scala – Shapeless:从副产品到不同副产品的地图
发布时间:2020-12-16 19:06:57 所属栏目:安全 来源:网络整理
导读:在下面,我试图做一个多态函数将RawFeatureValue转换为RefinedFeatureValue. import shapeless._object test { type RawFeatureValue = Int :+: Double :+: String :+: CNil type RefinedFeatureValue = Int :+: Double :+: CNil private object convert exte
在下面,我试图做一个多态函数将RawFeatureValue转换为RefinedFeatureValue.
import shapeless._ object test { type RawFeatureValue = Int :+: Double :+: String :+: CNil type RefinedFeatureValue = Int :+: Double :+: CNil private object convert extends Poly1 { implicit def caseInt = at[Int](i => i) implicit def caseDouble = at[Double](d => d) implicit def caseString = at[String](s => s.hashCode) } val a = Coproduct[RawFeatureValue](12) val b: RefinedFeatureValue = a map convert } 但是,所得到的类型是Int::Double::Int::CNil,与RefinedFeatureValue不兼容. [error] found : shapeless.:+:[Int,shapeless.:+:[Double,shapeless.:+:[Int,shapeless.CNil]]] [error] required: test.RefinedFeatureValue [error] (which expands to) shapeless.:+:[Int,shapeless.CNil]] [error] val b: RefinedFeatureValue = a map convert [error] ^ 如何告诉无形的两个Ints应该被视为一个? 解决方法
我可以想到的最简单的方法是将每个元素映射到目标副产品中,然后统一结果:
import shapeless._ type RawFeatureValue = Int :+: Double :+: String :+: CNil type RefinedFeatureValue = Int :+: Double :+: CNil object convert extends Poly1 { implicit val caseInt = at[Int](Coproduct[RefinedFeatureValue](_)) implicit val caseDouble = at[Double](Coproduct[RefinedFeatureValue](_)) implicit val caseString = at[String](s => Coproduct[RefinedFeatureValue](s.hashCode)) } 这样可以预期: scala> val a = Coproduct[RawFeatureValue](12) a: RawFeatureValue = 12 scala> val b: RefinedFeatureValue = a.map(convert).unify b: RefinedFeatureValue = 12 scala> val c = Coproduct[RawFeatureValue]("foo") c: RawFeatureValue = foo scala> val d: RefinedFeatureValue = c.map(convert).unify d: RefinedFeatureValue = 101574 这个解决方案并不错,但是看起来它似乎是一个单一的操作足够有用. (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |