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?
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
>>> 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'
其他回答
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
以下是我认为前面例子中最好的方面:
class Struct:
"""The recursive class for building and representing objects with."""
def __init__(self, obj):
for k, v in obj.items():
if isinstance(v, dict):
setattr(self, k, Struct(v))
else:
setattr(self, k, v)
def __getitem__(self, val):
return self.__dict__[val]
def __repr__(self):
return '{%s}' % str(', '.join('%s : %s' % (k, repr(v)) for (k, v) in self.__dict__.items()))
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')])
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']