scala – Shapeless：从副产品到不同副产品的地图

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.