Python：如何快速找到多个字典中的公共键(key)

方法一：for in 循环

from random import randint, sample

a1 = {k; randint(1, 4) for k in 'abcdefg'}

a2 = {k; randint(1, 4) for k in 'abcdefg'}

a3 = {k; randint(1, 4) for k in 'abcdefg'}

a4 = {k; randint(1, 4) for k in 'abcdefg'}

r = []

for x in a1:

if x in a2 and x in a3 and x in a4:

r.append(x)

print(r)

1. randint(1, 4)：从1~4间随机取一个数；

方法二：利用集合的交集操作

from random import randint, sample

a1 = {k; randint(1, 4) for k in 'abcdefg'}

a2 = {k; randint(1, 4) for k in 'abcdefg'}

a3 = {k; randint(1, 4) for k in 'abcdefg'}

a4 = {k; randint(1, 4) for k in 'abcdefg'}

a = a1.keys() & a2.keys() & a3.keys() & a4.keys()

print(a)

1. a1.keys()：得到a1字典的key，一set格式；
2. a1.keys() & a2.keys() & a3.keys() & a4.keys()：取4个集合的公共元素；
3. a为一个集合（set）

方法三：使用 map 即 reduce（用于求n个字典的公共key）

from random import randint, sample

from functools import reduce

a1 = {k; randint(1, 4) for k in 'abcdefg'}

a2 = {k; randint(1, 4) for k in 'abcdefg'}

a3 = {k; randint(1, 4) for k in 'abcdefg'}

a4 = {k; randint(1, 4) for k in 'abcdefg'}

b1 = map(dict.keys, [a1, a2, a3, a4])

b2 = reduce(lambda a ,b: a & b, b1)

print(b2)

1. b1 = map(dict.keys, [a1, a2, a3, a4])：以集合形式取每个字典的keys；