re

Regular Expression

import re
regex = r"([a-zA-Z]+) (\d+)"
match = re.search(regex, "I was born on June 24") 

Splitting String to List

Re.split([reExpression], [string]) OR Re.findall([reExpression], [string])

• '\s' #single white space

• '\s+' #split on one or more whitespace

• '\S+' #one or more non whitespace character

• '\w+') #one or more words

• '\W+' #any non word character

Replace

regex = r"[0-9+\. *]"
re.sub(regex, "", '1. Interdimensional \n2. Aliens')

'\\' as the pattern string, because the regular expression must be \, and each backslash must be expressed as \ inside a regular Python string literal.

The solution is to use Python’s raw string notation for regular expression patterns; backslashes are not handled in any special way in a string literal prefixed with 'r'