在Swift中,您可以定义@dynamicMemberLookup (see documentation)来直接访问嵌套在另一个类型中的属性。有没有Python的等价物?
我想用Python实现的示例
假设我有一个包含成员的类,例如:
c = OuterClass()
c.inner_class = ClassWithManyMembers()
c.inner_class.member1 = "1"
c.inner_class.member2 = "2"
c.inner_class.member3 = "3"我希望能够获取/设置这些成员,而不必每次都键入inner_class:
print(c.member1) # prints "1"
c.member1 = 3
print(c.member1) # prints "3"Swift (Source)中的示例:
按成员名称进行动态成员查找
@dynamicMemberLookup
struct DynamicStruct {
let dictionary = ["someDynamicMember": 325,
"someOtherMember": 787]
subscript(dynamicMember member: String) -> Int {
return dictionary[member] ?? 1054
}
}
let s = DynamicStruct()
// Use dynamic member lookup.
let dynamic = s.someDynamicMember
print(dynamic)
// Prints "325"按键路径动态查找成员
struct Point { var x, y: Int }
@dynamicMemberLookup
struct PassthroughWrapper<Value> {
var value: Value
subscript<T>(dynamicMember member: KeyPath<Value, T>) -> T {
get { return value[keyPath: member] }
}
}
let point = Point(x: 381, y: 431)
let wrapper = PassthroughWrapper(value: point)
print(wrapper.x)我在Python中唯一的想法是将所有嵌套属性直接monkey-patch到外部类。
发布于 2019-11-15 07:40:06
通常,当您想要重复访问内部对象时,可以只保存对它的引用。
c = OuterClass()
c.inner_class = ClassWithManyMembers()
ic = c.inner_class
print(ic.member1)
print(ic.member2)
print(ic.member3)
ic.member1 = "5"发布于 2019-11-15 05:03:49
我建议不要将类嵌套在另一个类中,但如果您必须这样做,请尝试这样做:
class MetaOuter(type):
def __getattr__(cls, attr):
for member in cls.__dict__.values():
if hasattr(member, attr):
return getattr(member, attr)
raise AttributeError(attr)
def __setattr__(cls, attr, value):
for member in cls.__dict__.values():
if hasattr(member, attr):
setattr(member, attr, value)
return
super().__setattr__(attr, value)
class Outer(metaclass=MetaOuter):
a = 0
class Inner:
x = 1
y = 2现在,Outer中嵌套类的任何属性都可以作为Outer的属性使用(并且可以写入
>>> Outer.x, Outer.y
(1, 2)
>>> Outer.a # Accessing regular attributes still works as usual
0
>>> Outer.x = True
>>> Outer.Inner.x
True如果您需要嵌套多个级别,请对任何内部封装类使用相同的元类:
class Outer(metaclass=MetaOuter):
a = 0
class Inner(metaclass=MetaOuter):
x = 1
y = 2
class Innerer:
z = 42>>> Outer.a, Outer.x, Outer.y, Outer.z
(0, 1, 2, 42)
>>> Outer.z = -1
>>> Outer.z
-1注意:请注意,如果您试图访问在多个嵌套类中找到的属性,则无法确定该属性将来自哪个类。在这种情况下,一种更可预测的实现将是处理将被查找的某种关键路径,但这本质上与Python默认提供的(例如,Outer.Inner.Innerer.z)相同。
https://stackoverflow.com/questions/58864852
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