python进阶之——如何快速找到多个字典中的公共键

方法1：

from random import randint,sample

#sample('abcdef',3)#加入只有6个球员，sample函数是随机取出3个球员
#sample('abcdef',randint(3,6))#每组进球球员的数目也是随机的。
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))}

res=[]
for k in s1:
	if k in s2 and k in s3:
		res.append(k)

方法1执行效率并不高，我们可以用另外一种集合的操作，如果遇到多轮操作，效率更慢

方法2：

from random import randint,sample

#sample('abcdef',3)#加入只有6个球员，sample函数是随机取出3个球员
#sample('abcdef',randint(3,6))#每组进球球员的数目也是随机的。
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))}

res=s1.viewkeys() & s2.viewkeys() & s3.viewkeys()#如果是前3轮可以用这种方法

方法3：如果是前n轮的话可以用map函数和reduce函数来做

from random import randint,sample

#sample('abcdef',3)#加入只有6个球员，sample函数是随机取出3个球员
#sample('abcdef',randint(3,6))#每组进球球员的数目也是随机的。
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))}

#map(dict.viewkeys(),[s1,s2,s3])
reduce(lambda a,b:a&b ,map(dict.viewkeys(),[s1,s2,s3]))