Group items with defaultdict¶

The question. You have a flat sequence and you want it bucketed into a {key: [items]} mapping — records by category, words by first letter, files by extension, students by grade.

The answer is defaultdict(list): append each item to its bucket and let missing keys create themselves. Below are the variations — lists, sets, computed keys, an index, and nesting — plus when a different tool fits.

Group into lists¶

The core pattern. Pick the grouping key for each item and append.

In [ ]:

from collections import defaultdict

people = [
    ('engineering', 'Ada'), ('sales', 'Bo'),
    ('engineering', 'Cleo'), ('sales', 'Dev'), ('engineering', 'Eve'),
]

by_team = defaultdict(list)
for team, name in people:
    by_team[team].append(name)

print(dict(by_team))
# {'engineering': ['Ada', 'Cleo', 'Eve'], 'sales': ['Bo', 'Dev']}

from collections import defaultdict people = [ ('engineering', 'Ada'), ('sales', 'Bo'), ('engineering', 'Cleo'), ('sales', 'Dev'), ('engineering', 'Eve'), ] by_team = defaultdict(list) for team, name in people: by_team[team].append(name) print(dict(by_team)) # {'engineering': ['Ada', 'Cleo', 'Eve'], 'sales': ['Bo', 'Dev']}

Group by a computed key¶

The key can be anything derived from the item — its length, a category, the result of a function.

In [ ]:

from collections import defaultdict

words = ['hi', 'cat', 'dog', 'a', 'tree', 'ok', 'sun']

by_length = defaultdict(list)
for word in words:
    by_length[len(word)].append(word)

print(dict(by_length))        # {2: ['hi', 'ok'], 3: ['cat', 'dog', 'sun'], 1: ['a'], 4: ['tree']}

from collections import defaultdict words = ['hi', 'cat', 'dog', 'a', 'tree', 'ok', 'sun'] by_length = defaultdict(list) for word in words: by_length[len(word)].append(word) print(dict(by_length)) # {2: ['hi', 'ok'], 3: ['cat', 'dog', 'sun'], 1: ['a'], 4: ['tree']}

Group into sets to drop duplicates¶

Swap the factory to set when each bucket should hold unique members.

In [ ]:

from collections import defaultdict

pairs = [('a', 1), ('a', 2), ('a', 1), ('b', 3)]

unique = defaultdict(set)
for key, value in pairs:
    unique[key].add(value)

print(dict(unique))           # {'a': {1, 2}, 'b': {3}}

from collections import defaultdict pairs = [('a', 1), ('a', 2), ('a', 1), ('b', 3)] unique = defaultdict(set) for key, value in pairs: unique[key].add(value) print(dict(unique)) # {'a': {1, 2}, 'b': {3}}

Build an index (item to positions)¶

A grouping where the value accumulates where each item appeared — the basis of a search index.

In [ ]:

from collections import defaultdict

text = 'the cat sat on the mat'.split()
index = defaultdict(list)
for position, word in enumerate(text):
    index[word].append(position)

print(dict(index))            # {'the': [0, 4], 'cat': [1], 'sat': [2], 'on': [3], 'mat': [5]}

from collections import defaultdict text = 'the cat sat on the mat'.split() index = defaultdict(list) for position, word in enumerate(text): index[word].append(position) print(dict(index)) # {'the': [0, 4], 'cat': [1], 'sat': [2], 'on': [3], 'mat': [5]}

Nested grouping¶

A lambda factory that returns another defaultdict builds two-level groups — here, names grouped by team and then by role.

In [ ]:

from collections import defaultdict

records = [
    ('eng', 'senior', 'Ada'), ('eng', 'junior', 'Bo'),
    ('eng', 'senior', 'Cleo'), ('sales', 'junior', 'Dev'),
]

grouped = defaultdict(lambda: defaultdict(list))
for team, role, name in records:
    grouped[team][role].append(name)

print(grouped['eng']['senior'])   # ['Ada', 'Cleo']

from collections import defaultdict records = [ ('eng', 'senior', 'Ada'), ('eng', 'junior', 'Bo'), ('eng', 'senior', 'Cleo'), ('sales', 'junior', 'Dev'), ] grouped = defaultdict(lambda: defaultdict(list)) for team, role, name in records: grouped[team][role].append(name) print(grouped['eng']['senior']) # ['Ada', 'Cleo']

When another tool fits better¶

• itertools.groupby groups consecutive equal keys, so it only matches defaultdict grouping if the data is already sorted by the key. For unsorted data, defaultdict is simpler and doesn't need a sort.
• dict.setdefault does the same job for a one-off without leaving a factory on the dict: groups.setdefault(key, []).append(item).
• For grouping and counting rather than collecting, reach for Counter (recipe).

In short¶

• defaultdict(list) + append is the grouping workhorse; the key can be any value you compute.
• Use set to dedupe within buckets, a nested lambda factory for multi-level groups.
• itertools.groupby needs sorted input; setdefault is the plain-dict one-off.