strs = [
"I like to run head first into a wall (not)"
"klsjlsk klsjdkls ,m s,mdn,mnsd,m (123)"
"a b c d e f g h i (j?)"
]
I want to remove the " (not), (123), (j?)"
why is
re.sub(r' (\([*]\))$', '', strs(0))
not doing it and what's the proper way?
You're using a capture group (the outside parentheses) to capture everything inside the parentheses. However, since you're not reusing that captured data, that's unnecessary. If you just want to remove all parentheses and the enclosed text at the end of a line (my guess based on what you supplied), you can do
re.sub(r'\([^\)]*\)$', '', strs[0])
Example: https://regex101.com/r/FOTTfi/1
If it's important to remove the space before the parentheses as well, just use a \s or \s+ at the start.
Yours doesn't work because [*] isn't doing what you think. It's looking for the literal *. If you want to find any number of characters, use .* instead.
Consider the following:
import re
strs = [
"I like to run head first into a wall (not)",
"klsjlsk klsjdkls ,m s,mdn,mnsd,m (123)",
"a b c d e f g h i (j?)"
]
# space(\s)
# openbracket(\()
# anychar(.*)
# smallestpossiblematch(?)
# closebracket(\))
pattern = r'\s*\(.*?\)'
# list comprehension with new strings
new = [re.sub(pattern, '', strs[i]) for i in range(len(strs))]
This is the regex you need to use print re.sub(r'\([^)]*\)', '', strs[0])
\( followed by anything that is not a bracket [^)] repeated 0 or more time * then followed by a bracket \)
re.sub(r'((\.+)\)$', '', strs[0])by the way, you have (0) in your code.