CODEWITHSUNDEEP
pythonregexparentheses
Cleb
Cleb

Reputation: 25997

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

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)?

Upvotes: 9

Views: 14730

Answers (7)

root
root

Reputation: 2428

No Python loops. No recursion. No regexes.

Just computing the difference between the cumulative counts of '(' and ')':

import numpy as np
s = '()a(x(x)x)b(x)c()d()'
s_array = np.array(list(s))
mask_open = s_array=='('
mask_close = s_array==')'
# Compute in how many parentheses each character is nested,
# while considering ')' as not nested:
nestedness_except_close = np.cumsum(mask_open) - np.cumsum(mask_close)
# ... and while considering ')' as nested:
nestedness = nestedness_except_close + mask_close
# Select only characters that aren't in any parentheses
result = ''.join(s_array[nestedness < 1])

This might be faster than other solutions.

Optional validity checks for the string:

# Check whether the number of `'('`s and `')'`s is the same
assert(nestedness_except_close[-1] == 0)
# Check whether some parentheses get closed before they got opened
assert((nestedness_except_close >= 0).all())

If you don't want to use NumPy, you can use itertools.accumulate() to compute the cumulative sums.

Upvotes: 0

ozdemir
ozdemir

Reputation: 389

I have found a solution here:

http://rachbelaid.com/recursive-regular-experession/

which says:

>>> import regex
>>> regex.search(r"^(\((?1)*\))(?1)*$", "()()") is not None
True
>>> regex.search(r"^(\((?1)*\))(?1)*$", "(((()))())") is not None
True
>>> regex.search(r"^(\((?1)*\))(?1)*$", "()(") is not None
False
>>> regex.search(r"^(\((?1)*\))(?1)*$", "(((())())") is not None
False

Upvotes: 0

Wiktor Stribiżew
Wiktor Stribiżew

Reputation: 626748

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

Upvotes: 20

bobble bubble
bobble bubble

Reputation: 18490

As mentioned before, you'd need a recursive regex for matching arbitrary levels of nesting but if you know there can only be a maximum of one level of nesting have a try with this pattern:

\((?:[^)(]|\([^)(]*\))*\)
  • [^)(] matches a character, that is not a parenthesis (negated class).
  • |\([^)(]*\) or it matches another ( ) pair with any amount of non )( inside.
  • (?:...)* all this any amount of times inside ( )

Here is a demo at regex101

Before the alternation used [^)(] without + quantifier to fail faster if unbalanced.
You need to add more levels of nesting that might occure. Eg for max 2 levels:

\((?:[^)(]|\((?:[^)(]|\([^)(]*\))*\))*\)

Another demo at regex101

Upvotes: 6

Bryce Drew
Bryce Drew

Reputation: 6729

https://regex101.com/r/kQ2jS3/1

'(\(.*\))'

This captures the furthest parentheses, and everything in between the parentheses.

Your old regex captures the first parentheses, and everything between to the next parentheses.

Upvotes: 1

Serge Ballesta
Serge Ballesta

Reputation: 148890

If you are sure that the parentheses are initially balanced, just use the greedy version:

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

Upvotes: 1

alexamici
alexamici

Reputation: 784

re matches are eager so they try to match as much text as possible, for the simple test case you mention just let the regex run:

>>> re.sub(r'\(.*\)', '', 'stuff(remove(me))')
'stuff'

Upvotes: 1

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