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
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')])
其他回答
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
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
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
我不满意那些被标记和点赞的答案,所以这里有一个简单而通用的解决方案,用于将json风格的嵌套数据结构(由字典和列表组成)转换为普通对象的层次结构:
# tested in: Python 3.8
from collections import abc
from typings import Any, Iterable, Mapping, Union
class DataObject:
def __repr__(self):
return str({k: v for k, v in vars(self).items()})
def data_to_object(data: Union[Mapping[str, Any], Iterable]) -> object:
"""
Example
-------
>>> data = {
... "name": "Bob Howard",
... "positions": [{"department": "ER", "manager_id": 13}],
... }
... data_to_object(data).positions[0].manager_id
13
"""
if isinstance(data, abc.Mapping):
r = DataObject()
for k, v in data.items():
if type(v) is dict or type(v) is list:
setattr(r, k, data_to_object(v))
else:
setattr(r, k, v)
return r
elif isinstance(data, abc.Iterable):
return [data_to_object(e) for e in data]
else:
return data
X.__dict__.update (d)应该没问题。