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
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