Scala：如何根据标签过滤不精打细的HList

Question

我有以下几点：

sealed trait baseData {
  def weight: Int
  def priority: Int
} 

sealed trait moreData {
  def weight: Int
  def priority: Int
  def t: String
  def id: String
} 

case class data1(override val weight: Int, override val priority: Int) extends baseData 
case class moreData1 (override val weight:Int, override val priority: Int, override val t: String, override val id: String)extends moreData

并从诸如此类的案例类生成HList：

val h1 = LabelledGeneric[data1].to(data1(1,2))
val h2 = LabelledGeneric[moreData1].to(moreData1(3,4,"a","b"))

如何修剪或过滤h2 ，使其仅保留h1存在的字段？ 我感觉我需要某种类似val filtered = h2.foldRight(HNil)(keepFunc) ，但是还无法弄清楚如何编写keepFunc 。 有任何想法吗？

Answer 1

您可以根据键进行过滤。 该代码不是通用的，但希望足以说明这一概念。

val gen1 = LabelledGeneric[data1]
val gen2 = LabelledGeneric[moreData1]

val h1 = gen1.to(data1(1,2))
val h2 = gen2.to(moreData1(3,4,"a","b"))

val keys1 = Keys[gen1.Repr].apply
val keys2 = Keys[gen2.Repr].apply

object pair extends Poly2 {
  implicit def default[T, U] = at[T, U]((_, _))
}

object keep extends Poly2 {
  implicit def keepFunc[T, K, L <: HList] =
    at[(T, K), L] { case ((t, key), l) =>
      if (keys1.toList.contains(key)) t :: l else l
    }
}

val filtered = h2.zipWith(keys2)(pair).foldRight(HNil)(keep)