我如何使Python字典成员访问通过点“。”?
例如,我想写mydict.val而不是mydict['val']。
我还想以这种方式访问嵌套字典。例如
mydict.mydict2.val
会提到
mydict = { 'mydict2': { 'val': ... } }
我如何使Python字典成员访问通过点“。”?
例如,我想写mydict.val而不是mydict['val']。
我还想以这种方式访问嵌套字典。例如
mydict.mydict2.val
会提到
mydict = { 'mydict2': { 'val': ... } }
当前回答
@derek73的答案非常简洁,但它不能被pickle或(深度)复制,并且它在缺少键时返回None。下面的代码修复了这个问题。
编辑:我没有看到上面的答案解决了完全相同的问题(点赞)。我把答案留在这里供参考。
class dotdict(dict):
__setattr__ = dict.__setitem__
__delattr__ = dict.__delitem__
def __getattr__(self, name):
try:
return self[name]
except KeyError:
raise AttributeError(name)
其他回答
def dict_to_object(dick):
# http://stackoverflow.com/a/1305663/968442
class Struct:
def __init__(self, **entries):
self.__dict__.update(entries)
return Struct(**dick)
如果一个人决定永久地将字典转换为对象,这应该做到。您可以在访问之前创建一个丢弃对象。
d = dict_to_object(d)
您可以使用SimpleNamespace来实现这一点
from types import SimpleNamespace
# Assign values
args = SimpleNamespace()
args.username = 'admin'
# Retrive values
print(args.username) # output: admin
派生自dict和并实现__getattr__和__setattr__。
或者你也可以用Bunch,非常相似。
我不认为这是可能的monkeypatch内置字典类。
基于Kugel的回答,并考虑到Mike Graham的警告,如果我们制作一个包装器呢?
class DictWrap(object):
""" Wrap an existing dict, or create a new one, and access with either dot
notation or key lookup.
The attribute _data is reserved and stores the underlying dictionary.
When using the += operator with create=True, the empty nested dict is
replaced with the operand, effectively creating a default dictionary
of mixed types.
args:
d({}): Existing dict to wrap, an empty dict is created by default
create(True): Create an empty, nested dict instead of raising a KeyError
example:
>>>dw = DictWrap({'pp':3})
>>>dw.a.b += 2
>>>dw.a.b += 2
>>>dw.a['c'] += 'Hello'
>>>dw.a['c'] += ' World'
>>>dw.a.d
>>>print dw._data
{'a': {'c': 'Hello World', 'b': 4, 'd': {}}, 'pp': 3}
"""
def __init__(self, d=None, create=True):
if d is None:
d = {}
supr = super(DictWrap, self)
supr.__setattr__('_data', d)
supr.__setattr__('__create', create)
def __getattr__(self, name):
try:
value = self._data[name]
except KeyError:
if not super(DictWrap, self).__getattribute__('__create'):
raise
value = {}
self._data[name] = value
if hasattr(value, 'items'):
create = super(DictWrap, self).__getattribute__('__create')
return DictWrap(value, create)
return value
def __setattr__(self, name, value):
self._data[name] = value
def __getitem__(self, key):
try:
value = self._data[key]
except KeyError:
if not super(DictWrap, self).__getattribute__('__create'):
raise
value = {}
self._data[key] = value
if hasattr(value, 'items'):
create = super(DictWrap, self).__getattribute__('__create')
return DictWrap(value, create)
return value
def __setitem__(self, key, value):
self._data[key] = value
def __iadd__(self, other):
if self._data:
raise TypeError("A Nested dict will only be replaced if it's empty")
else:
return other
这是我对@derek73的回答。我用字典。__getitem__作为__getattr__,因此它仍然抛出KeyError,并且im重命名字典公共方法以“”前缀(“”包围导致特殊方法名称冲突,如__get__将被视为一个描述符方法)。无论如何,由于关键的dict基方法,您无法将键作为属性获得完全清晰的命名空间,因此解决方案并不完美,但您可以拥有键属性,如get, pop, items等。
class DotDictMeta(type):
def __new__(
cls,
name,
bases,
attrs,
rename_method=lambda n: f'__{n}__',
**custom_methods,
):
d = dict
attrs.update(
cls.get_hidden_or_renamed_methods(rename_method),
__getattr__=d.__getitem__,
__setattr__=d.__setitem__,
__delattr__=d.__delitem__,
**custom_methods,
)
return super().__new__(cls, name, bases, attrs)
def __init__(self, name, bases, attrs, **_):
super().__init__(name, bases, attrs)
@property
def attribute_error(self):
raise AttributeError
@classmethod
def get_hidden_or_renamed_methods(cls, rename_method=None):
public_methods = tuple(
i for i in dict.__dict__.items() if not i[0].startswith('__')
)
error = cls.attribute_error
hidden_methods = ((k, error) for k, v in public_methods)
yield from hidden_methods
if rename_method:
renamed_methods = ((rename_method(k), v) for k, v in public_methods)
yield from renamed_methods
class DotDict(dict, metaclass=DotDictMeta):
pass
你可以从DotDict命名空间中删除dict方法,并继续使用dict类方法,当你想操作其他dict实例并希望使用相同的方法而不需要额外检查它是否为DotDict时,它也很有用。
dct = dict(a=1)
dot_dct = DotDict(b=2)
foo = {c: i for i, c in enumerate('xyz')}
for d in (dct, dot_dct):
# you would have to use dct.update and dot_dct.__update methods
dict.update(d, foo)
assert dict.get(dot, 'foo', 0) is 0