Remove text between () and []

Question

I have a very long string of text with () and [] in it. I’m trying to remove the characters between the parentheses and brackets but I cannot figure out how.

The list is similar to this:

x = "This is a sentence. (once a day) [twice a day]"

This list isn’t what I’m working with but is very similar and a lot shorter.

Asked By: Tic

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Source

Answer 1

This should work for parentheses. Regular expressions will "consume" the text it has matched so it won’t work for nested parentheses.

import re
regex = re.compile(".*?((.*?))")
result = re.findall(regex, mystring)

or this would find one set of parentheses, simply loop to find more:

start = mystring.find("(")
end = mystring.find(")")
if start != -1 and end != -1:
  result = mystring[start+1:end]

Answered By: mbowden

Answer 2

Run this script, it works even with nested brackets.
Uses basic logical tests.

def a(test_str):
    ret = ''
    skip1c = 0
    skip2c = 0
    for i in test_str:
        if i == '[':
            skip1c += 1
        elif i == '(':
            skip2c += 1
        elif i == ']' and skip1c > 0:
            skip1c -= 1
        elif i == ')'and skip2c > 0:
            skip2c -= 1
        elif skip1c == 0 and skip2c == 0:
            ret += i
    return ret

x = "ewq[a [(b] ([c))]] This is a sentence. (once a day) [twice a day]"
x = a(x)
print x
print repr(x)

Just incase you don’t run it,
Here’s the output:

>>> 
ewq This is a sentence.  
'ewq This is a sentence.  '

Answered By: pradyunsg

Answer 3

You can use re.sub function.

>>> import re 
>>> x = "This is a sentence. (once a day) [twice a day]"
>>> re.sub("([([]).*?([)]])", "g<1>g<2>", x)
'This is a sentence. () []'

If you want to remove the [] and the () you can use this code:

>>> import re 
>>> x = "This is a sentence. (once a day) [twice a day]"
>>> re.sub("[([].*?[)]]", "", x)
'This is a sentence.  '

Important: This code will not work with nested symbols

Explanation

The first regex groups ( or [ into group 1 (by surrounding it with parentheses) and ) or ] into group 2, matching these groups and all characters that come in between them. After matching, the matched portion is substituted with groups 1 and 2, leaving the final string with nothing inside the brackets. The second regex is self explanatory from this -> match everything and substitute with the empty string.

— modified from comment by Ajay Thomas

Answered By: jvallver

Answer 4

Here’s a solution similar to @pradyunsg’s answer (it works with arbitrary nested brackets):

def remove_text_inside_brackets(text, brackets="()[]"):
    count = [0] * (len(brackets) // 2) # count open/close brackets
    saved_chars = []
    for character in text:
        for i, b in enumerate(brackets):
            if character == b: # found bracket
                kind, is_close = divmod(i, 2)
                count[kind] += (-1)**is_close # `+1`: open, `-1`: close
                if count[kind] < 0: # unbalanced bracket
                    count[kind] = 0  # keep it
                else:  # found bracket to remove
                    break
        else: # character is not a [balanced] bracket
            if not any(count): # outside brackets
                saved_chars.append(character)
    return ''.join(saved_chars)

print(repr(remove_text_inside_brackets(
    "This is a sentence. (once a day) [twice a day]")))
# -> 'This is a sentence.  '

Answered By: jfs

Answer 5

You can split, filter, and join the string again. If your brackets are well defined the following code should do.

import re
x = "".join(re.split("(|)|[|]", x)[::2])

Answered By: user3592579

Answer 6

You can try this. Can remove the bracket and the content exist inside it.

 import re
    x = "This is a sentence. (once a day) [twice a day]"
    x = re.sub("(.*?)|[.*?]","",x)
    print(x)

Expected ouput :

This is a sentence.

Answered By: Avinash Raut

Answer 7

The RegEx (.*?)|[.*?] removes bracket content by finding pairs, first it remove paranthesis and then square brackets. I also works fine for the nested brackets as it acts in sequence. Ofcourse, it would break in case of bad brackets scenario.


    _brackets = re.compile("(.*?)|[.*?]")
    _spaces = re.compile("s+")
    
    _b = _brackets.sub(" ", "microRNAs (miR) play a role in cancer ([1], [2])")
    _s = _spaces.sub(" ", _b.strip())
    print(_s)
    
    # OUTPUT: microRNAs play a role in cancer

Answered By: miRastic

Answer 8

For anyone who appreciates the simplicity of the accepted answer by jvallver, and is looking for more readability from their code:

>>> import re
>>> x = 'This is a sentence. (once a day) [twice a day]'
>>> opening_braces = '(['
>>> closing_braces = ')]'
>>> non_greedy_wildcard = '.*?'
>>> re.sub(f'[{opening_braces}]{non_greedy_wildcard}[{closing_braces}]', '', x)
'This is a sentence.  '

Most of the explanation for why this regex works is included in the code. Your future self will thank you for the 3 additional lines.

(Replace the f-string with the equivalent string concatenation for Python2 compatibility)

Answered By: Dave Trost

Answer 9

"Place the cursor at the beginning of the parenthesis". then

dib OR di(

Answered By: islamux

Remove text between () and []

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