【Python基础教程】快速找到多个字典中的公共键(key)的方法
2021/10/6 22:11:13
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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
''' 学习中遇到问题没人解答?小编创建了一个Python学习交流群:531509025 寻找有志同道合的小伙伴,互帮互助,群里还有不错的视频学习教程和PDF电子书! ''' 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;
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