函数未返回预期的列表

Question

我有3个清单：

members = ['Alex', 'Danny', 'Kieran', 'Zoe', 'Caroline']
visitors = ['Scott', 'Helen', 'Raj', 'Danny']
signups = ['Helen', 'Scott']

函数应返回既不是现有成员也不是新成员的visitors列表。如果没有，则返回None

这是我的代码：

def guests(member,signup,visitor):
    visit = []
    for person in visitor:
        if person not in member and signup:
            visit.append(person)
            return visit
        return print("None.")

但是我的输出返回["Scott"]。我似乎不知道我做得不对的地方是什么

Answer 1

首先，您在获得第一个值之后返回，其次，您可能需要更改 if person not in member and signup到if person not in member and person not in signup

members = ['Alex', 'Danny', 'Kieran', 'Zoe', 'Caroline']
visitors = ['Scott', 'Helen', 'Raj', 'Danny']
signups = ['Helen', 'Scott']
def guests(member,signup,visitor):
    visit = []
    for person in visitor:
        if person not in member and person not in signup:
            visit.append(person)
    return visit
print(guests(members,signups,visitors))

Answer 2

如果缩进错误，则在第一次出现已注册且不是成员的人员后返回列表的值，则应将缩进修正为：

def guests(member,signup,visitor):
    visit = []
    for person in visitor:
        if person not in member and person not in signup:
            visit.append(person)
    return visit if len(visit) != 0 else None

另外，请注意我是如何更改您的返回值的。您不应该返回Print('None')，因为虽然它最终将返回None（因为print返回None，这不是最佳实践）

Answer 3

person not in member and signup的计算结果为(person not in member) and (signup)

当person为“Scott”时，第一部分为True，第二部分（signup）为非空列表，因此在布尔上下文中它被计算为True，因此整个条件为True

然后，立即return输出，而无需进一步循环

你的意思是：

def guests(member,signup,visitor):
    visit = []
    for person in visitor:
        if person not in member and person not in signup:
            visit.append(person)
    return visit if len(visit) != 0 else None