我知道这里已经有很多答案了，我迟到了，但这个方法将递归和“就地”将字典转换为类对象结构……适用于3.x.x

def dictToObject(d):
    for k,v in d.items():
        if isinstance(v, dict):
            d[k] = dictToObject(v)
    return namedtuple('object', d.keys())(*d.values())

# Dictionary created from JSON file
d = {
    'primaryKey': 'id', 
    'metadata': 
        {
            'rows': 0, 
            'lastID': 0
        }, 
    'columns': 
        {
            'col2': {
                'dataType': 'string', 
                'name': 'addressLine1'
            }, 
            'col1': {
                'datatype': 'string', 
                'name': 'postcode'
            }, 
            'col3': {
                'dataType': 'string', 
                'name': 'addressLine2'
            }, 
            'col0': {
                'datatype': 'integer', 
                'name': 'id'
            }, 
            'col4': {
                'dataType': 'string', 
                'name': 'contactNumber'
            }
        }, 
        'secondaryKeys': {}
}

d1 = dictToObject(d)
d1.columns.col1 # == object(datatype='string', name='postcode')
d1.metadata.rows # == 0

老式问答，但我有更多的话题要谈。似乎没有人谈论递归字典。这是我的代码:

#!/usr/bin/env python

class Object( dict ):
    def __init__( self, data = None ):
        super( Object, self ).__init__()
        if data:
            self.__update( data, {} )

    def __update( self, data, did ):
        dataid = id(data)
        did[ dataid ] = self

        for k in data:
            dkid = id(data[k])
            if did.has_key(dkid):
                self[k] = did[dkid]
            elif isinstance( data[k], Object ):
                self[k] = data[k]
            elif isinstance( data[k], dict ):
                obj = Object()
                obj.__update( data[k], did )
                self[k] = obj
                obj = None
            else:
                self[k] = data[k]

    def __getattr__( self, key ):
        return self.get( key, None )

    def __setattr__( self, key, value ):
        if isinstance(value,dict):
            self[key] = Object( value )
        else:
            self[key] = value

    def update( self, *args ):
        for obj in args:
            for k in obj:
                if isinstance(obj[k],dict):
                    self[k] = Object( obj[k] )
                else:
                    self[k] = obj[k]
        return self

    def merge( self, *args ):
        for obj in args:
            for k in obj:
                if self.has_key(k):
                    if isinstance(self[k],list) and isinstance(obj[k],list):
                        self[k] += obj[k]
                    elif isinstance(self[k],list):
                        self[k].append( obj[k] )
                    elif isinstance(obj[k],list):
                        self[k] = [self[k]] + obj[k]
                    elif isinstance(self[k],Object) and isinstance(obj[k],Object):
                        self[k].merge( obj[k] )
                    elif isinstance(self[k],Object) and isinstance(obj[k],dict):
                        self[k].merge( obj[k] )
                    else:
                        self[k] = [ self[k], obj[k] ]
                else:
                    if isinstance(obj[k],dict):
                        self[k] = Object( obj[k] )
                    else:
                        self[k] = obj[k]
        return self

def test01():
    class UObject( Object ):
        pass
    obj = Object({1:2})
    d = {}
    d.update({
        "a": 1,
        "b": {
            "c": 2,
            "d": [ 3, 4, 5 ],
            "e": [ [6,7], (8,9) ],
            "self": d,
        },
        1: 10,
        "1": 11,
        "obj": obj,
    })
    x = UObject(d)


    assert x.a == x["a"] == 1
    assert x.b.c == x["b"]["c"] == 2
    assert x.b.d[0] == 3
    assert x.b.d[1] == 4
    assert x.b.e[0][0] == 6
    assert x.b.e[1][0] == 8
    assert x[1] == 10
    assert x["1"] == 11
    assert x[1] != x["1"]
    assert id(x) == id(x.b.self.b.self) == id(x.b.self)
    assert x.b.self.a == x.b.self.b.self.a == 1

    x.x = 12
    assert x.x == x["x"] == 12
    x.y = {"a":13,"b":[14,15]}
    assert x.y.a == 13
    assert x.y.b[0] == 14

def test02():
    x = Object({
        "a": {
            "b": 1,
            "c": [ 2, 3 ]
        },
        1: 6,
        2: [ 8, 9 ],
        3: 11,
    })
    y = Object({
        "a": {
            "b": 4,
            "c": [ 5 ]
        },
        1: 7,
        2: 10,
        3: [ 12 , 13 ],
    })
    z = {
        3: 14,
        2: 15,
        "a": {
            "b": 16,
            "c": 17,
        }
    }
    x.merge( y, z )
    assert 2 in x.a.c
    assert 3 in x.a.c
    assert 5 in x.a.c
    assert 1 in x.a.b
    assert 4 in x.a.b
    assert 8 in x[2]
    assert 9 in x[2]
    assert 10 in x[2]
    assert 11 in x[3]
    assert 12 in x[3]
    assert 13 in x[3]
    assert 14 in x[3]
    assert 15 in x[2]
    assert 16 in x.a.b
    assert 17 in x.a.c

if __name__ == '__main__':
    test01()
    test02()

你可以使用一个自定义对象钩子来利用标准库的json模块:

import json

class obj(object):
    def __init__(self, dict_):
        self.__dict__.update(dict_)

def dict2obj(d):
    return json.loads(json.dumps(d), object_hook=obj)

使用示例:

>>> d = {'a': 1, 'b': {'c': 2}, 'd': ['hi', {'foo': 'bar'}]}
>>> o = dict2obj(d)
>>> o.a
1
>>> o.b.c
2
>>> o.d[0]
u'hi'
>>> o.d[1].foo
u'bar'

而且它不像namedtuple那样是严格只读的，也就是说，你可以改变值-而不是结构:

>>> o.b.c = 3
>>> o.b.c
3

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。

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

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