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?

Solution 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

Solution 2

>>> def cmp(a, b):
...     return [c for c in a if c.isalpha()] == [c for c in b if c.isalpha()]
... 
>>> cmp('Hai, this is a test', 'Hai ! this is a test')
True
>>> cmp('Hai, this is a test', 'Hai this is a test')
True
>>> cmp('Hai, this is a test', 'other string')
False

This creates two temporary lists, but doesn't modify the original strings in any way.

Solution 3

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

Solution 4

To compare an arbitrary number of strings for alphabetic equivalence,

def samealphabetic(*args):
    return len(set(filter(lambda s: s.isalpha(), arg) for arg in args)) <= 1
print samealphabetic('Hai, this is a test',
                     'Hai ! this is a test',
                     'Hai this is a test')

Which prints True. Should change <= depending on what you want to return for no arguments.

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Comments

[Dan Lecocq]

What are 'special characters' in this context? Importantly, are spaces counted? Because then ignoring the '!' leaves two spaces as opposed to the one in the first string.

[DSM]

Instead of c in string.letters, you can use c.isalpha().

[Adam Dobrawy]

It is not Python3 compatible.

[Martijn Pieters]

@AdamDobrawy: Use str.maketrans(dict.fromkeys(remove)) as the map (first argument).