Question

我正在尝试将子类的字典与类继承合并

class Person(object):
    nametag = {
        "name": "Bob",
        "occupation": "Nobody"
    }

class Teacher(Person):
    nametag = {
        "occupation": "Professor",
        "Subject": "Python"
    }

    def __init__(self):
        nametag = dict(Person.nametag.items() + self.nametag.items())

最终我需要：

Teacher().nametag["name"] == "bob"
Teacher().nametag["occupation"] == "Professor"
Teacher().nametag["subject"] == "Python"

旁注，Teacher.nametag和Person.nametag将是非常大的词典，dict(d2.keys() + d1.keys())是最好的方法吗？

Answer 1

据推测，各个类的每个实例将会有不同的nametag。因此，最佳实现不是您当前具有的类属性字典，而是实例属性：

class Person(object):
    def __init__(self, name, occupation):
        self.name = name
        self.occupation = occupation

class Teacher(Person):
    def __init__(self, name, occupation, subject):
        super().__init__(name, occupation)
        self.subject = subject

bob = Teacher("Bob", "Professor", "Python")
bob.name == "Bob"

如果您确实需要nametag，则可以将其实现为：

class Person(object):

    def __init__(self, name, occupation):
        self.name = name
        self.occupation = occupation

    @property
    def nametag(self):
        return {"name": self.name,
                "occupation": self.occupation}

class Teacher(Person):

    def __init__(self, name, occupation, subject):
        super().__init__(name, occupation)
        self.subject = subject

    @property
    def nametag(self):
        tag = super().nametag
        tag["subject"] = self.subject
        return tag

bob = Teacher("Bob", "Professor", "Python")
bob.nametag["name"] == "Bob"

作为旁注，

dict(d2.keys() + d1.keys())

会给你一个错误。要合并两个词典，请执行以下操作：

d1.update(d2)

Answer 2

如果名称标签需要是Person和Teacher的属性，那么将它们作为彼此继承的单独类拉出来（可以使用数据库外键）：

class PersonNametag(object):
    name = 'Bob'
    occupation = 'Nobody'


class TeacherNametag(PersonNametag):
    occupation = 'Professor'
    Subject = 'Python'


class Person(object):
    def __init__(self):
        self.nametag = PersonNametag()


class Teacher(Person):
    def __init__(self):
        self.nametag = TeacherNametag()

合并继承的类的变量

2 个答案: