【Python基礎教程】快速找到多個字典中的公共鍵(key)的方法

方法一：for in循環

from random import randint, sample

a1 = {k: randint(1, 4) for k in 'abcdefg'}
a2 = {k: randint(1, 4) for k in 'abc123456789'}
a3 = {k: randint(1, 4) for k in 'abcinubububu'}
a4 = {k: randint(1, 4) for k in 'abc89898989'}
r = []

for x in a1:
    if x in a2 and x in a3 and x in a4:
        r.append(x)
print(r)
randint(1, 4)  # 從1~4間隨機取一個數

方法二：利用集合的交集操作

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'''
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)

a1.keys()：得到a1字典的key，一set格式；
a1.keys() & a2.keys() & a3.keys() & a4.keys()：取4個集合的公共元素；
a為一個集合（set）

方法三：使用map即reduce（用於求n個字典的公共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)

b1 = map(dict.keys, [a1, a2, a3, a4])  #以集合形式取每個字典的keys；