Scala List函数用于对连续的相同元素进行分组

Question

鉴于例如:

List(5, 2, 3, 3, 3, 5, 5, 3, 3, 2, 2, 2)

我想去:

List(List(5), List(2), List(3, 3, 3), List(5, 5), List(3, 3), List(2, 2, 2))

我会假设有一个简单的List函数来执行此操作,但我无法找到它.

Answer 1

这是我通常使用的技巧:

def split[T](list: List[T]) : List[List[T]] = list match {
  case Nil => Nil
  case h::t => val segment = list takeWhile {h ==}
    segment :: split(list drop segment.length)
}

实际上......不是,我通常会对集合类型进行抽象,并使用尾递归进行优化,但希望保持简单的答案.

Answer 2

val xs = List(5, 2, 3, 3, 3, 5, 5, 3, 3, 2, 2, 2)

这是另一种方式.

(List(xs.take(1)) /: xs.tail)((l,r) =>
  if (l.head.head==r) (r :: l.head) :: l.tail else List(r) :: l
).reverseMap(_.reverse)

Answer 3

该死的雷克斯克尔,写下我想要的答案.由于存在轻微的风格差异,这是我的看法:

list.tail.foldLeft(List(list take 1)) { 
    case (acc @ (lst @ hd :: _) :: tl, el) => 
        if (el == hd) (el :: lst) :: tl 
        else (el :: Nil) :: acc 
}

由于元素相同,我没有打扰逆转子列表.

Answer 4

list.foldRight(List[List[Int]]()){
  (e, l) => l match {
    case (`e` :: xs) :: fs => (e :: e :: xs) :: fs
    case _ => List(e) :: l
  }
}

要么

list.zip(false :: list.sliding(2).collect{case List(a,b) => a == b}.toList)
 .foldLeft(List[List[Int]]())((l,e) => if(e._2) (e._1 :: l.head) :: l.tail 
                                       else List(e._1) :: l ).reverse

[编辑]

//find the hidden way 
//the beauty must be somewhere
//when we talk scala

def split(l: List[Int]): List[List[Int]] = 
  l.headOption.map{x => val (h,t)=l.span{x==}; h::split(t)}.getOrElse(Nil)

Answer 5

我在处理集合方法时有这些实现.最后,我检查了更简单的inits和tails实现,并省略了集群.无论多么简单,每种新方法最终都会收取很难从外部看到的大税.但这是我没有使用的实现.

import generic._
import scala.reflect.ClassManifest
import mutable.ListBuffer
import annotation.tailrec
import annotation.unchecked.{ uncheckedVariance => uV }

def inits: List[Repr] = repSequence(x => (x, x.init), Nil)
def tails: List[Repr] = repSequence(x => (x, x.tail), Nil)
def cluster[A1 >: A : Equiv]: List[Repr] =
  repSequence(x => x.span(y => implicitly[Equiv[A1]].equiv(y, x.head)))

private def repSequence(
  f: Traversable[A @uV] => (Traversable[A @uV], Traversable[A @uV]),
  extras: Traversable[A @uV]*): List[Repr] = {

  def mkRepr(xs: Traversable[A @uV]): Repr = newBuilder ++= xs result
  val bb = new ListBuffer[Repr]

  @tailrec def loop(xs: Repr): List[Repr] = {
    val seq = toCollection(xs)
    if (seq.isEmpty)
      return (bb ++= (extras map mkRepr)).result

    val (hd, tl) = f(seq)
    bb += mkRepr(hd)
    loop(mkRepr(tl))
  }

  loop(self.repr)
}

[编辑:我忘了其他人不会知道内幕.此代码是从TraversableLike内部编写的,因此它不会开箱即用.]