Question

How do you make a regex that match when the beginning and the end is of the same length?
For example

>>> String = '[[A], [[B]], [C], [[D]]]'
>>> Result = re.findall(pattern, String)
>>> Result
>>> [ '[A]', '[[B]]', '[C]', '[[D]]' ]

Currently I use the pattern \[.*?\] but it resulted in

>>> ['[[A]', '[[B]', '[C]', '[[D]']

Thanks in advance.

Answer 1

You can define such a regular expression for a finite number of beginning/end characters (ie, something like “if it starts and ends with 1, or starts and ends with 2, or etc”). You, however, cannot do this for an unlimited number of characters. This is simply a fact of regular expressions. Regular expressions are the language of finite-state machines, and finite-state machines cannot do counting; at least the power of a pushdown-automaton (context-free grammar) is needed for that.

Put simply, a regular expression can say: “I saw x and then I saw y” but it cannot say “I saw x and then I saw y the same number of times” because it cannot remember how many times it saw x.

However, you can easily do this using the full power of the Python programming language, which is Turing-complete! Turing-complete languages can definitely do counting:

>>> string = '[[A], [[B]], [C], [[D]]]'
>>> sameBrackets = lambda s: len(re.findall('\[',s)) == len(re.findall('\]',s))
>>> filter(sameBrackets, string.split(", "))
['[[B]]', '[C]']