into a flat list with this little trick:

["".join(x) for x in re.findall(all_patterns, iterable)]

The expected output will be like so:

['pattern_1', 'pattern_2', 'pattern_1', 'pattern_3']

It was tested on Python 3.7. Hope it helps!

Solution 3

Your regular expression has 2 groups, just look at the number of parenthesis you are using :). One group would be ([a-z]+) and the other one (1([a-z]+)2|[a-z]). The key is that you can have groups inside other groups. So, if possible, you should build a regular expression with only one group, so that you don't have to post-process the result.

An example of regular expression with only one group would be:

>>> import re
>>> s = 'ab1cd2efg1hij2k'
>>> re.findall('((?<=1)[a-z]+(?=2)|[a-z])', s)
['a', 'b', 'cd', 'e', 'f', 'g', 'hij', 'k']

Solution 4

Look at this answer for similar question: https://bugs.python.org/issue6663 Just drop the parenthesis if you are using findall:

import re
s = 'ab1cd2efg1hij2k'
re.findall( r'(?<=1)[a-z]+(?=2)|[a-z]', s )

Solution 5

If you want to have an 'or' match without having the split into match groups just add a '?:' to the beginning of the 'or' match.

Without '?:'

re.findall('(test (word1|word2))', 'test word1')

Output:
[('test word1', 'word1')]

With '?:'

re.findall('(test (?:word1|word2))', 'test word1')

Output:
['test word1']

Further explanation: https://www.ocpsoft.org/tutorials/regular-expressions/or-in-regex/

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Comments

[Blckknght]

Your pattern is pretty odd. You don't need the non-capturing groups around 1 and 2, or the group around the whole pattern (which you expend a bunch of effort to skip in the output). Instead, just accept that the findall call will return 2-tuples and turn them into single values with a = [x or y for x, y in a].

[A. Rabus]

"non-capturing group" is the keyword here... (added solely for the search engine)

[grantr]

saved me-thanks