Source: Python Morsels

Notes

Exercise is to detect an anagram, with bonuses for ignoring spaces, punctuation, and diacritical marks on Latin characters (but not the characters themselves).

My solution

Using isalnum() to retain alphanumeric characters and discard everything else, and unicodedata.normalize() to remove diacritical marks.

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import unicodedata

def fmt(s):
    s = unicodedata.normalize('NFKD', s)\
       .encode('ascii', 'ignore')\
       .decode('utf-8')
    return ''.join(sorted(c.lower() for c in s if c.isalnum()))


def is_anagram(s1, s2):
    return False if len(s1) != len(s1) else fmt(s1) == fmt(s2)

Provided solutions

Were mostly identical to mine, except for a useful variation using Counter to tabulate characters and use the tabulation as a basis for comparison.

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from collections import Counter

def is_anagram(word1, word2):
    return Counter(word1) == Counter(word2)