Scala List function for grouping consecutive ident

问题:

Given e.g.:

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

I'd like to get to:

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

I would assume there is a simple List function that does this, but am unable to find it.

回答1:

This is the trick that I normally use:

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)
}

Actually... It's not, I usually abstract over the collection type and optimize with tail recursion as well, but wanted to keep the answer simple.

回答2:

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

Here's another way.

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

回答3:

Damn Rex Kerr, for writing the answer I'd go for. Since there are minor stylistic differences, here's my take:

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

Since the elements are identical, I didn't bother reversing the sublists.

回答4:

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

Or

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

[Edit]

//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)

回答5:

I have these implementations lying around from working on collections methods. In the end I checked in simpler implementations of inits and tails and left out cluster. Every new method no matter how simple ends up collecting a big tax which is hard to see from the outside. But here's the implementation I didn't use.

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)
}

[Edit: I forget other people won't know the internals. This code is written from inside of TraversableLike, so it wouldn't run out of the box.]

回答6:

Here's a tail-recursive solution inspired by @Kevin Wright and @Landei:

@tailrec
def sliceEqual[A](s: Seq[A], acc: Seq[Seq[A]] = Seq()): Seq[Seq[A]] = {
  s match {
    case fst :: rest =>
      val (l, r) = s.span(fst==)
      sliceEqual(r, acc :+ l)
    case Nil => acc
  }
}

回答7:

Here's a slightly cleaner one:

def groupConsequtive[A](list: List[A]): List[List[A]] = list match {
  case head :: tail =>
    val (t1, t2) = tail.span(_ == head)
    (head :: t1) :: groupConsequtive(t2)
  case _ => Nil  
}

tail-recursive version

@tailrec
def groupConsequtive[A](list: List[A], acc: List[List[A]] = Nil): List[List[A]] = list match {
  case head :: tail =>
    val (t1, t2) = tail.span(_ == head)
    groupConsequtive(t2, acc :+ (head :: t1))
  case _ => acc
}

回答8:

this could be simpler:

val input = List(5, 2, 3, 3, 3, 5, 5, 3, 3, 2, 2, 2)
input groupBy identity values