Group anagrams in Python

Two words are anagrams when they use exactly the same letters, like "listen" and "silent". If you sort the letters of each word, all anagrams give the same result, so the sorted letters make a good dictionary key for each group.

The program

from collections import defaultdict

words = ["listen", "silent", "enlist", "google", "tea", "eat", "Ate", "bat", "tab", "python"]

groups = defaultdict(list)
for word in words:
    key = "".join(sorted(word.lower()))
    groups[key].append(word)

for key, group in groups.items():
    print(f"{key:8} {group}")

eilnst   ['listen', 'silent', 'enlist']
eggloo   ['google']
aet      ['tea', 'eat', 'Ate']
abt      ['bat', 'tab']
hnopty   ['python']

How it works

• sorted(word.lower()) gives a list of the letters in alphabetical order: sorted("tea") is ['a', 'e', 't']. "".join(...) turns that list back into the string "aet".
• word.lower() comes first, so "Ate" goes into the same group as "tea" and "eat".
• groups is a defaultdict(list). When a key is used for the first time, it starts with an empty list, so groups[key].append(word) works without checking whether the key exists.
• A dictionary keeps keys in the order they were first added, so the groups print in the order their first word appeared.

Only the real groups

Words with no anagram partner, like "google" and "python", form groups of one. To keep only the groups with two or more words:

from collections import defaultdict

words = ["listen", "silent", "enlist", "google", "tea", "eat", "Ate", "bat", "tab", "python"]

groups = defaultdict(list)
for word in words:
    groups["".join(sorted(word.lower()))].append(word)

anagram_groups = [group for group in groups.values() if len(group) > 1]
print(anagram_groups)

[['listen', 'silent', 'enlist'], ['tea', 'eat', 'Ate'], ['bat', 'tab']]