这个怎么样:

from functools import partial
d2o=partial(type, "d2o", ())

然后可以这样使用:

>>> o=d2o({"a" : 5, "b" : 3})
>>> print o.a
5
>>> print o.b
3

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

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

>>> def dict2obj(d):
        if isinstance(d, list):
            d = [dict2obj(x) for x in d]
        if not isinstance(d, dict):
            return d
        class C(object):
            pass
        o = C()
        for k in d:
            o.__dict__[k] = dict2obj(d[k])
        return o


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

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"

我认为字典由数字、字符串和字典组成，大多数时候就足够了。 所以我忽略了元组、列表和其他类型没有出现在字典的最后一个维度的情况。

考虑了继承，结合递归，方便地解决了打印问题，并提供了两种数据查询方式，一种数据编辑方式。

请看下面的例子，这是一个描述学生信息的词典:

group=["class1","class2","class3","class4",]
rank=["rank1","rank2","rank3","rank4","rank5",]
data=["name","sex","height","weight","score"]

#build a dict based on the lists above
student_dic=dict([(g,dict([(r,dict([(d,'') for d in data])) for r in rank ]))for g in group])

#this is the solution
class dic2class(dict):
    def __init__(self, dic):
        for key,val in dic.items():
            self.__dict__[key]=self[key]=dic2class(val) if isinstance(val,dict) else val


student_class=dic2class(student_dic)

#one way to edit:
student_class.class1.rank1['sex']='male'
student_class.class1.rank1['name']='Nan Xiang'

#two ways to query:
print student_class.class1.rank1
print student_class.class1['rank1']
print '-'*50
for rank in student_class.class1:
    print getattr(student_class.class1,rank)

结果:

{'score': '', 'sex': 'male', 'name': 'Nan Xiang', 'weight': '', 'height': ''}
{'score': '', 'sex': 'male', 'name': 'Nan Xiang', 'weight': '', 'height': ''}
--------------------------------------------------
{'score': '', 'sex': '', 'name': '', 'weight': '', 'height': ''}
{'score': '', 'sex': '', 'name': '', 'weight': '', 'height': ''}
{'score': '', 'sex': 'male', 'name': 'Nan Xiang', 'weight': '', 'height': ''}
{'score': '', 'sex': '', 'name': '', 'weight': '', 'height': ''}
{'score': '', 'sex': '', 'name': '', 'weight': '', 'height': ''}