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Python:如何使用super()初始化2个超类?

superclass subclass inheritance class python

Adriano_Pinaffo · 技术社区 · 8 年前

我有两个超类,父亲和母亲,他们将由孩子继承。

class Father:
  def __init__(self, **kwargs):
    self.fathername = kwargs["ffn"] + " " + kwargs["fln"]
    self.fatherage = kwargs["fa"]

class Mother:
  def __init__(self, **kwargs):
    self.mothername = kwargs["mfn"] + " " + kwargs["mln"]
    self.motherage = kwargs["ma"]

class Child(Father, Mother):
  def __init__(self, **kwargs):
    self.name = kwargs["name"] + " " + kwargs["lastname"]
    self.age = kwargs["age"]

如果我打电话给他们的父母 __init__()

    Father.__init__(self, **kwargs)
    Mother.__init__(self, **kwargs)

但是如何使用super()实现同样的效果呢?如果我像下面这样称呼它,它只初始化父亲而不是母亲(因为父亲是MRO中的下一个)

    super().__init__(**kwargs)

下面就是 __str__() 重写以显示分配的内容。

def __str__(self):
    return \
    "Im {}, {} years old".format(self.name, self.age) + "\n" + \
    "My dad is {} and he is {} years old".format(self.fathername, self.fatherage) + "\n" + \
    "My mom is {} and she is {} years old".format(self.mothername, self.motherage)

familyname = "Simpson"
child = Child(**{"name": "Bart", "lastname": familyname, "age": 15, "ffn": "Hommer", "fln": familyname, "fa": 54, "mfn": "Marggie", "mln": familyname, "ma": 46})

当我尝试打印对象时,它将失败,因为母超类从未初始化(当我使用 super() __初始化() )

print(child)

The program raises a runtime error

Traceback (most recent call last):
  File "/tmp/pyadv.py", line 225, in <module>
    print(child)
  File "/tmp/pyadv.py", line 217, in __str__
    return     "Im {}, {} years old".format(self.name, self.age) + "\n" +     "My dad is {} and he is {} years old".format(self.fathername, self.fatherage) + "\n" +     "My mom is {} and she is {} years old".format(self.mothername, self.motherage)
AttributeError: 'Child' object has no attribute 'mothername'

那么,如何使用super初始化两个超类呢?

编辑: 我试图补充 super(Father, self).__init__(**kwargs) 和 super(Mother, self).__init__(**kwargs) 去超级班级方法,但我得到以下错误:

Traceback (most recent call last):
  File "/tmp/pyadv.py", line 225, in <module>
    child = Child(**{"name": "Bart", "lastname": familyname, "age": 15, "ffn": "Hommer", "fln": familyname, "fa": 54, "mfn": "Marggie", "mln": familyname, "ma": 46})
  File "/tmp/pyadv.py", line 217, in __init__
    super().__init__(**kwargs)
  File "/tmp/pyadv.py", line 199, in __init__
    super(Father, self).__init__(**kwargs)
  File "/tmp/pyadv.py", line 208, in __init__
    super(Mother, self).__init__(**kwargs)
TypeError: object.__init__() takes no parameters

我还试图补充 super(Father, self).__init__() 和 super(Mother, self).__init__() (内部没有参数) )去超级班级 __初始化()

Traceback (most recent call last):
  File "/tmp/pyadv.py", line 225, in <module>
    child = Child(**{"name": "Bart", "lastname": familyname, "age": 15, "ffn": "Hommer", "fln": familyname, "fa": 54, "mfn": "Marggie", "mln": familyname, "ma": 46})
  File "/tmp/pyadv.py", line 217, in __init__
    super().__init__(**kwargs)
  File "/tmp/pyadv.py", line 199, in __init__
    super(Father, self).__init__()
  File "/tmp/pyadv.py", line 206, in __init__
    self.mothername = kwargs["mfn"] + " " + kwargs["mln"]
KeyError: 'mfn'

解决方案1:

解决方案2: here .

