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
不是对OP问题的直接回答，但受到启发，也许对一些人有用。我已经创建了一个基于对象的解决方案使用内部__dict__(在任何方式优化代码)
payload = {
"name": "John",
"location": {
"lat": 53.12312312,
"long": 43.21345112
},
"numbers": [
{
"role": "home",
"number": "070-12345678"
},
{
"role": "office",
"number": "070-12345679"
}
]
}
class Map(object):
"""
Dot style access to object members, access raw values
with an underscore e.g.
class Foo(Map):
def foo(self):
return self.get('foo') + 'bar'
obj = Foo(**{'foo': 'foo'})
obj.foo => 'foobar'
obj._foo => 'foo'
"""
def __init__(self, *args, **kwargs):
for arg in args:
if isinstance(arg, dict):
for k, v in arg.iteritems():
self.__dict__[k] = v
self.__dict__['_' + k] = v
if kwargs:
for k, v in kwargs.iteritems():
self.__dict__[k] = v
self.__dict__['_' + k] = v
def __getattribute__(self, attr):
if hasattr(self, 'get_' + attr):
return object.__getattribute__(self, 'get_' + attr)()
else:
return object.__getattribute__(self, attr)
def get(self, key):
try:
return self.__dict__.get('get_' + key)()
except (AttributeError, TypeError):
return self.__dict__.get(key)
def __repr__(self):
return u"<{name} object>".format(
name=self.__class__.__name__
)
class Number(Map):
def get_role(self):
return self.get('role')
def get_number(self):
return self.get('number')
class Location(Map):
def get_latitude(self):
return self.get('lat') + 1
def get_longitude(self):
return self.get('long') + 1
class Item(Map):
def get_name(self):
return self.get('name') + " Doe"
def get_location(self):
return Location(**self.get('location'))
def get_numbers(self):
return [Number(**n) for n in self.get('numbers')]
# Tests
obj = Item({'foo': 'bar'}, **payload)
assert type(obj) == Item
assert obj._name == "John"
assert obj.name == "John Doe"
assert type(obj.location) == Location
assert obj.location._lat == 53.12312312
assert obj.location._long == 43.21345112
assert obj.location.latitude == 54.12312312
assert obj.location.longitude == 44.21345112
for n in obj.numbers:
assert type(n) == Number
if n.role == 'home':
assert n.number == "070-12345678"
if n.role == 'office':
assert n.number == "070-12345679"
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)
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
不是对OP问题的直接回答，但受到启发，也许对一些人有用。我已经创建了一个基于对象的解决方案使用内部__dict__(在任何方式优化代码)
payload = {
"name": "John",
"location": {
"lat": 53.12312312,
"long": 43.21345112
},
"numbers": [
{
"role": "home",
"number": "070-12345678"
},
{
"role": "office",
"number": "070-12345679"
}
]
}
class Map(object):
"""
Dot style access to object members, access raw values
with an underscore e.g.
class Foo(Map):
def foo(self):
return self.get('foo') + 'bar'
obj = Foo(**{'foo': 'foo'})
obj.foo => 'foobar'
obj._foo => 'foo'
"""
def __init__(self, *args, **kwargs):
for arg in args:
if isinstance(arg, dict):
for k, v in arg.iteritems():
self.__dict__[k] = v
self.__dict__['_' + k] = v
if kwargs:
for k, v in kwargs.iteritems():
self.__dict__[k] = v
self.__dict__['_' + k] = v
def __getattribute__(self, attr):
if hasattr(self, 'get_' + attr):
return object.__getattribute__(self, 'get_' + attr)()
else:
return object.__getattribute__(self, attr)
def get(self, key):
try:
return self.__dict__.get('get_' + key)()
except (AttributeError, TypeError):
return self.__dict__.get(key)
def __repr__(self):
return u"<{name} object>".format(
name=self.__class__.__name__
)
class Number(Map):
def get_role(self):
return self.get('role')
def get_number(self):
return self.get('number')
class Location(Map):
def get_latitude(self):
return self.get('lat') + 1
def get_longitude(self):
return self.get('long') + 1
class Item(Map):
def get_name(self):
return self.get('name') + " Doe"
def get_location(self):
return Location(**self.get('location'))
def get_numbers(self):
return [Number(**n) for n in self.get('numbers')]
# Tests
obj = Item({'foo': 'bar'}, **payload)
assert type(obj) == Item
assert obj._name == "John"
assert obj.name == "John Doe"
assert type(obj.location) == Location
assert obj.location._lat == 53.12312312
assert obj.location._long == 43.21345112
assert obj.location.latitude == 54.12312312
assert obj.location.longitude == 44.21345112
for n in obj.numbers:
assert type(n) == Number
if n.role == 'home':
assert n.number == "070-12345678"
if n.role == 'office':
assert n.number == "070-12345679"
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。
我只需要使用虚线路径字符串访问字典，所以我想到了:
def get_value_from_path(dictionary, parts):
""" extracts a value from a dictionary using a dotted path string """
if type(parts) is str:
parts = parts.split('.')
if len(parts) > 1:
return get_value_from_path(dictionary[parts[0]], parts[1:])
return dictionary[parts[0]]
a = {'a':{'b':'c'}}
print(get_value_from_path(a, 'a.b')) # c
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