计算多个字典键的 Pythonic 方法

Question

假设有 3 个字典first, second,third具有以下值first = {'a': 0.2, 'b': 0.001}second = {'a': 0.99, 'c': 0.78}third = {'c': 1, 'd': 0.1}total = {'_first': first, '_second': second, '_third':third}有没有一种方法可以快速获得一个数据结构，它可以使用而不是多个字典来保存每个键（a, b, c, ）的计数信息。例如，答案应该返回类似since , , key在这些字典中出现两次而只出现一次的内容。dtotal{'a':2, 'b':2, 'c':2, 'd':1}abcd

Answer 1

from collections import Counter

from itertools import chain

first = {'a': 0.2, 'b': 0.001}

second = {'a': 0.99, 'c': 0.78}

third = {'c': 1, 'd': 0.1}

print(Counter(chain(first, second, third)))

以存储在字典中的可变数量的字典来解释已编辑的问题total

total = {'_first': first, '_second': second, '_third':third}

print(Counter(chain.from_iterable(total.values())))

Answer 2

没有任何进口：

def dict_keys_count(*dicts):

    keys = []

    for _dict in dicts:

        keys += _dict.keys()

    keys_counts = {}

    for key in set(keys):

        keys_counts[key] = keys.count(key)

    return keys_counts

print(dict_keys_count(first,second,third))

# {'b': 1, 'c': 2, 'd': 1, 'a': 2}

速度比较：我与接受的答案

from time import time

t0 = time()

for _ in range(int(1e7)):

    dict_keys_count(first,second,third)

print("No-import solution {}:".format(time() - t0))

# 17.77

t0 = time()

for _ in range(int(1e7)):

    Counter(chain(first, second, third))

print("Accepted solution {}:".format(time() - t0))

# 24.01

Answer 3

您可以使用collections.Counter：

import collections

first = {'a': 0.2, 'b': 0.001}

second = {'a': 0.99, 'c': 0.78}

third = {'c': 1, 'd': 0.1}

r = dict(collections.Counter([*first, *second, *third]))

输出：

{'a': 2, 'c': 2, 'b': 1, 'd': 1}