How can i remove only the last bracket from a string in python?

Question

How can i remove only the last bracket from a string ?

For example, INPUT 1:

"hell(h)o(world)" 

i want this result:

"hell(h)o"

Input 2 :-

hel(lo(wor)ld)

i want :-

hel

as you can see the middle brackets remain intact only the last bracket got removed.

I tried :-

import re
string = 'hell(h)o(world)' 
print(re.sub('[()]', '', string))

output :-

hellhoworld

i figured out a solution :-

i did it like this

string = 'hell(h)o(world)' 
if (string[-1] == ")"):
    add=int(string.rfind('(', 0))
    print(string[:add])

output :-

hell(h)o

looking for other optimised solutions/suggestions..

Answer 1

Please see the below if this is useful, Let me know I will optimize further.

string = 'hell(h)o(world)'
count=0
r=''
for i in reversed(string):
    if count <2 and (i == ')' or i=='('):
        count+=1
        pass
    else:
        r+=i
for i in reversed(r):
    print(i, end='')

Answer 2

If you want to remove the last bracket from the string even if it's not at the end of the string, you can try something like this. This will only work if you know you have a substring beginning and ending with parentheses somewhere in the string, so you may want to implement some sort of check for that. You will also need to modify if you are dealing with nested parenthesis.

str = "hell(h)o(world)"
r_str = str[::-1]    # creates reverse copy of string
for i in range(len(str)):
    if r_str[i] == ")":
        start = i
    elif r_str[i] == "(":
        end = i+1
        break
x = r_str[start:end][::-1]    # substring that we want to remove
str = str.replace(x,'')
print(str)

output:

hell(h)o

If the string is not at the end:

str = "hell(h)o(world)blahblahblah"

output:

hell(h)oblahblahblah

Edit: Here is a modified version to detect nested parenthesis. However, please keep in mind that this will not work if there are unbalanced parenthesis in the string.

str = "hell(h)o(w(orld))"
r_str = str[::-1]
p_count = 0
for i in range(len(str)):
    if r_str[i] == ")":
        if p_count == 0:
            start = i
        p_count = p_count+1
    elif r_str[i] == "(":
        if p_count == 1:
            end = i+1
            break
        else:
            p_count = p_count - 1
x = r_str[start:end][::-1]
print("x:", x)
str = str.replace(x,'')
print(str)

output:

hell(h)o

Answer 3

Something like this?

string = 'hell(h)o(w(orl)d)23'
new_str = ''
escaped = 0
for char in reversed(string):
    if escaped is not None and char == ')':
        escaped += 1

    if not escaped:
        new_str = char + new_str

    if escaped is not None and char == '(':
        escaped -= 1
        if escaped == 0:
            escaped = None

print(new_str)

This starts escaping when a ) and stops when it's current level is closed with (. So a nested () would not effect it.

Answer 4

Using re.sub('[()]', '', string) will replace any parenthesis in the string with an empty string.

To match the last set of balanced parenthesis, and if you can make use of the regex PyPi module, you can use a recursive pattern repeating the first sub group, and assert that to the right there are no more occurrences of either ( or )

(\((?:[^()\n]++|(?1))*\))(?=[^()\n]*$)

The pattern matches:

• ( Capture group 1
  • \( Match (
  • (?:[^()\n]++|(?1))* Repeat 0+ times matching either any char except ( ) or a newline. If you do, recurse group 1 using (?1)
  • \) Match )
• ) Close group 1
• (?=[^()\n]*$) Positive lookahead, assert till the end of the string no ( or ) or newline

See a regex demo and a Python demo.

For example

import regex

strings = [
    "hell(h)o(world)",
    "hel(lo(wor)ld)",
    "hell(h)o(world)blahblahblah"
]

pattern = r"(\((?:[^()]++|(?1))*\))(?=[^()]*$)"

for s in strings:
    print(regex.sub(pattern, "", s))

Output

hell(h)o
hel
hell(h)oblahblahblah