Great question! Let's fix this up to efficiently group anagrams with Pandas, ditching those slow row-by-row loops. First, let's break down the issues with your current code:

• Using iterrows() is inefficient (especially for larger datasets) — Pandas is built for vectorized operations, not manual row iteration.
• Grouping by the original word column won't group anagrams together, since "acb" and "bca" are treated as distinct strings.
• The split column approach doesn't create a consistent identifier for matching anagrams.

The Core Idea: Create a Unique Anagram Key #

For any set of anagrams, sorting the letters of the word will produce the same string (e.g., "acb" → sorted to "abc", "bca" → also "abc"). This sorted string acts as a reliable grouping key that all anagrams share.

Here's the optimized, loop-free code:

import pandas as pd

# Step 1: Read your word list correctly
# Adjust sep if your file uses a different delimiter (use '\s+' for spaces, or omit sep if one word per line)
wordlist = pd.read_csv('data/example.txt', sep='\s+', header=None, names=['word'])

# Step 2: Remove duplicates (you had this part right!)
wordlist = wordlist.drop_duplicates(keep='first')

# Step 3: Generate a consistent anagram key (sorted letters joined into a string)
wordlist['anagram_key'] = wordlist['word'].apply(lambda x: ''.join(sorted(x)))

# Step 4: Group by the key and collect all anagrams in each group
anagram_groups = wordlist.groupby('anagram_key')['word'].apply(list).reset_index(name='anagrams')

print(anagram_groups)

What This Does: #

• Reading the file: sep='\s+' handles spaces between quoted words in your sample. If your file has one word per line, just remove the sep parameter entirely.
• Anagram key: The lambda function sorts each word's characters and joins them back into a string. This creates a unique, shared identifier for every anagram set.
• Grouping: groupby('anagram_key') clusters all anagrams together, and apply(list) collects them into a clean list for each group.

Example Output for Your Sample Data: #

anagram_key      anagrams
0         abc      [acb, bca]
1         foo      [foo, oof]
2     aeilnps      [spaniel]

Alternative: Using Frozensets (Less Intuitive but Valid) #

If you prefer using sets instead of sorted strings, you can use a frozenset (regular sets aren't hashable and can't be used as group keys):

wordlist['anagram_key'] = wordlist['word'].apply(lambda x: frozenset(x))

That said, sorted strings are more readable and easier to debug, so they're the better default choice.

Why This Beats Loops: #

• Vectorized operations like apply run on optimized C-level code under the hood, making them drastically faster than iterrows() for large datasets.
• The logic is clean, maintainable, and avoids manual row-by-row processing.

内容的提问来源于stack exchange，提问作者Davtho1983