的Python：多重继承，拷贝构造函数，类的初始化和（）超载

Question

我使用拷贝构造函数寻找一种方法来初始化派生类和（）运算符类似于C++

class Rectangle { 
    int width, height; 
    public: 
    Rectangle (int,int); 
    int area() {return (width*height);} 
}; 

Rectangle::Rectangle (int a, int b) { 
    width = a; 
    height = b; 
} 

r = Rectangle(2,3) 
s = Rectangle(r) /* <--using copy constructor to initialize*/ 

，然后我就在想我将如何实现intitialisation的情况下，这种方式我已经从自己的另外两个多名会员和派生类想出了如下：

class MyBase1(object): 
    def __init__(self, *args, **kwargs): 
     self.x = kwargs.get('x') 
     self.y = kwargs.get('y') 
     print("mybase1 {}".format(kwargs)) 

    def print_base1(self): 
     pass 


class MyBase2(object): 
    def __init__(self, *args, **kwargs): 
     self.s = kwargs.get('s') 
     self.p = kwargs.get('p') 
     print("mybase2 {}".format(kwargs)) 

    def print_base2(self): 
     pass 


class MyChild(MyBase1, MyBase2): 

    def __init__(self, **kwargs): 
     MyBase1.__init__(self, **kwargs) 
     MyBase2.__init__(self, **kwargs) 
     self.function_name = kwargs.get('function') 


    def __call__(self, my_base1, my_base2, **kwargs): 
     initialization_dictionary = dict(vars(my_base1), **vars(my_base2)) 
     initialization_dictionary = dict(initialization_dictionary, **kwargs) 
     newInstance = MyChild(**initialization_dictionary) 
     return newInstance 

调用，则：

base1 = MyBase1(x=1, y=2) 
base2 = MyBase2(s=3, p=4) 

child = MyChild()(base1, base2, function='arcsine') #<--initialising 

[stm for stm in dir(child) if not stm.startswith('__')] 
# gives:['function_name', 'p', 'print_base1', 'print_base2', 's', 'x', 'y'] 

vars(child) 
# gives:{'function_name': 'arcsine', 'p': 4, 's': 3, 'x': 1, 'y': 2} 

所以我想知道这是多么非pythonic的方式？如果有更好的方法（或者无法做到这一点）？

Answer 1

那么，你不想创建一个实例来创建一个新的实例，所以你可能需要一个classmethod或staticmethod。这不是使用__call__的地方。

我可以这样做：

class MyChild(MyBase1, MyBase2): 
    @classmethod 
    def build_from_bases(klass, base1, base2, **kwargs): 
     kwargs.update(base1.__dict__) 
     # Note if base2 has values for x and y, they will clobber what was in base1 
     kwargs.update(base2.__dict__) 
     return klass(**kwargs) 

但在使用和基础1的和Base2实例建立MyChild的实例并不觉得喜欢的事我会在Python做。更可能使用明显：

mychild = MyChild(x=base1.x, y=base1.y, s=base2.s, p=base2.p, function='foo') 

真的，我宁愿如此，现在我不必关心重挫值，或其他古怪。

你可以结合两种，如果你真的想要的快捷方法：

class MyChild(MyBase1, MyBase2): 
    @classmethod 
    def build_from_bases(klass, base1, base2, **kwargs): 
     return klass(x=base1.x, y=base1.y, s=base2.s, p=base2.p, **kwargs) 

在蟒蛇少“聪明”经常“更好”