Question

我正在寻找一种在给定的成对条件下分裂的集合方法，例如

val x = List("a" -> 1, "a" -> 2, "b" -> 3, "c" -> 4, "c" -> 5)

implicit class RichIterableLike[A, CC[~] <: Iterable[~]](it: CC[A]) {
  def groupWith(fun: (A, A) => Boolean): Iterator[CC[A]] = new Iterator[CC[A]] {
    def hasNext: Boolean = ???

    def next(): CC[A] = ???
  }
}

assert(x.groupWith(_._1 != _._1).toList == 
  List(List("a" -> 1, "a" -> 2), List("b" -> 3), List("c" -> 4, "c" -> 5))
)

所以这是一种递归的span。

虽然我能够实现???，但我想知道

如果我正在监督的馆藏中已存在某些内容
该方法应该被称为什么; groupWith听起来不对劲。它应该简洁，但不知何故反映出函数参数对成对进行操作。我想，groupWhere会更接近，但仍然不清楚。
实际上我想在使用groupWith时，谓词逻辑应该被反转，所以我会使用x.groupWith(_._1 == _._1)
关于类型的想法。返回Iterator[CC[A]]看起来对我来说是合理的。也许应该花费CanBuildFrom并返回Iterator[To]？

Answer 1

所以这是我的建议。我坚持groupWith，因为spans在我看来不是很具描述性。确实groupBy具有非常不同的语义，但是grouped(size: Int)是相似的。

我试图纯粹基于现有迭代器的组合来创建我的迭代器，但这变得很混乱，所以这里是更低级别的版本：

import scala.collection.generic.CanBuildFrom
import scala.annotation.tailrec
import language.higherKinds

object Extensions {
  private final class GroupWithIterator[A, CC[~] <: Iterable[~], To](
      it: CC[A], p: (A, A) => Boolean)(implicit cbf: CanBuildFrom[CC[A], A, To])
    extends Iterator[To] {

    private val peer      = it.iterator
    private var consumed  = true
    private var elem      = null.asInstanceOf[A]

    def hasNext: Boolean = !consumed || peer.hasNext

    private def pop(): A = {
      if (!consumed) return elem
      if (!peer.hasNext)
        throw new NoSuchElementException("next on empty iterator")

      val res   = peer.next()
      elem      = res
      consumed  = false
      res
    }

    def next(): To = {
      val b = cbf()

      @tailrec def loop(pred: A): Unit = {
        b       += pred
        consumed = true
        if (!peer.isEmpty) {
          val succ = pop()
          if (p(pred, succ)) loop(succ)
        }
      }

      loop(pop())
      b.result()
    }
  }

  implicit final class RichIterableLike[A, CC[~] <: Iterable[~]](val it: CC[A])
    extends AnyVal {
    /** Clumps the collection into groups based on a predicate which determines
      * if successive elements belong to the same group.
      *
      * For example:
      * {{
      *   val x = List("a", "a", "b", "a", "b", "b")
      *   x.groupWith(_ == _).to[Vector]
      * }}
      *
      * produces `Vector(List("a", "a"), List("b"), List("a"), List("b", "b"))`.
      *
      * @param p    a function which is evaluated with successive pairs of
      *             the input collection. As long as the predicate holds
      *             (the function returns `true`), elements are lumped together.
      *             When the predicate becomes `false`, a new group is started.
      *
      * @param cbf  a builder factory for the group type
      * @tparam To  the group type
      * @return     an iterator over the groups.
      */
    def groupWith[To](p: (A, A) => Boolean)
                     (implicit cbf: CanBuildFrom[CC[A], A, To]): Iterator[To] =
      new GroupWithIterator(it, p)
  }
}

也就是说，谓词被反转而不是问题。

import Extensions._
val x = List("a" -> 1, "a" -> 2, "b" -> 3, "c" -> 4, "c" -> 5)
x.groupWith(_._1 == _._1).to[Vector]
// -> Vector(List((a,1), (a,2)), List((b,3)), List((c,4), (c,5)))

Answer 2

你可以通过折叠来实现它吗？这是一个未经优化的版本：

  def groupWith[A](ls: List[A])(p: (A, A) => Boolean): List[List[A]] = 
    ls.foldLeft(List[List[A]]()) { (acc, x) => 
      if(acc.isEmpty)
        List(List(x))
      else
        if(p(acc.last.head, x))
          acc.init ++ List(acc.last ++ List(x))
        else 
          acc ++ List(List(x))
    }

  val x = List("a" -> 1, "a" -> 2, "b" -> 3, "c" -> 4, "c" -> 5, "a" -> 4)    
  println(groupWith(x)(_._1 == _._1))
  //List(List((a,1), (a,2)), List((b,3)), List((c,4), (c,5)), List((a,4)))

Answer 3

您还可以编写使用tailrec / pattern匹配的版本：

  def groupWith[A](s: Seq[A])(p: (A, A) => Boolean): Seq[Seq[A]] = {
    @tailrec
    def rec(xs: Seq[A], acc: Seq[Seq[A]] = Vector.empty): Seq[Seq[A]] = {
      (xs.headOption, acc.lastOption) match {
        case (None, _) => acc
        case (Some(a), None) => rec(xs.tail, acc :+ Vector(a))
        case (Some(a), Some(group)) if p(a, group.last) => rec(xs.tail, acc.init :+ (acc.last :+ a))
        case (Some(a), Some(_)) => rec(xs.tail, acc :+ Vector(a))
      }
    }

    rec(s)
  }

关于在给定谓词的情况下分割多次的集合方法的想法

3 个答案: