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sorted() 関数の計算量

sorted() 関数は、イテラブルの要素を並べ替えた新しいリストを返します。

計算量の分析

場合 時間 空間 備考
基本的なソート O(n log n) O(n) Timsort / Powersort
キー関数を使う場合 O(n log n + n*k) O(n) k はキー関数の実行時間、キーは要素ごとに一度だけ計算される
降順のソート O(n log n) O(n) 追加のコストはない
すでに整列済み O(n) O(n) 最良の場合

基本的な使い方

単純なソート

# O(n log n) - Timsort (≤3.10) or Powersort (3.11+)
numbers = [3, 1, 4, 1, 5, 9, 2, 6]
result = sorted(numbers)
# [1, 1, 2, 3, 4, 5, 6, 9]

# Works with any iterable
result = sorted((3, 1, 4))  # Tuple input
# [1, 3, 4]

result = sorted({3, 1, 4})  # Set input
# [1, 3, 4]

result = sorted("cadb")     # String input
# ['a', 'b', 'c', 'd']

降順のソート

# O(n log n) - same complexity
numbers = [3, 1, 4, 1, 5, 9, 2, 6]
result = sorted(numbers, reverse=True)
# [9, 6, 5, 4, 3, 2, 1, 1]

# Works with strings
words = ["apple", "pie", "cat"]
result = sorted(words, reverse=True)
# ["pie", "cat", "apple"]

キー関数を使う

独自の比較

# O(n log n + n*k) where k = key function time
# Key is computed once per element, then comparisons use cached keys
words = ["apple", "pie", "cat", "banana"]
result = sorted(words, key=len)  # Sort by length
# ["pie", "cat", "apple", "banana"]

# Sort by last character
result = sorted(words, key=lambda x: x[-1])
# ["apple", "banana", "pie", "cat"]

オブジェクトのソート

# O(n log n) - simple key extraction
class Person:
    def __init__(self, name, age):
        self.name = name
        self.age = age

    def __repr__(self):
        return f"Person({self.name}, {self.age})"

people = [
    Person("Alice", 30),
    Person("Bob", 25),
    Person("Charlie", 35),
]

# Sort by age
result = sorted(people, key=lambda p: p.age)
# [Person(Bob, 25), Person(Alice, 30), Person(Charlie, 35)]

# Using operator module (more efficient)
from operator import attrgetter
result = sorted(people, key=attrgetter('age'))  # Same O(n log n)

タプルのソート

# O(n log n) - lexicographic comparison
coords = [(1, 5), (3, 2), (2, 8)]
result = sorted(coords)
# [(1, 5), (2, 8), (3, 2)]

# Sort by second element
result = sorted(coords, key=lambda c: c[1])
# [(3, 2), (1, 5), (2, 8)]

ソートアルゴリズム

仕組み

Python uses Timsort (Python 2.3-3.10) or Powersort (Python 3.11+).
Both are hybrid algorithms combining merge sort and insertion sort:
1. Divide array into small chunks (runs) - ~32-64 elements
2. Sort each run with insertion sort - O(k²) per run
3. Merge runs together - O(n log n) overall
4. Already sorted data: O(n) - detects and uses it

Powersort uses an improved merge policy but has the same complexity.

性能の特性

# Best case - O(n) - already sorted or reverse sorted
numbers = list(range(1000000))
result = sorted(numbers)  # Nearly O(n) for nearly sorted data

# Average case - O(n log n)
import random
numbers = list(range(1000))
random.shuffle(numbers)
result = sorted(numbers)  # O(n log n)

# Worst case - O(n log n) - still guaranteed
numbers = [1000 - i for i in range(1000)]  # Reverse sorted
result = sorted(numbers)  # O(n log n) - handles well

性能に関するパターン

sorted() と sort()

# sorted() - creates new list, O(n log n) time, O(n) space
original = [3, 1, 4, 1, 5]
result = sorted(original)  # [1, 1, 3, 4, 5]
# original unchanged

# list.sort() - in-place, O(n log n) time, O(n) space
original = [3, 1, 4, 1, 5]
original.sort()  # [1, 1, 3, 4, 5]
# original modified

# Both use same algorithm, same complexity but sorted() makes copy

コストの高いキー関数

# O(n*k + n log n) - key computed once per element, then cached
def expensive_key(x):
    # O(m) - expensive computation
    return sum(range(x))

numbers = list(range(1000))
result = sorted(numbers, key=expensive_key)
# Complexity: O(n*m + n log n) - key called n times, then n log n comparisons

# Better: pre-compute keys
from operator import itemgetter
keys = [(x, expensive_key(x)) for x in numbers]  # O(n*m)
result = sorted(keys, key=itemgetter(1))         # O(n log n)
# Total: O(n*m + n log n)

Decorate-Sort-Undecorate (DSU)

# O(n log n) - when computing key is expensive
def get_sort_key(item):
    # Some expensive computation
    return complex_calculation(item)

# With key: O(n*k + n log n) - key computed once per element
result = sorted(items, key=get_sort_key)

# Faster: O(n*k + n log n)
decorated = [(get_sort_key(item), item) for item in items]  # O(n*k)
sorted_decorated = sorted(decorated)                          # O(n log n)
result = [item for _, item in sorted_decorated]              # O(n)

ソートの安定性

# sorted() is stable - preserves order of equal elements
data = [(1, 'a'), (2, 'b'), (1, 'c'), (2, 'd')]
result = sorted(data, key=lambda x: x[0])
# [(1, 'a'), (1, 'c'), (2, 'b'), (2, 'd')]
# Among equal keys, original order preserved

よくあるパターン

複数のソート基準

# Sort by multiple attributes - O(n log n)
students = [
    ('Alice', 85),
    ('Bob', 85),
    ('Charlie', 90),
]

# Sort by score descending, then name ascending
result = sorted(students, key=lambda s: (-s[1], s[0]))
# [('Charlie', 90), ('Alice', 85), ('Bob', 85)]

大文字小文字を無視したソート

# O(n log n) - with case conversion
words = ["Apple", "banana", "Cherry", "date"]
result = sorted(words, key=str.lower)
# ["Apple", "banana", "Cherry", "date"]

独自の順序でのソート

# O(n log n) - custom comparison key
priority = {'high': 0, 'medium': 1, 'low': 2}
tasks = [
    {'name': 'A', 'priority': 'low'},
    {'name': 'B', 'priority': 'high'},
    {'name': 'C', 'priority': 'medium'},
]

result = sorted(tasks, key=lambda t: priority[t['priority']])
# B (high), C (medium), A (low)

ほかのソート手段との比較

sorted() と list.sort()

# sorted() - returns new list, original unchanged
original = [3, 1, 4, 1, 5]
result = sorted(original)

# list.sort() - modifies in-place, returns None
original = [3, 1, 4, 1, 5]
original.sort()

# Both: O(n log n) time, O(n) space for Timsort/Powersort
# Choose based on whether you need original

sorted() と heapq.nsmallest()

# sorted() - O(n log n), entire list sorted
numbers = list(range(1000000))
all_sorted = sorted(numbers)

# heapq.nsmallest() - O(n log k) for k items
import heapq
k_smallest = heapq.nsmallest(10, numbers)  # Much faster if k << n

端の場合

空のリスト

# O(1) - no sorting needed
result = sorted([])
# []

要素が 1 つ

# O(1) - nothing to sort
result = sorted([42])
# [42]

すでに整列済み

# O(n) - Timsort/Powersort detects and handles efficiently
numbers = list(range(1000000))
result = sorted(numbers)  # Nearly O(n)

逆順に整列済み

# O(n) - also handled efficiently
numbers = list(range(1000000, 0, -1))
result = sorted(numbers)  # Nearly O(n)

ベストプラクティス

✅ 推奨:

  • 並べ替えた新しいリストを作るには sorted() を使う
  • 独自の順序には key 引数を使う
  • 属性で並べ替えるときは lambda ではなく operator.attrgetter() を使う
  • 同じキーで何度も並べ替えるなら、コストの高いキーは事前に計算しておく

❌ 避けるべきこと:

  • sorted() を何度も呼ぶ（結果をキャッシュする）
  • 複雑な lambda 関数（代わりに関数を定義する）
  • キーの中でコストの高い計算をして並べ替える（事前に計算する）
  • sorted() が新しいリストを作ること（メモリを使うこと）を忘れる

関連する関数

バージョン別の注記

  • Python 2.3-3.10: Timsort アルゴリズムを使う
  • Python 3.11+: Powersort を使う（マージ方針が改善されたが、計算量は同じ）