让我来解释一下不久前我几乎用过的一个解决方案。但首先，我没有这样做的原因可以通过以下代码来说明:

d = {'from': 1}
x = dict2obj(d)

print x.from

给出这个错误:

  File "test.py", line 20
    print x.from == 1
                ^
SyntaxError: invalid syntax

因为“from”是Python关键字，所以某些字典键是不允许的。

现在我的解决方案允许直接使用字典项的名称来访问它们。但是它也允许你使用“字典语义”。下面是使用示例的代码:

class dict2obj(dict):
    def __init__(self, dict_):
        super(dict2obj, self).__init__(dict_)
        for key in self:
            item = self[key]
            if isinstance(item, list):
                for idx, it in enumerate(item):
                    if isinstance(it, dict):
                        item[idx] = dict2obj(it)
            elif isinstance(item, dict):
                self[key] = dict2obj(item)

    def __getattr__(self, key):
        return self[key]

d = {'a': 1, 'b': {'c': 2}, 'd': ["hi", {'foo': "bar"}]}

x = dict2obj(d)

assert x.a == x['a'] == 1
assert x.b.c == x['b']['c'] == 2
assert x.d[1].foo == x['d'][1]['foo'] == "bar"

X.__dict__.update (d)应该没问题。

我不满意那些被标记和点赞的答案，所以这里有一个简单而通用的解决方案，用于将json风格的嵌套数据结构(由字典和列表组成)转换为普通对象的层次结构:

# tested in: Python 3.8
from collections import abc
from typings import Any, Iterable, Mapping, Union

class DataObject:
    def __repr__(self):
        return str({k: v for k, v in vars(self).items()})

def data_to_object(data: Union[Mapping[str, Any], Iterable]) -> object:
    """
    Example
    -------
    >>> data = {
    ...     "name": "Bob Howard",
    ...     "positions": [{"department": "ER", "manager_id": 13}],
    ... }
    ... data_to_object(data).positions[0].manager_id
    13
    """
    if isinstance(data, abc.Mapping):
        r = DataObject()
        for k, v in data.items():
            if type(v) is dict or type(v) is list:
                setattr(r, k, data_to_object(v))
            else:
                setattr(r, k, v)
        return r
    elif isinstance(data, abc.Iterable):
        return [data_to_object(e) for e in data]
    else:
        return data

将字典转换为对象

from types import SimpleNamespace

def dict2obj(data):
    """将字典对象转换为可访问的对象属性"""
    if not isinstance(data, dict):
        raise ValueError('data must be dict object.')

    def _d2o(d):
        _d = {}
        for key, item in d.items():
            if isinstance(item, dict):
                _d[key] = _d2o(item)
            else:
                _d[key] = item
        return SimpleNamespace(**_d)

    return _d2o(data)

参考答案

最简单的方法是使用collections.namedtuple。

我发现下面的4行代码是最漂亮的，它支持嵌套字典:

def dict_to_namedtuple(typename, data):
    return namedtuple(typename, data.keys())(
        *(dict_to_namedtuple(typename + '_' + k, v) if isinstance(v, dict) else v for k, v in data.items())
    )

输出看起来也会很好:

>>> nt = dict_to_namedtuple('config', {
...     'path': '/app',
...     'debug': {'level': 'error', 'stream': 'stdout'}
... })

>>> print(nt)
config(path='/app', debug=config_debug(level='error', stream='stdout'))

>>> print(nt.debug.level)
'error'

x = type('new_dict', (object,), d)

然后再加上递归，就完成了。

编辑这是我如何实现它:

>>> d
{'a': 1, 'b': {'c': 2}, 'd': ['hi', {'foo': 'bar'}]}
>>> def obj_dic(d):
    top = type('new', (object,), d)
    seqs = tuple, list, set, frozenset
    for i, j in d.items():
        if isinstance(j, dict):
            setattr(top, i, obj_dic(j))
        elif isinstance(j, seqs):
            setattr(top, i, 
                type(j)(obj_dic(sj) if isinstance(sj, dict) else sj for sj in j))
        else:
            setattr(top, i, j)
    return top

>>> x = obj_dic(d)
>>> x.a
1
>>> x.b.c
2
>>> x.d[1].foo
'bar'