【Python】リストの要素ごとの数を速く数えるcollections.Counterの使い方

# Counterの基本: インポートと最小例
from collections import Counter

data = ["a", "b", "a", "c", "b", "a"]  # 文字列のリスト
counts = Counter(data)                  # 出現回数をまとめて集計

print(counts)           # Counterオブジェクト(辞書風)
print(counts["a"])      # キー"a"の個数
print(counts["x"])      # 存在しない要素は0を返す(例外にならない)

from collections import Counter

fruits = ["apple", "banana", "apple", "orange", "banana", "apple"]
c = Counter(fruits)

print(c)
for name, n in c.items():
    print(f"{name}: {n}")

from collections import Counter

text = "Time flies like an arrow; fruit flies like a banana."
words = [w.strip(".,;").lower() for w in text.split()]
counts = Counter(words)

print(counts.most_common(3))  # 上位3件

from collections import Counter

codes = [200, 200, 404, 500, 200, 404, 301]
c = Counter(codes)

print(c)
print("200の件数:", c[200])
print("302の件数(存在しない):", c[302])  # 0

from collections import Counter

c = Counter(["x", "y", "x"])
need = "z"

# 例: 個数が一定以上か判定し、そのまま加算更新
if c[need] < 2:             # 存在しなくても0扱い
    c[need] += 1            # そのままインクリメントできる

print(c)
print("zの個数:", c["z"])

from collections import Counter

events = ["login", "view", "view", "click", "view", "login", "purchase"]
c = Counter(events)

print("上位2件:", c.most_common(2))
print("全件(降順):", c.most_common())

from collections import Counter

c = Counter("mississippi")  # 文字の頻度
# 降順(頻度が高い順)
desc = sorted(c.items(), key=lambda kv: kv[1], reverse=True)
# 昇順(頻度が低い順)
asc = sorted(c.items(), key=lambda kv: kv[1])

print("降順:", desc)
print("昇順:", asc)

from collections import Counter
import json

c = Counter(["A", "B", "A", "C"])
as_dict = dict(c)  # 通常の辞書に変換

# ループ
for key, count in as_dict.items():
    print(f"{key} => {count}")

# JSON文字列に変換
print(json.dumps(as_dict, ensure_ascii=False, indent=2))

from collections import Counter

raw = ["Apple", "apple", "APPLE", "Äpfel", "äpfel"]
lower_counts = Counter(w.lower() for w in raw)
casefold_counts = Counter(w.casefold() for w in raw)  # より強力な正規化

print("lower:", lower_counts)
print("casefold:", casefold_counts)

from collections import Counter
from itertools import chain

nested = [["a", "b"], ["a", "c", "b"], ["a"]]

# 1) 内包表記でフラット化
flat1 = [x for row in nested for x in row]
# 2) chain.from_iterableでフラット化
flat2 = list(chain.from_iterable(nested))

c1 = Counter(flat1)
c2 = Counter(flat2)

print(c1)
print(c2)

from collections import Counter

items = [["x", 1], ["x", 1], ["y", 2]]

try:
    Counter(items)  # リストをキーにしようとするとTypeError
except TypeError as e:
    print("エラー:", e)

# 対策: タプルに変換してイミュータブル化
items_as_tuple = [tuple(x) for x in items]
c = Counter(items_as_tuple)
print(c)

# 性能目安のデモ(参考値)
from collections import Counter
from timeit import timeit
import random

random.seed(0)
data = [random.randint(0, 1000) for _ in range(200_000)]  # 20万件

def using_count():
    # 例: 100種類のキーを数える
    keys = range(100)
    return {k: data.count(k) for k in keys}

def using_counter():
    c = Counter(data)
    return {k: c[k] for k in range(100)}

t1 = timeit(using_count, number=3)
t2 = timeit(using_counter, number=3)

print(f"list.countの合計時間: {t1:.3f}s")
print(f"Counterの合計時間   : {t2:.3f}s")

from collections import Counter

day1 = ["A", "B", "A"]
day2 = ["B", "C"]
day3 = ["A", "C", "C"]

total = Counter()
total.update(day1)            # リストを追加集計
total.update(day2)
total.update(day3)

print("累積:", total)

# 既存のCounterをさらに合算することも可能
part = Counter({"A": 2, "D": 5})
total.update(part)

print("合算後:", total)

from collections import Counter

base = Counter("aabbb")
base.subtract("abbbbb")  # bを多く減らす

print(base)              # 負の値も保持される