Question

如何用一个Scala流压缩HList来生成成对的HList？具体来说：

import shapeless._

val a = 1.3 :: true :: "z" :: HNil
val b = Stream.from(1)

val ab: (Double, Int) :: (Boolean, Int) :: (String, Int) :: HNil = ???

Answer 1

trait Zipper[A <: HList, B] {
  type Out <: HList
  def zip(a: A, bs: Stream[B]): Out
}

implicit def hnilZipper[B] = new Zipper[HNil, B] {
  type Out = HNil
  def zip(a: HNil, bs: Stream[B]): HNil = HNil
}

implicit def consZipper[Head, Tail <: HList, B](implicit z: Zipper[Tail, B]) = new Zipper[Head :: Tail, B] {
  type Out = (Head, B) :: z.Out
  def zip(a: Head :: Tail, bs: Stream[B]): Out = {
    (a.head, bs.head) :: z.zip(a.tail, bs.tail)
  }
}

def zip[A <: HList, B](a: A, b: Stream[B])(implicit z: Zipper[A, B]): z.Out = z.zip(a, b)

正如Marth所说，存在安全问题，如果Stream短于HList，则这将失败。但是如果您担心的话，您可以很容易地修改它以返回Option[(A1, B) :: (A2, B) :: ...]或(A1, Option[B]) :: (A2, Option[B]) :: ...。

Answer 2

有可能，但是有限制。通常，您必须指定要压缩的Scala集合的所需长度。结果当然是两个HList中较短者的长度。

import shapeless._
import syntax.std.traversable._  // toHList

val a = 1.3 :: true :: "z" :: HNil  // 3 elements

val short_b = Stream.from(1).take(2).toHList[Int::Int::HNil]
val long_b  = Stream.from(7).take(5).toHList[Int::Int::Int::Int::Int::HNil]

toHList返回一个Option[HList]，因此我们将对结果进行map以提取HList进行压缩。

short_b.map(a zip _)  // 2 element result
//res0: Option[(Double, Int) :: (Boolean, Int) :: HNil] =
// Some((1.3,1) :: (true,2) :: HNil)

long_b.map(a zip _)  // 3 element result
//res1: Option[(Double, Int) :: (Boolean, Int) :: (String, Int) :: HNil] =
// Some((1.3,7) :: (true,8) :: (z,9) :: HNil)

如何用简单的Scala集合压缩HList？

2 个答案: