Scala List function for grouping consecutive identical elements
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(开发者_StackOverflow中文版3, 3), List(2, 2, 2))
I would assume there is a simple List function that does this, but am unable to find it.
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.
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)
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.
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)
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.]
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
}
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
}
}
this could be simpler:
val input = List(5, 2, 3, 3, 3, 5, 5, 3, 3, 2, 2, 2)
input groupBy identity values
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