python - 在 python 中使用 str.format() 女巫类

Question

我有一个带有 __getitem__() 函数的类，它可以像字典一样订阅。但是，当我尝试将它传递给 str.format() 时，我得到了一个 TypeError。我如何在 python 中使用带有 format() 函数的类？

>>> class C(object):
      id=int()
      name=str()

      def __init__(self, id, name):
        self.id=id
        self.name=name

      def __getitem__(self, key):
        return getattr(self, key)

>>> d=dict(id=1, name='xyz')
>>> c=C(id=1, name='xyz')
>>>
>>> #Subscription works for both objects
>>> print(d['id'])
1
>>> print(c['id'])
1
>>>
>>> s='{id} {name}'
>>> #format() only works on dict()
>>> print(s.format(**d))
1 xyz
>>> print(s.format(**c))
Traceback (most recent call last):
  File "<pyshell#13>", line 1, in <module>
    print(s.format(**c))
TypeError: format() argument after ** must be a mapping, not C

Answer 1

正如一些评论提到您可以从 dict 继承，它不起作用的原因是:

If the syntax **expression appears in the function call, the expression must evaluate to a mapping, the contents of which are treated as additional keyword arguments. In the case of a keyword appearing in both expression and as an explicit keyword argument, a TypeError exception is raised.

要使其正常工作，您需要实现映射 ABC。类似这样的事情:

from collections.abc import Mapping


class C(Mapping):

    id=int()
    name=str()

    def __init__(self, id, name):
        self.id = id
        self.name = name

    def __iter__(self):
        for x in self.__dict__.keys():
            yield x

    def __len__(self):
        return len(self.__dict__)

    def __getitem__(self, key):
        return self.__dict__[key]

这样你就可以使用 s = '{id}{name}'.format(**c)
而不是 s = '{id}{name}'.format(**c.__dict__)

如果您希望能够像在字典中那样更改您的类变量，您还可以使用 collections.abc 模块中的 MutableMapping。 MutableMapping 还需要实现 __setitem__ 和 __delitem__