Extracting the digit between parentheses in Python-CodePudding

import re    
print(re.search('\(\d{1,}\)', "b'   1. Population, 2016 (1)'"))

I am trying to extract the digits (one or more) between parentheses in strings. The above codes show my attempted solution. I checked my regular expression on https://regex101.com/ and expected the codes to return True. However, the returned value is None. Can someone let me know what happened?

CodePudding user response：

Your current regex pattern is only valid if you make it a raw string:

inp = "b'   1. Population, 2016 (1)'"
nums = re.findall(r'\((\d{1,})\)', inp)
print(nums)  # ['1']

Otherwise, you would have to double escape the \\d in the pattern.

CodePudding user response：

Below RE will help you grab the digits inside the brackets only when the digits are present.

r"\((?P<digits_inside_brackets>\d )\)

For your scenario, the above RE will match 1 under the group "digits_inside_brackets". It can be executed through below snippet

import re
user_string = "b'   1. Population, 2016 (1)"
comp = re.compile(r"\((?P<digits_inside_brackets>\d )\)") # Captures when digits are the only 
for i in re.finditer(comp, user_string):
    print(i.group("digits_inside_brackets"))

Output for the above snippet

Grab digits even when white space are provided:

r"\(\s*(?P<digits_inside_brackets>\d )\s*\)

Grab digits inside brackets at any condition:

r"\(\D*(?P<digits_inside_brackets>\d )\D*\)

Output when applied with above RE