I'm trying to convert string to a list
str = "ab(1234)bcta(45am)in23i(ab78lk)"
Expected Output
res_str = ["ab","bcta","in23i"]
I tried removing brackets from str.
re.sub(r'\([^)]*\)', '', str)
5 Answers
You may use a negated character class with a lookahead:
>>> s = "ab(1234)bcta(45am)in23i(ab78lk)"
>>> print (re.findall(r'[^()]+(?=\()', s))
['ab', 'bcta', 'in23i']
RegEx Details:
[^()]+: Match 1 of more of any character that is not(and)(?=\(): Lookahead to assert that there is a(ahead
Comments
So many options here. One possibility would be using split:
import re
str = "ab(1234)bcta(45am)in23i(ab78lk)"
print(re.split(r'\(.*?\)', str)[:-1])
Returns:
['ab', 'bcta', 'in23i']
A second option would be to split by all paranthesis and slice your resulting array:
import re
str = "ab(1234)bcta(45am)in23i(ab78lk)"
print(re.split('[()]', str)[0:-1:2])
Where [0:-1:2] means to start at index 0, to stop at second to last index, and step two indices.
Comments
Use re.split
import re
str = "ab(1234)bcta(45am)in23i(ab78lk)"
print(re.split('\(.*?\)', str))
Returns:
['ab', 'bcta', 'in23i', '']
If you want to get rid of empty strings in your list, you may use a filter:
print(list(filter(None, re.split('\(.*?\)', str))))
Returns:
['ab', 'bcta', 'in23i']
Comments
You may match all alphanumeric characters followed by a ( :
>>> re.findall('\w+(?=\()',str)
['ab', 'bcta', 'in23i']
or using re.sub as you were:
>>> re.sub('\([^)]+\)',' ',str).split()
['ab', 'bcta', 'in23i']
Comments
Just for the sake of complexity :
>>>> str = "ab(1234)bcta(45am)in23i(ab78lk)"
>>>> res_str = [y[-1] for y in [ x.split(')') for x in str.split('(')]][0:-1]
['ab', 'bcta', 'in23i']
Comments
