我的字典是这样的:

addr_bk = {
    'person': [
        {'name': 'Andrew', 'id': 123, 'email': 'andrew@mailserver.com',
         'phone': [{'type': 2, 'number': '633311122'},
                   {'type': 0, 'number': '97788665'}]
        },
        {'name': 'Tom', 'id': 456,
         'phone': [{'type': 0, 'number': '91122334'}]}, 
        {'name': 'Jack', 'id': 7788, 'email': 'jack@gmail.com'}
    ]
}

可以看到，我已经嵌套了字典和字典列表。 这是因为addr_bk是从使用lwpb.codec转换为python dict的协议缓冲区数据解码的。有可选字段(例如email =>，其中键可能不可用)和重复字段(例如phone =>转换为dict列表)。

我尝试了上述所有建议的解决方案。有些不能很好地处理嵌套字典。其他的则不容易打印对象的详细信息。

只有Dawie Strauss的dic2obj (dict)解决方案最有效。

我已经增强了一点，当找不到钥匙时处理:

# Work the best, with nested dictionaries & lists! :)
# Able to print out all items.
class dict2obj_new(dict):
    def __init__(self, dict_):
        super(dict2obj_new, 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_new(it)
            elif isinstance(item, dict):
                self[key] = dict2obj_new(item)

    def __getattr__(self, key):
        # Enhanced to handle key not found.
        if self.has_key(key):
            return self[key]
        else:
            return None

然后，我用:

# Testing...
ab = dict2obj_new(addr_bk)

for person in ab.person:
  print "Person ID:", person.id
  print "  Name:", person.name
  # Check if optional field is available before printing.
  if person.email:
    print "  E-mail address:", person.email

  # Check if optional field is available before printing.
  if person.phone:
    for phone_number in person.phone:
      if phone_number.type == codec.enums.PhoneType.MOBILE:
        print "  Mobile phone #:",
      elif phone_number.type == codec.enums.PhoneType.HOME:
        print "  Home phone #:",
      else:
        print "  Work phone #:",
      print phone_number.number

这是另一个实现:

class DictObj(object):
    def __init__(self, d):
        self.__dict__ = d

def dict_to_obj(d):
    if isinstance(d, (list, tuple)): return map(dict_to_obj, d)
    elif not isinstance(d, dict): return d
    return DictObj(dict((k, dict_to_obj(v)) for (k,v) in d.iteritems()))

[编辑]遗漏了在列表中处理字典的部分，而不仅仅是其他字典。添加修复。

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'

这是另一种将字典列表转换为对象的替代方法:

def dict2object(in_dict):
    class Struct(object):
        def __init__(self, in_dict):
            for key, value in in_dict.items():
                if isinstance(value, (list, tuple)):
                    setattr(
                        self, key,
                        [Struct(sub_dict) if isinstance(sub_dict, dict)
                         else sub_dict for sub_dict in value])
                else:
                    setattr(
                        self, key,
                        Struct(value) if isinstance(value, dict)
                        else value)
    return [Struct(sub_dict) for sub_dict in in_dict] \
        if isinstance(in_dict, list) else Struct(in_dict)

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

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')])