python進階之——如何快速找到多個字典中的公共鍵

方法1：

from random import randint,sample

#sample('abcdef',3)#加入只有6個球員，sample函數是隨機取出3個球員
#sample('abcdef',randint(3,6))#每組進球球員的數目也是隨機的。
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執行效率並不高，我們可以用另外一種集合的操作，如果遇到多輪操作，效率更慢

方法2：

from random import randint,sample

#sample('abcdef',3)#加入只有6個球員，sample函數是隨機取出3個球員
#sample('abcdef',randint(3,6))#每組進球球員的數目也是隨機的。
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輪可以用這種方法

方法3：如果是前n輪的話可以用map函數和reduce函數來做

from random import randint,sample

#sample('abcdef',3)#加入只有6個球員，sample函數是隨機取出3個球員
#sample('abcdef',randint(3,6))#每組進球球員的數目也是隨機的。
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]))