Here is my code:
import re
string = r"('Option A' | 'Option B') & ('Option C' | 'Option D')"
word_list = re.split(r"[\(.\)]", string)
-> ['', "'Option A' | 'Option B'", ' & ', "'Option C' | 'Option D'", '']
I want the following result:
-> ["('Option A' | 'Option B')", ' & ', "('Option C' | 'Option D')"]
You can use re.findall to capture each parenthesis group:
import re
string = r"('Option A' | 'Option B') & ('Option C' | 'Option D')"
pattern = r"(\([^\)]+\))"
re.findall(pattern, string)
# ["('Option A' | 'Option B')", "('Option C' | 'Option D')"]
This also works with re.split
re.split(pattern, string)
# ['', "('Option A' | 'Option B')", ' & ', "('Option C' | 'Option D')", '']
If you want to remove empty elements from using re.split you can:
[s for s in re.split(pattern, string) if s]
# ["('Option A' | 'Option B')", ' & ', "('Option C' | 'Option D')"]
How the pattern works:
( begin capture group\( matches the character ( literally[^\)]+ Match between one and unlimited characters that are not )\) matches the character ) literally) end capture group
word_list.remove('') next to re.split(pattern, string) eliminated empty elements. Please update it in your answer as well.