【Python基础教程】快速找到多个字典中的公共键(key)的方法

我不是女神ヾ 2022-09-12 14:56 125阅读 0赞

**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 '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)  # 从1~4间随机取一个数

**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个集合的公共元素；  
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)
    
    b1 = map(dict.keys, [a1, a2, a3, a4])  #以集合形式取每个字典的keys；