python 实现namedtuple(元类)

PS:非原创,网友的代码

代码:

def namedtuple(tuple_name, attrs):
    class Meta(type):
        def __call__(self, *args):
            return type.__call__(self, args)
    attrs_set = set(attrs)
    def __init__(self, args):
        for key, value in zip(attrs, args):
            self.__dict__[key]=value
    def __str__(self):
        values = [str(x) for x in self.__dict__.values()]
        return tuple_name+'(' + ', '.join(values) + ')'
    def to_dict(self):
        return self.__dict__
    return Meta(tuple_name, (tuple,),{'attr_keys':attrs_set,'__init__':__init__,'__str__':__str__,'to_dict':to_dict})

说明:
最关键的是

        def __call__(self, *args):
            return type.__call__(self, args)

通过拦截类的实例化过程,将参数list传入元祖,args再传入下面的__init__中,确保只有一个参数。

    def __init__(self, args):
        for key, value in zip(attrs, args):
            self.__dict__[key]=value

__init__中,通过迭代将参数传入self.__dict__字典中,之后即可用p.x的方式调用属性。
使用:

In [1]: from namedtuple_like import namedtuple

In [2]: Point = namedtuple('p',['x','y'])

In [3]: p = Point(1,2)

In [4]: p.x
Out[4]: 1

In [5]: p.y
Out[5]: 2

In [8]: p.to_dict()
Out[8]: {'x': 1, 'y': 2}

In [12]: print(p)
p(1, 2)

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