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

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

# Applies to Python-3 Standard Library
class Struct(object):
    def __init__(self, data):
        for name, value in data.items():
            setattr(self, name, self._wrap(value))

    def _wrap(self, value):
        if isinstance(value, (tuple, list, set, frozenset)): 
            return type(value)([self._wrap(v) for v in value])
        else:
            return Struct(value) if isinstance(value, dict) else value


# Applies to Python-2 Standard Library
class Struct(object):
    def __init__(self, data):
        for name, value in data.iteritems():
            setattr(self, name, self._wrap(value))

    def _wrap(self, value):
        if isinstance(value, (tuple, list, set, frozenset)): 
            return type(value)([self._wrap(v) for v in value])
        else:
            return Struct(value) if isinstance(value, dict) else value

可以用于任何深度的任何序列/字典/值结构。

通常情况下，您希望将字典层次结构镜像到对象中，而不是列表或元组，它们通常处于最低级别。我是这样做的:

class defDictToObject(object):

    def __init__(self, myDict):
        for key, value in myDict.items():
            if type(value) == dict:
                setattr(self, key, defDictToObject(value))
            else:
                setattr(self, key, value)

所以我们这样做:

myDict = { 'a': 1,
           'b': { 
              'b1': {'x': 1,
                    'y': 2} },
           'c': ['hi', 'bar'] 
         }

并获得:

x.b.b1。* 1

X.c ['hi'， 'bar']

这也很有效

class DObj(object):
    pass

dobj = Dobj()
dobj.__dict__ = {'a': 'aaa', 'b': 'bbb'}

print dobj.a
>>> aaa
print dobj.b
>>> bbb

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'