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})

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': ''}

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。

下面的代码来自这里，适用于嵌套字典和ide，如VS code能够提示现有的属性:

class Struct:
    def __init__(self, **kwargs):
        for key, value in kwargs.items():
            if isinstance(value, dict):
                self.__dict__[key] = Struct(**value)
            else:
                self.__dict__[key] = value


my_dict = {
    'name': 'bobbyhadz',
    'address': {
        'country': 'Country A',
        'city': 'City A',
        'codes': [1, 2, 3]
    },
}

obj = Struct(**my_dict)

如果您想了解如何加载YAML文件并将其转换为Python对象，请参阅以下要点。

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

更新:在Python 2.6及以上版本中，考虑namedtuple数据结构是否适合您的需求:

>>> from collections import namedtuple
>>> MyStruct = namedtuple('MyStruct', 'a b d')
>>> s = MyStruct(a=1, b={'c': 2}, d=['hi'])
>>> s
MyStruct(a=1, b={'c': 2}, d=['hi'])
>>> s.a
1
>>> s.b
{'c': 2}
>>> s.c
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: 'MyStruct' object has no attribute 'c'
>>> s.d
['hi']

替代方案(原答案内容)为:

class Struct:
    def __init__(self, **entries):
        self.__dict__.update(entries)

然后，你可以使用:

>>> args = {'a': 1, 'b': 2}
>>> s = Struct(**args)
>>> s
<__main__.Struct instance at 0x01D6A738>
>>> s.a
1
>>> s.b
2