Regular expression to return string split up respecting nested parentheses

Using regex only for the task might work but it wouldn't be straightforward.

Another possibility is writing a simple algorithm to track the parentheses in the string:

1. Split the string at all parentheses, while returning the delimiter (e.g. using re.split)
2. Keep a counters tracking the parentheses: start_parens_count for ( and end_parens_count for ).
3. Using the counters, proceed by either splitting at white spaces or adding the current data into a temp var ( term)
4. When the left most parenthesis has been closed, append term to the list of values & reset the counters/temp vars.

Here's an example:

import re

string = "1 2 3 (test 0, test 0) (test (0 test) 0)"


result, start_parens_count, end_parens_count, term = [], 0, 0, ""
for x in re.split(r"([()])", string):
    if not x.strip():
        continue
    elif x == "(":
        if start_parens_count > 0:
            term += "("
        start_parens_count += 1
    elif x == ")":
        end_parens_count += 1
        if end_parens_count == start_parens_count:
            result.append(term)
            end_parens_count, start_parens_count, term = 0, 0, ""
        else:
            term += ")"
    elif start_parens_count > end_parens_count:
        term += x
    else:
        result.extend(x.strip(" ").split(" "))


print(result)
# ['1', '2', '3', 'test 0, test 0', 'test (0 test) 0']

Not very elegant, but works.

You can use pip install regex and use

import regex
text = "1 2 3 (test 0, test 0) (test (0 test) 0)"
matches = [match.group() for match in regex.finditer(r"(?:(\((?>[^()]+|(?1))*\))|\S)+", text)]
print(matches)
# => ['1', '2', '3', '(test 0, test 0)', '(test (0 test) 0)']

See the online Python demo. See the regex demo. The regex matches:

• (?: - start of a non-capturing group:
  • (\((?>[^()]+|(?1))*\)) - a text between any nested parentheses
• | - or
  • \S - any non-whitespace char
• )+ - end of the group, repeat one or more times

Alternatively, you can use pyparsing as well.

import pyparsing as pp

pattern = pp.ZeroOrMore(pp.Regex(r'\S+') ^ pp.original_text_for(pp.nested_expr('(', ')')))

# Tests
string = '1 2 3 (test 0, test 0) (test (0 test) 0)'
result = pattern.parse_string(string).as_list()
answer = ['1', '2', '3', '(test 0, test 0)', '(test (0 test) 0)']
assert result == answer

string = ''
result = pattern.parse_string(string).as_list()
answer = []
assert result == answer

string = 'a'
result = pattern.parse_string(string).as_list()
answer = ['a']
assert result == answer

string = ' a (1) ! '
result = pattern.parse_string(string).as_list()
answer = ['a', '(1)', '!']
assert result == answer

string = ' a (b) cd (e f) g hi (j (k l) m) (o p (qr (s t) u v) w (x y) z)'
result = pattern.parse_string(string).as_list()
answer = ['a', '(b)', 'cd', '(e f)', 'g', 'hi', '(j (k l) m)', '(o p (qr (s t) u v) w (x y) z)']
assert result == answer

* pyparsing can be installed by pip install pyparsing

In addition, you can directly parse all the nested parentheses at once:

pattern = pp.ZeroOrMore(pp.Regex(r'\S+') ^ pp.nested_expr('(', ')'))

string = '1 2 3 (test 0, test 0) (test (0 test) 0)'
result = pattern.parse_string(string).as_list()
answer = ['1', '2', '3', ['test', '0,', 'test', '0'], ['test', ['0', 'test'], '0']]
assert result == answer

* Whitespace is a delimiter in this case.

Note:

If a pair of parentheses gets broken inside () (for example a(b(c), a(b)c), etc), an unexpected result is obtained or IndexError is raised. So be careful to use. (See: Python extract string in a phrase)