如果你想让它递归的话，在之前接受的答案所做的基础上。

class FullStruct:
    def __init__(self, **kwargs):
        for key, value in kwargs.items():
            if isinstance(value, dict):
                f = FullStruct(**value)
                self.__dict__.update({key: f})
            else:
                self.__dict__.update({key: value})

在2021年，使用pydantic BaseModel -将嵌套字典和嵌套json对象转换为python对象，反之亦然:

https://pydantic-docs.helpmanual.io/usage/models/

>>> class Foo(BaseModel):
...     count: int
...     size: float = None
... 
>>> 
>>> class Bar(BaseModel):
...     apple = 'x'
...     banana = 'y'
... 
>>> 
>>> class Spam(BaseModel):
...     foo: Foo
...     bars: List[Bar]
... 
>>> 
>>> m = Spam(foo={'count': 4}, bars=[{'apple': 'x1'}, {'apple': 'x2'}])

对象to dict

>>> print(m.dict())
{'foo': {'count': 4, 'size': None}, 'bars': [{'apple': 'x1', 'banana': 'y'}, {'apple': 'x2', 'banana': 'y'}]}

对象转换为JSON

>>> print(m.json())
{"foo": {"count": 4, "size": null}, "bars": [{"apple": "x1", "banana": "y"}, {"apple": "x2", "banana": "y"}]}

反对的词典

>>> spam = Spam.parse_obj({'foo': {'count': 4, 'size': None}, 'bars': [{'apple': 'x1', 'banana': 'y'}, {'apple': 'x2', 'banana': 'y2'}]})
>>> spam
Spam(foo=Foo(count=4, size=None), bars=[Bar(apple='x1', banana='y'), Bar(apple='x2', banana='y2')])

JSON到对象

>>> spam = Spam.parse_raw('{"foo": {"count": 4, "size": null}, "bars": [{"apple": "x1", "banana": "y"}, {"apple": "x2", "banana": "y"}]}')
>>> spam
Spam(foo=Foo(count=4, size=None), bars=[Bar(apple='x1', banana='y'), Bar(apple='x2', banana='y')])

在@max-sirwa的代码上更新了递归数组展开

class Objectify:
    def __init__(self, **kwargs):
        for key, value in kwargs.items():
            if isinstance(value, dict):
                f = Objectify(**value)
                self.__dict__.update({key: f})
            elif isinstance(value, list):
                t = []
                for i in value:
                    t.append(Objectify(**i)) if isinstance(i, dict) else t.append(i)
                self.__dict__.update({key: t})
            else:
                self.__dict__.update({key: value})

以下是我认为前面例子中最好的方面:

class Struct:
    """The recursive class for building and representing objects with."""

    def __init__(self, obj):
        for k, v in obj.items():
            if isinstance(v, dict):
                setattr(self, k, Struct(v))
            else:
                setattr(self, k, v)

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

    def __repr__(self):
        return '{%s}' % str(', '.join('%s : %s' % (k, repr(v)) for (k, v) in self.__dict__.items()))

如果只是将dict赋值给一个空对象的__dict__呢?

class Object:
    """If your dict is "flat", this is a simple way to create an object from a dict

    >>> obj = Object()
    >>> obj.__dict__ = d
    >>> d.a
    1
    """
    pass

当然，这在你嵌套的dict例子上失败了，除非你递归地遍历dict:

# For a nested dict, you need to recursively update __dict__
def dict2obj(d):
    """Convert a dict to an object

    >>> d = {'a': 1, 'b': {'c': 2}, 'd': ["hi", {'foo': "bar"}]}
    >>> obj = dict2obj(d)
    >>> obj.b.c
    2
    >>> obj.d
    ["hi", {'foo': "bar"}]
    """
    try:
        d = dict(d)
    except (TypeError, ValueError):
        return d
    obj = Object()
    for k, v in d.iteritems():
        obj.__dict__[k] = dict2obj(v)
    return obj

你的例子列表元素可能是一个映射，一个(键，值)对的列表，像这样:

>>> d = {'a': 1, 'b': {'c': 2}, 'd': [("hi", {'foo': "bar"})]}
>>> obj = dict2obj(d)
>>> obj.d.hi.foo
"bar"

如果你想让它递归的话，在之前接受的答案所做的基础上。

class FullStruct:
    def __init__(self, **kwargs):
        for key, value in kwargs.items():
            if isinstance(value, dict):
                f = FullStruct(**value)
                self.__dict__.update({key: f})
            else:
                self.__dict__.update({key: value})