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itertools.groupby

itertools

Groups consecutive equal items. The data must be sorted by the key first!

Syntax

from itertools import groupby
groupby(iterable, key=None)

Beyond the classic "forgot to sort" mistake, a group stays valid only until you move on to the next one. All groups share the same underlying iterator, so list(groupby(data)) hands you empty groups: by the time the loop ends, the earlier ones have been consumed. If you need everything at once, call list(group) inside the loop body, before advancing.

Examples

from itertools import groupby
data = ['A', 'A', 'B', 'B', 'B', 'A']
for key, group in groupby(data):
    print(key, list(group))  # → A ['A', 'A'] / B ['B', 'B', 'B'] / A ['A']

# Группирует только соседей — для полной группировки сортируют заранее
from itertools import groupby
words = ['apple', 'apricot', 'banana', 'blueberry', 'cherry']
for letter, group in groupby(words, key=lambda w: w[0]):
    print(letter, list(group))  # → a ['apple', 'apricot'] / b ['banana', 'blueberry'] / c ['cherry']

from itertools import groupby
data = sorted([1, 1, 2, 2, 2, 3])
print({k: list(v) for k, v in groupby(data)})  # → {1: [1, 1], 2: [2, 2, 2], 3: [3]}
nums = sorted([1, 2, 4, 6, 3, 7, 8], key=lambda x: x % 2)
for parity, grp in groupby(nums, key=lambda x: x % 2):
    print('odd' if parity else 'even', list(grp))  # → even [2, 4, 6, 8] / odd [1, 3, 7]

# Длины серий
from itertools import groupby
runs = [(k, len(list(v))) for k, v in groupby('AAABBBCCDDDDEE')]
print(runs)  # → [('A', 3), ('B', 3), ('C', 2), ('D', 4), ('E', 2)]

See also

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