from random import randint, sample?
# 隨機(jī)取樣進(jìn)球球員信息
print(sample('abcdef', randint(3, 6)))
# 字典解析產(chǎn)生每輪數(shù)據(jù)
s1 = {x: randint(1, 4) for x in sample('abcdef', randint(3, 6))}
s2 = {x: randint(1, 4) for x in sample('abcdef', randint(3, 6))}
s3 = {x: randint(1, 4) for x in sample('abcdef', randint(3, 6))}
print(s1, s2, s3)
# 方法1：
# 迭代第1輪中鍵，之后再去判斷這個(gè)key是否在s2和s3當(dāng)中，如果在就表示這個(gè)鍵是公共鍵
res = []
for k in s1:
? ? if k in s2 and k in s3:
? ? ? ? res.append(k)
print(res)
?
# 方法2：
# 獲取每一個(gè)字典中的所有key,python2為viewkeys()
print(s1.keys(), s2.keys(), s3.keys())
# 取所有集合的&交集，就是公共鍵
print(s1.keys() & s2.keys() & s3.keys())
?
# 方法3：
# 使用map得到每一輪的keys集合
print(list(map(dict.keys, [s1, s2, s3])))
?
from functools import reduce
# 使用reduce函數(shù)，取每一輪的keys集合的交集
print(reduce(lambda a, b: a & b, map(dict.keys, [s1, s2, s3])))

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(jī)取一個(gè)數(shù)

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)

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(gè)字典的keys；