python - 虚拟类(class):做得对吗？

我一直在阅读描述类继承、抽象基类甚至Python接口的文档。但没有接缝是我想要的。也就是说，一种构建虚拟类的简单方法。当虚拟类被调用时，我希望它根据给定的参数实例化一些更具体的类，并将其返回调用函数。目前，我有一个将调用重新路由到虚拟类的概要方法，将其向下路由到基础类。
想法如下：

class Shape:
    def __init__(self, description):
        if   description == "It's flat":  self.underlying_class = Line(description)
        elif description == "It's spiky": self.underlying_class = Triangle(description)
        elif description == "It's big":   self.underlying_class = Rectangle(description)
    def number_of_edges(self, parameters):
        return self.underlying_class(parameters)

class Line:
    def __init__(self, description):
        self.desc = description
    def number_of_edges(self, parameters):
        return 1

class Triangle:
    def __init__(self, description):
        self.desc = description
    def number_of_edges(self, parameters):
        return 3

class Rectangle:
    def __init__(self, description):
        self.desc = description
    def number_of_edges(self, parameters):
        return 4

shape_dont_know_what_it_is = Shape("It's big")
shape_dont_know_what_it_is.number_of_edges(parameters)

我的重新路由远不是最佳的，因为只传递了对_edges（）函数个数的调用。在形状上添加这样的东西也不需要缝合来完成这个技巧：

def __getattr__(self, *args):
    return underlying_class.__getattr__(*args)

我做错了什么？整个想法实施得不好吗？非常感谢您的帮助。

最佳答案

我更喜欢在工厂里做：

def factory(description):
    if   description == "It's flat":  return Line(description)
    elif description == "It's spiky": return Triangle(description)
    elif description == "It's big":   return Rectangle(description)

或：

def factory(description):
    classDict = {"It's flat":Line("It's flat"), "It's spiky":Triangle("It's spiky"), "It's big":Rectangle("It's big")}
    return classDict[description]

从形状继承类

class Line(Shape):
    def __init__(self, description):
        self.desc = description
    def number_of_edges(self, parameters):
        return 1