python数据结构-如何快速找到多个字典中的公共键

如何快速找到多个字典中的公共键

问题举例

统计每轮都进球的球员：

第1轮{‘tom’:1, 'meixi':2}

第2轮{‘coco’:3, 'meixi':4, 'marton':2}

第3轮{'coco':2, 'meixi':1, 'david':1}

 

for循环、列表解析和set交集

from random import randint, sample
d1 = {k: randint(1, 5) for k in sample('abcdefg', randint(3, 6))}
d2 = {k: randint(1, 5) for k in sample('abcdefg', randint(3, 6))}
d3 = {k: randint(1, 5) for k in sample('abcdefg', randint(3, 6))}
print(d1, d2, d3)

print("\n(1)for loop")
res1 = []
for key in d1:
    if key in d2 and key in d3:
        res1.append(key)
print(res1)

print("\n(2)list comprehension")
res = [k for k in d1 if k in d2 and k in d3]
print(res)

print("\n(3)set intersection")
s1 = d1.keys()
s2 = d2.keys()
s3 = d3.keys()
res3 = s1 & s2 & s3
print(res3)

分析：在实际场景中我们并不知道有几轮，我们需要更通用的一种方法

我们把字典放在列表中，这里仅以三个字典为例，当然列表中可以放n个字典

 

通用方法（两种）

from random import randint, sample
from functools import reduce
d1 = {k: randint(1, 5) for k in sample('abcdefg', randint(3, 6))}
d2 = {k: randint(1, 5) for k in sample('abcdefg', randint(3, 6))}
d3 = {k: randint(1, 5) for k in sample('abcdefg', randint(3, 6))}
print(d1, d2, d3)

d_list = [d1, d2, d3]
print("\nsolution1:")
res1 = [k for k in d_list[0] if all(map(lambda d: k in d, d_list[1:]))]
print(res1)

print("\nsolution2:")
res2 = reduce(lambda a, b: a & b, map(dict.keys, d_list))
print(res2)

 

参考资料：python3实用编程技巧进阶