2016-02-12 27 views
6

Le collezioni Scala hanno il metodo sortBy. Questo metodo è stabile?L'ordinamento scala è stabile?

def sortList(source : List[Int]) : List[Int] = 
    source.sortBy(_ % 2) 

Questo esempio conserva sempre l'ordine?

risposta

12

Sì, è stabile. Riferimento dal codice sorgente Scala:

https://github.com/scala/scala/blob/2.11.x/src/library/scala/collection/SeqLike.scala#L627

def sortBy[B](f: A => B)(implicit ord: Ordering[B]): Repr = sorted(ord on f) 

/** Sorts this $coll according to an Ordering. 
* 
* The sort is stable. That is, elements that are equal (as determined by 
* `lt`) appear in the same order in the sorted sequence as in the original. 
* 
* @see [[scala.math.Ordering]] 
* 
* @param ord the ordering to be used to compare elements. 
* @return  a $coll consisting of the elements of this $coll 
*    sorted according to the ordering `ord`. 
*/ 
def sorted[B >: A](implicit ord: Ordering[B]): Repr = { 
    val len = this.length 
    val b = newBuilder 
    if (len == 1) b ++= this 
    else if (len > 1) { 
    b.sizeHint(len) 
    val arr = new Array[AnyRef](len) // Previously used ArraySeq for more compact but slower code 
    var i = 0 
    for (x <- this) { 
     arr(i) = x.asInstanceOf[AnyRef] 
     i += 1 
    } 
    java.util.Arrays.sort(arr, ord.asInstanceOf[Ordering[Object]]) 
    i = 0 
    while (i < arr.length) { 
     b += arr(i).asInstanceOf[A] 
     i += 1 
    } 
    } 
    b.result() 
} 
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