Question

我有一组以一种模式重复出现的整数：

val repeatingSequence = List(1,2,3,1,2,3,4,1,2,1,2,3,4,5)

我希望在模式重复时将List列出来;在这种情况下，当序列返回1时：

val groupedBySequence = List(List(1,2,3), List(1,2,3,4), List(1,2), List(1,2,3,4,5))

请注意，当序列跳回到1时，我正在分组，但序列可以是任意长度。我和我的同事通过添加一个名为'groupWhen'

的附加方法解决了这个问题

class IteratorW[A](itr: Iterator[A]) {
  def groupWhen(fn: A => Boolean): Iterator[Seq[A]] = {
    val bitr = itr.buffered
    new Iterator[Seq[A]] {
      override def hasNext = bitr.hasNext
      override def next = {
        val xs = collection.mutable.ListBuffer(bitr.next)
        while (bitr.hasNext && !fn(bitr.head)) xs += bitr.next
        xs.toSeq
      }
    }
  }
}
implicit def ToIteratorW[A](itr: Iterator[A]): IteratorW[A] = new IteratorW(itr)

> repeatingSequence.iterator.groupWhen(_ == 1).toSeq
List(List(1,2,3), List(1,2,3,4), List(1,2), List(1,2,3,4,5))

但是，我们都觉得这里有一个更优雅的解决方案潜伏在收藏库中。

Answer 1

给定一个迭代器itr，这将解决问题：

val head = iter.next()
val out = (
  Iterator continually {iter takeWhile (_ != head)}
  takeWhile {!_.isEmpty}
  map {head :: _.toList}
).toList

Answer 2

众所周知，折叠可以做任何事情......;）

  val rs = List(1,2,3,1,2,3,4,1,2,1,2,3,4,5)
  val res = (rs++List(1)).foldLeft((List[List[Int]](),List[Int]()))((acc,e) => acc match {
    case (res,subl) => {
      if (e == 1) ((subl.reverse)::res,1::Nil) else (res, e::subl)
    }
  })
  println(res._1.reverse.tail)

请将此视为模糊Scala竞赛的参赛作品而不是真正的答案。

Answer 3

这是我使用span打击的一个不太优雅的解决方案：

def groupWhen[A](fn: A => Boolean)(xs: List[A]): List[List[A]] = {
  xs.span(!fn(_)) match {
    case (Nil, Nil) => Nil
    case (Nil, z::zs) => groupWhen(fn)(zs) match {
      case ys::yss => (z::ys) :: yss
      case Nil => List(List(z))
    }
    case (ys, zs) => ys :: groupWhen(fn)(zs)
  }
}

scala> groupWhen[Int](_==1)(List(1,2,3,1,2,3,4,1,2,3,4,5))
res39: List[List[Int]] = List(List(1, 2, 3), List(1, 2, 3, 4), List(1, 2, 3, 4, 5))

scala> groupWhen[Int](_==1)(List(5,4,3,2,1,2,3,1,2,3,4,1,2,3,4,5))
res40: List[List[Int]] = List(List(5, 4, 3, 2), List(1, 2, 3), List(1, 2, 3, 4), List(1, 2, 3, 4, 5))

Answer 4

import scala.collection.mutable.ListBuffer
import scala.collection.breakOut

val repeatingSequence = List(1,2,3,1,2,3,4,1,2,1,2,3,4,5)
val groupedBySequence: List[List[Int]] = repeatingSequence.foldLeft(ListBuffer[ListBuffer[Int]]()) {
  case (acc, 1) => acc += ListBuffer(1)
  case (acc, n) => acc.last += n; acc
}.map(_.toList)(breakOut)

如何在Scala中对可变长度的重复序列进行分组

4 个答案: