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
这是另一种将字典列表转换为对象的替代方法:
def dict2object(in_dict):
class Struct(object):
def __init__(self, in_dict):
for key, value in in_dict.items():
if isinstance(value, (list, tuple)):
setattr(
self, key,
[Struct(sub_dict) if isinstance(sub_dict, dict)
else sub_dict for sub_dict in value])
else:
setattr(
self, key,
Struct(value) if isinstance(value, dict)
else value)
return [Struct(sub_dict) for sub_dict in in_dict] \
if isinstance(in_dict, list) else Struct(in_dict)
其他回答
# 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。
这是另一种将字典列表转换为对象的替代方法:
def dict2object(in_dict):
class Struct(object):
def __init__(self, in_dict):
for key, value in in_dict.items():
if isinstance(value, (list, tuple)):
setattr(
self, key,
[Struct(sub_dict) if isinstance(sub_dict, dict)
else sub_dict for sub_dict in value])
else:
setattr(
self, key,
Struct(value) if isinstance(value, dict)
else value)
return [Struct(sub_dict) for sub_dict in in_dict] \
if isinstance(in_dict, list) else Struct(in_dict)
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']
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')])