class Root:
  def __init__(self, **kwargs):
    pass

class Father(Root):
  def __init__(self, **kwargs):
    self.fathername = kwargs["ffn"] + " " + kwargs["fln"]
    self.fatherage = kwargs["fa"]
    super().__init__(**kwargs)

class Mother(Root):
  def __init__(self, **kwargs):
    self.mothername = kwargs["mfn"] + " " + kwargs["mln"]
    self.motherage = kwargs["ma"]
    super().__init__(**kwargs)

class Child(Father, Mother):
  def __init__(self, **kwargs):
    self.name = kwargs["name"] + " " + kwargs["lastname"]
    self.age = kwargs["age"]
    super().__init__(**kwargs)
  def __str__(self):
    return \
    "Im {}, {} years old".format(self.name, self.age) + "\n" + \
    "My dad is {} and he is {} years old".format(self.fathername, self.fatherage) + "\n" + \
    "My mom is {} and she is {} years old".format(self.mothername, self.motherage)

familyname = "Simpson"
child = Child(**{"name": "Bart", "lastname": familyname, "age": 15, "ffn": "Homer", "fln": familyname, "fa": 54, "mfn": "Marge", "mln": familyname, "ma": 46})

print(child)

2 回复 | 直到 8 年前

Blckknght 8 年前

Python中的多重继承需要协作。也就是说,两个父类需要知道彼此存在的可能性(尽管它们不需要知道彼此的任何细节)。那么,任何一个先被命名的父母都可以调用另一个父母的名字 __init__ 方法。就是这样 super

kwargs 听写在你的字典里电话。当第二个家长试图调用MRO中的最后一个类时,这就成了一个问题, object ,它不希望接收任何关键字参数。相反,每个班级 super().__init__ (除非它知道它的一个父类也需要参数)。

试试这个:

class Father:
  def __init__(self, ffn, fln, fa, **kwargs): # name the parameters we expect
    super().__init__(**kwargs)           # pass on any unknown arguments
    self.fathername = ffn + " " + fln    # use parameters by name, rather than via kwargs
    self.fatherage = fa

class Mother:
  def __init__(self, mfn, mln, ma, **kwargs):
    super().__init__(**kwargs)
    self.mothername = mfn + " " + mln
    self.motherage = ma

class Child(Father, Mother):
  def __init__(self, name, lastname, age, **kwargs):
    super().__init__(**kwargs)
    self.name = name " " + lastname
    self.age = age

    def __str__(self):
        return \
        "Im {}, {} years old".format(self.name, self.age) + "\n" + \
        "My dad is {} and he is {} years old".format(self.fathername, self.fatherage) + "\n" + \
        "My mom is {} and she is {} years old".format(self.mothername, self.motherage)


familyname = "Simpson"
child = Child(name="Bart", lastname=familyname, age=15, # you can use keyword syntax here
              ffn="Homer", fln=familyname, fa=54,
              mfn="Marge", mln=familyname, ma=46)
print(child)

请注意,在Python3中,通常不需要将任何参数传递给 super()

最后一个音符。虽然我确信你的代码只是一个例子,但在面向对象设计方面,类的名称是非常糟糕的。继承意味着 IS-A公司 Mother 或者 Father ,但代码说他是,因为他是和班级。描述与父母之间人际关系的更好方法是 . 每个孩子 有一个 有-A 使用封装的关系。这意味着,子对象对属性中每个父对象都有一个引用。有趣的是,只需一节课就可以做到这一点(这可能就是为什么你没有学习这一点,如果你被教导继承的话):

class Person:
    def __init__(self, firstname, lastname, age, father=None, mother=None):
        self.name = firstname + " " + lastname
        self.age = age
        self.father = father  #  set the attributes for our parents here
        self.mother = mother

fn = "Simpson"
bart = Person("Bart", fn, 15, Person("Homer", fn, 54), Person("Marge", fn, 46))

user10287394 user10287394 8 年前

从这本书的公认答案来看 link :

通过super调用不会调用所有的父级,而是调用下一个在MRO链中的作用。要使其正常工作,您需要使用在所有的比赛中都很出色 __init__s

class Parent1(object):
    def __init__(self):
        super(Parent1, self).__init__()
        self.var1 = 1

class Parent2(object):
    def __init__(self):
        super(Parent2, self).__init__()
        self.var2 = 2

class Child(Parent1, Parent2):
    def __init__(self):
        super(Child, self).__init__()