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我正在寻找一种方法来使用复制构造函数和 () 运算符初始化派生类,就像在 C++ 中一样

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*/

然后我在考虑如何实现这种初始化方式,以防我有一个派生自其他两个加上它自己的成员的类并提出以下内容:

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

然后调用:

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}

所以我想知道这有多少是非pythonic方式?如果有更好的方法(或没有方法)来做同样的事情?

4

1 回答 1

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好吧,你不想创建一个实例来创建一个新实例,所以你可能想要一个classmethodor staticmethod。这也不是使用的地方__call__

我可能会这样做:

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)

但是使用 Base1 和 Base2 的实例来构建 MyChild 的实例并不像我在 python 中所做的那样。更有可能使用明显的:

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

真的我更喜欢这样,现在我不必担心破坏价值或其他怪异。

如果你真的想要捷径方法,你可以将两者结合起来:

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 python 中,更少的“聪明”通常是“更好”

于 2017-03-27T22:23:25.907 回答