How to remove all text between the outer parentheses in a string?

Question

When I have a string like this:

s1 = 'stuff(remove_me)'

I can easily remove the parentheses and the text within using

# returns 'stuff'
res1 = re.sub(r'\([^)]*\)', '', s1)

as explained here.

But I sometimes encounter nested expressions like this:

s2 = 'stuff(remove(me))'

When I run the command from above, I end up with

'stuff)'

I also tried:

re.sub('\(.*?\)', '', s2)

which gives me the same output.

How can I remove everything within the outer parentheses - including the parentheses themselves - so that I also end up with 'stuff' (which should work for arbitrarily complex expressions)?

Answer 1

NOTE: \(.*\) matches the first ( from the left, then matches any 0+ characters (other than a newline if a DOTALL modifier is not enabled) up to the last ), and does not account for properly nested parentheses.

To remove nested parentheses correctly with a regular expression in Python, you may use a simple \([^()]*\) (matching a (, then 0+ chars other than ( and ) and then a )) in a while block using re.subn:

def remove_text_between_parens(text):
    n = 1  # run at least once
    while n:
        text, n = re.subn(r'\([^()]*\)', '', text)  # remove non-nested/flat balanced parts
    return text

Bascially: remove the (...) with no ( and ) inside until no match is found. Usage:

print(remove_text_between_parens('stuff (inside (nested) brackets) (and (some(are)) here) here'))
# => stuff   here

A non-regex way is also possible:

def removeNestedParentheses(s):
    ret = ''
    skip = 0
    for i in s:
        if i == '(':
            skip += 1
        elif i == ')'and skip > 0:
            skip -= 1
        elif skip == 0:
            ret += i
    return ret

x = removeNestedParentheses('stuff (inside (nested) brackets) (and (some(are)) here) here')
print(x)              
# => 'stuff   here'

See another Python demo