您的测试无法进行，因为您正在将密码(类型为str:字符串)与list进行比较.由于对象是不可比较的，因此结果只是False(即使它们是 可比较的，此处也不存在相等性，而是要检查的非空交集)

Your test cannot work because you're comparing your password (of type str: string) against a list. Since objects are non-comparable, the result is just False (even if they were comparable there is no equality here, but a non-empty intersection to check)

您需要检查每个列表中密码中至少有1个成员

You need to check for each list that there's at least 1 member of the list in the password

使用any定义一个辅助功能，该功能检查列表中的字母是否在密码中(lambda可能太多):

Define an aux function which checks if a letter of the list is in the password (a lambda would be too much maybe) using any:

def ok(passwd,l):
    return any(k in passwd for k in l)

然后使用all在这种情况下测试所有四个列表:

Then test all your four lists against this condition using all:

elif len(ww) >= 6 and len(ww)<= 12:
    sww = set(ww)
    if all(ok(sww,l) for l in [klein,groot,nummers,symbolen]):
        print ("uw wachtwoord is Sterk")

请注意通过将密码(这是一个列表，对于in运算符为O(n))转换为字符set的字符(其中存在in运算符，但速度更快)的略微优化.此外，集合将删除重复的字符，这对于本示例来说是完美的.

Note the slight optimization by converting the password (which is kind of a list so O(n) for in operator) by a set of characters (where the in operator exists but is way faster). Besides, a set will remove duplicate characters, which is perfect for this example.

更紧凑的版本，没有aux功能，使用lambda毕竟不是那么难理解:

More compact version without the aux function and using a lambda which is not so difficult to understand after all:

elif len(ww) >= 6 and len(ww)<= 12:
    sww = set(ww)
    if all(lambda l: any(k in sww for k in l) for l in [klein,groot,nummers,symbolen]):
        print ("uw wachtwoord is Sterk")