@derek73的答案非常简洁，但它不能被pickle或(深度)复制，并且它在缺少键时返回None。下面的代码修复了这个问题。

编辑:我没有看到上面的答案解决了完全相同的问题(点赞)。我把答案留在这里供参考。

class dotdict(dict):
    __setattr__ = dict.__setitem__
    __delattr__ = dict.__delitem__

    def __getattr__(self, name):
        try:
            return self[name]
        except KeyError:
            raise AttributeError(name)

用于无限级别的字典、列表、字典的列表和列表的字典的嵌套。

它还支持酸洗

这是这个答案的延伸。

class DotDict(dict):
    # https://stackoverflow.com/a/70665030/913098
    """
    Example:
    m = Map({'first_name': 'Eduardo'}, last_name='Pool', age=24, sports=['Soccer'])

    Iterable are assumed to have a constructor taking list as input.
    """

    def __init__(self, *args, **kwargs):
        super(DotDict, self).__init__(*args, **kwargs)

        args_with_kwargs = []
        for arg in args:
            args_with_kwargs.append(arg)
        args_with_kwargs.append(kwargs)
        args = args_with_kwargs

        for arg in args:
            if isinstance(arg, dict):
                for k, v in arg.items():
                    self[k] = v
                    if isinstance(v, dict):
                        self[k] = DotDict(v)
                    elif isinstance(v, str) or isinstance(v, bytes):
                        self[k] = v
                    elif isinstance(v, Iterable):
                        klass = type(v)
                        map_value: List[Any] = []
                        for e in v:
                            map_e = DotDict(e) if isinstance(e, dict) else e
                            map_value.append(map_e)
                        self[k] = klass(map_value)



    def __getattr__(self, attr):
        return self.get(attr)

    def __setattr__(self, key, value):
        self.__setitem__(key, value)

    def __setitem__(self, key, value):
        super(DotDict, self).__setitem__(key, value)
        self.__dict__.update({key: value})

    def __delattr__(self, item):
        self.__delitem__(item)

    def __delitem__(self, key):
        super(DotDict, self).__delitem__(key)
        del self.__dict__[key]

    def __getstate__(self):
        return self.__dict__

    def __setstate__(self, d):
        self.__dict__.update(d)


if __name__ == "__main__":
    import pickle
    def test_map():
        d = {
            "a": 1,
            "b": {
                "c": "d",
                "e": 2,
                "f": None
            },
            "g": [],
            "h": [1, "i"],
            "j": [1, "k", {}],
            "l":
                [
                    1,
                    "m",
                    {
                        "n": [3],
                        "o": "p",
                        "q": {
                            "r": "s",
                            "t": ["u", 5, {"v": "w"}, ],
                            "x": ("z", 1)
                        }
                    }
                ],
        }
        map_d = DotDict(d)
        w = map_d.l[2].q.t[2].v
        assert w == "w"

        pickled = pickle.dumps(map_d)
        unpickled = pickle.loads(pickled)
        assert unpickled == map_d

        kwargs_check = DotDict(a=1, b=[dict(c=2, d="3"), 5])
        assert kwargs_check.b[0].d == "3"

        kwargs_and_args_check = DotDict(d, a=1, b=[dict(c=2, d="3"), 5])
        assert kwargs_and_args_check.l[2].q.t[2].v == "w"
        assert kwargs_and_args_check.b[0].d == "3"



    test_map()

Fabric有一个非常好的、最小的实现。将其扩展为允许嵌套访问，我们可以使用defaultdict，结果看起来像这样:

from collections import defaultdict

class AttributeDict(defaultdict):
    def __init__(self):
        super(AttributeDict, self).__init__(AttributeDict)

    def __getattr__(self, key):
        try:
            return self[key]
        except KeyError:
            raise AttributeError(key)

    def __setattr__(self, key, value):
        self[key] = value

可以这样使用它:

keys = AttributeDict()
keys.abc.xyz.x = 123
keys.abc.xyz.a.b.c = 234

这详细阐述了Kugel的回答“从dict和派生并实现__getattr__和__setattr__”。现在你知道怎么做了!

基于epool的答案，这个版本允许你通过点操作符访问任何字典:

foo = {
    "bar" : {
        "baz" : [ {"boo" : "hoo"} , {"baba" : "loo"} ]
    }
}

例如，foo.bar.baz[1]。爸爸回答“loo”。

class Map(dict):
    def __init__(self, *args, **kwargs):
        super(Map, self).__init__(*args, **kwargs)
        for arg in args:
            if isinstance(arg, dict):
                for k, v in arg.items():
                    if isinstance(v, dict):
                        v = Map(v)
                    if isinstance(v, list):
                        self.__convert(v)
                    self[k] = v

        if kwargs:
            for k, v in kwargs.items():
                if isinstance(v, dict):
                    v = Map(v)
                elif isinstance(v, list):
                    self.__convert(v)
                self[k] = v

    def __convert(self, v):
        for elem in range(0, len(v)):
            if isinstance(v[elem], dict):
                v[elem] = Map(v[elem])
            elif isinstance(v[elem], list):
                self.__convert(v[elem])

    def __getattr__(self, attr):
        return self.get(attr)

    def __setattr__(self, key, value):
        self.__setitem__(key, value)

    def __setitem__(self, key, value):
        super(Map, self).__setitem__(key, value)
        self.__dict__.update({key: value})

    def __delattr__(self, item):
        self.__delitem__(item)

    def __delitem__(self, key):
        super(Map, self).__delitem__(key)
        del self.__dict__[key]

这也适用于嵌套字典，并确保后面追加的字典行为相同:

class DotDict(dict):

    def __init__(self, *args, **kwargs):
        super().__init__(*args, **kwargs)
        # Recursively turn nested dicts into DotDicts
        for key, value in self.items():
            if type(value) is dict:
                self[key] = DotDict(value)

    def __setitem__(self, key, item):
        if type(item) is dict:
            item = DotDict(item)
        super().__setitem__(key, item)

    __setattr__ = __setitem__
    __getattr__ = dict.__getitem__

我试了一下:

class dotdict(dict):
    def __getattr__(self, name):
        return self[name]

你也可以尝试__getattribute__。

使每个字典都是一种类型的dotdict就足够了，如果你想从多层字典初始化它，也可以尝试实现__init__。