如何将类的每个实例与另一个类的每个实例相关联?

编程入门行业动态更新时间:2024-10-15 04:28:06

本文介绍了如何将类的每个实例与另一个类的每个实例相关联?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！问题描述

我正在使用Protégé4.3开发一个简单的本体，并尝试创建一个模型，使类 C 的每个实例都具有特定的值 v ，该值是给定属性 p 的类 V 的实例.我可以使用一个公理得到这个推论

I'm working on a simple ontology, using Protégé 4.3 and trying to create a model such that every instance of a class C has some particular value v, which is an instance of a class V, for a given property p. I can get this inference using an axiom

C subClassOf (p 值 v)

但是我想我希望能够更像a

but I think I want to be able to state this more along the lines of a

C subClassOf (p 一些 V)

因为我实际上希望每个 C 实例都与 V 的每个实例相关.

because I'd actually like each instance of C to be related to every instance of V.

我的示例是汽车和动作-我想为汽车创建一组动作.当我创建汽车的新实例时，我希望每个实例都有一套动作.如果我添加了新的动作实例，我希望所有汽车实例都能反映出它们所有相关的动作.

My example is automobiles and actions — I want to create a set of actions for automobiles. When I create new instances of automobiles I want each to have a set of actions. If I add new instances of actions I want all the instances of automobiles to reflect all their related actions.

为了添加更多细节，我定义了 Auto 类和两个子类: Car 和 Truck .我还定义了一个 Action 类和一个子类 TruckAction .我定义了一个 hasAction 属性，其域为 Auto ，范围为 Action ，用于关联自动操作和操作.我进一步创建了几个不同类型的人:

To add further detail, I have defined the Auto class and two subclasses: Car and Truck. I have also defined an Action class and a subclass, TruckAction. I have defined a hasAction property with domain Auto and range Action to relate autos and actions. I have further created several individuals of different types:

自动
- 卡车{F150}
- 汽车{ChevyMalibu}
- Auto
  - Truck {F150}
  - Car {ChevyMalibu}
  - TruckAction {LoadCargo}
  当我将 Car 设为(具有动作值Accelerate)的子类时，我可以看到 ChevyMalibu 具有推断的对象属性 hasAction Accelerate ，但这似乎类似于对对象属性声明进行硬编码.我想为 ChevyMalibu 推断出 all 的汽车动作，同样也为 F150 推断出 all 的TruckActions.我不太确定为什么做诸如使 Car 成为(具有Action动作)的子类之类的事情不会使我失望的原因.
  
  When I make Car a subclass of (hasAction value Accelerate), I can see that ChevyMalibu has the inferred object property hasAction Accelerate, but this seems akin to hard-coding the object property assertion. I would like to have all car actions inferred for the ChevyMalibu and likewise all TruckActions inferred for F150. I'm not quite sure why doing something like making Car a subclass of (hasAction some Action) won't get me there.
  
  我认为对 OWL类型推断有限制的雄辩答案与我的问题有关，但我不能完全将其加起来.
  
  I think that the eloquent answer to OWL type inference with a restriction is related to my question, but I can't quite add this up.
  推荐答案
  编码规则
  您正确地注意到，如果单个 x 是 p值y 的实例，那么您将推断断言 p(x，y).但是，简单地说 x 是 p的一个实例，有些Y 不会(没有更多信息)不会让您推断 p(x， y i )用于任何特定的 y i ，即使 y i em>是您已声明的 Y 的唯一特定实例.这是因为OWL做出了开放世界假设；仅仅因为您没有说某件事是正确的(或错误的)，并不表示它不可能是正确的(或错误的). (当然，您可能已经说了其他的话，可以让您确定是真还是假.)这在另一个问题为什么此DL查询不返回任何个人?
  
  Encoding the rules
  You correctly note that if an individual x is an instance of p value y, then you'll infer the assertion p(x,y). However, simply saying that x is an instance of p some Y doesn't (without some more information) won't let you infer that p(x,yi) for any particular yi, even if yi is the only particular instance of Y that you've declared. This is because OWL makes the open world assumption; just because you haven't said something is true (or false) yet doesn't mean that it couldn't be true (or false). (Of course, you might have already said other things that would let you figure out whether something is true or false.) This is described in more detail in another question, Why is this DL-Query not returning any individuals?
  
  如果我对它的理解正确，那么您尝试做的就是这样:
  
  If I've understood it correctly, what you're trying to do is say something like:
- 每一个车辆与 hasAction 关联到每个 Action
- 每一个 Car 通过 hasAction 与每个 CarAction
- 每一个 Truck 与 hasAction 相关联，每个 TruckAction
- every Vehicle is related by hasAction to every Action
- every Car is related by hasAction to every CarAction
- every Truck is related by hasAction to every TruckAction
- 子财产链公理不需要推理程序来获得SWRL支持；和
- 次财产链公理适用于所有个人，但SWRL规则仅适用于具名个人，因此您可以获得更好的覆盖范围.