我正在寻找一种优雅的方式来获得数据使用属性访问字典与一些嵌套的字典和列表(即javascript风格的对象语法)。

例如:

>>> d = {'a': 1, 'b': {'c': 2}, 'd': ["hi", {'foo': "bar"}]}

应该以这样的方式访问:

>>> x = dict2obj(d)
>>> x.a
1
>>> x.b.c
2
>>> x.d[1].foo
bar

我想,如果没有递归,这是不可能的,但是有什么更好的方法来获得字典的对象样式呢?


当前回答

我想上传我对这个小范例的看法。

class Struct(dict):
  def __init__(self,data):
    for key, value in data.items():
      if isinstance(value, dict):
        setattr(self, key, Struct(value))
      else:   
        setattr(self, key, type(value).__init__(value))

      dict.__init__(self,data)

它保留导入到类中的类型的属性。我唯一关心的是从解析的字典中覆盖方法。但其他方面似乎很可靠!

其他回答

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})

令人惊讶的是,没有人提到邦奇。这个库专门用于提供对dict对象的属性样式访问,并完全符合OP的要求。一个示范:

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

Python 3的库可以在https://github.com/Infinidat/munch上获得-来源是codyzu

>>> from munch import DefaultMunch
>>> d = {'a': 1, 'b': {'c': 2}, 'd': ["hi", {'foo': "bar"}]}
>>> obj = DefaultMunch.fromDict(d)
>>> obj.b.c
2
>>> obj.a
1
>>> obj.d[1].foo
'bar'
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'

如果你想让它递归的话,在之前接受的答案所做的基础上。

class FullStruct:
    def __init__(self, **kwargs):
        for key, value in kwargs.items():
            if isinstance(value, dict):
                f = FullStruct(**value)
                self.__dict__.update({key: f})
            else:
                self.__dict__.update({key: value})

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')])