Python : How to compare strings and ignore white space and special characters

Question

I want to compare two strings such that the comparison should ignore differences in the special characters. That is,

Hai, this is a test

Should match with

Hai ! this is a test "or" Hai this is a test

Is there any way to do this without modifying the original strings?

Answer 1

This removes punctuation and whitespace before doing the comparison:

In [32]: import string

In [33]: def compare(s1, s2):
    ...:     remove = string.punctuation + string.whitespace
    ...:     return s1.translate(None, remove) == s2.translate(None, remove)

In [34]: compare('Hai, this is a test', 'Hai ! this is a test')
Out[34]: True

Answer 2

Generally, you'd replace the characters you wish to ignore, and then compare them:

import re
def equal(a, b):
    # Ignore non-space and non-word characters
    regex = re.compile(r'[^\s\w]')
    return regex.sub('', a) == regex.sub('', b)

>>> equal('Hai, this is a test', 'Hai this is a test')
True
>>> equal('Hai, this is a test', 'Hai this@#)($! i@#($()@#s a test!!!')
True