Python – Why does re.findall return a list of tuples when the pattern only contains one group

findallpythonregex

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.

Best Solution

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']

Related Solutions

Regex – Greedy vs. Reluctant vs. Possessive Qualifiers

I'll give it a shot.

A greedy quantifier first matches as much as possible. So the .* matches the entire string. Then the matcher tries to match the f following, but there are no characters left. So it "backtracks", making the greedy quantifier match one less character (leaving the "o" at the end of the string unmatched). That still doesn't match the f in the regex, so it backtracks one more step, making the greedy quantifier match one less character again (leaving the "oo" at the end of the string unmatched). That still doesn't match the f in the regex, so it backtracks one more step (leaving the "foo" at the end of the string unmatched). Now, the matcher finally matches the f in the regex, and the o and the next o are matched too. Success!

A reluctant or "non-greedy" quantifier first matches as little as possible. So the .* matches nothing at first, leaving the entire string unmatched. Then the matcher tries to match the f following, but the unmatched portion of the string starts with "x" so that doesn't work. So the matcher backtracks, making the non-greedy quantifier match one more character (now it matches the "x", leaving "fooxxxxxxfoo" unmatched). Then it tries to match the f, which succeeds, and the o and the next o in the regex match too. Success!

In your example, it then starts the process over with the remaining unmatched portion of the string, "xxxxxxfoo", following the same process.

A possessive quantifier is just like the greedy quantifier, but it doesn't backtrack. So it starts out with .* matching the entire string, leaving nothing unmatched. Then there is nothing left for it to match with the f in the regex. Since the possessive quantifier doesn't backtrack, the match fails there.

Python – Split string based on a regular expression

By using (,), you are capturing the group, if you simply remove them you will not have this problem.

>>> str1 = "a    b     c      d"
>>> re.split(" +", str1)
['a', 'b', 'c', 'd']

However there is no need for regex, str.split without any delimiter specified will split this by whitespace for you. This would be the best way in this case.

>>> str1.split()
['a', 'b', 'c', 'd']

If you really wanted regex you can use this ('\s' represents whitespace and it's clearer):

>>> re.split("\s+", str1)
['a', 'b', 'c', 'd']

or you can find all non-whitespace characters

>>> re.findall(r'\S+',str1)
['a', 'b', 'c', 'd']

Best Solution

Related Solutions

Regex – Greedy vs. Reluctant vs. Possessive Qualifiers

Python – Split string based on a regular expression

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