SortedSet映射并不总是保留结果中的元素排序？

Question

鉴于以下Scala 2.9.2代码:

更新了非工作示例

import collection.immutable.SortedSet

case class Bar(s: String)

trait Foo {
  val stuff: SortedSet[String]
  def makeBars(bs: Map[String, String])
    = stuff.map(k => Bar(bs.getOrElse(k, "-"))).toList
}

case class Bazz(rawStuff: List[String]) extends Foo {
  val stuff = SortedSet(rawStuff: _*)
}


// test it out....
val b = Bazz(List("A","B","C"))
b.makeBars(Map("A"->"1","B"->"2","C"->"3"))
// List[Bar] = List(Bar(1), Bar(2), Bar(3))
// Looks good?

// Make a really big list not in order. This is why we pass it to a SortedSet...    
val data =  Stream.continually(util.Random.shuffle(List("A","B","C","D","E","F"))).take(100).toList
val b2 = Bazz(data.flatten)

// And how about a sparse map...?
val bs = util.Random.shuffle(Map("A" -> "1", "B" -> "2", "E" -> "5").toList).toMap
b2.makeBars(bs)
// res24: List[Bar] = List(Bar(1), Bar(2), Bar(-), Bar(5))

我发现,在某些情况下,makeBars延伸类的方法Foo并没有返回排序列表.事实上,列表排序并不能反映排序SortedSet

关于上面的代码我不知道什么,Scala不会总是将a映射SortedSet到List具有SortedSet排序顺序的元素？

Answer 1

你对隐含的解决方案感到惊讶.

该map方法需要一个CanBuildFrom与目标集合类型兼容的实例(在简单​​情况下,与源集合类型相同)和mapper函数的返回类型.

在特定情况下SortedSet,其隐式CanBuildFrom要求可以使用Ordering[A](其中A是映射器函数的返回类型).当你的map函数返回编译器已经知道如何找到Orderingfor的东西时,你就是好的:

scala> val ss = collection.immutable.SortedSet(10,9,8,7,6,5,4,3,2,1)
ss: scala.collection.immutable.SortedSet[Int] = TreeSet(1, 2, 3, 4, 5, 
                                                        6, 7, 8, 9, 10)

scala> val result1 = ss.map(_ * 2)
result1: scala.collection.immutable.SortedSet[Int] = TreeSet(2, 4, 6, 8, 10, 
                                                            12, 14, 16, 18, 20) 
                 // still sorted because Ordering[Int] is readily available

scala> val result2 = ss.map(_ + " is a number")
result2: scala.collection.immutable.SortedSet[String] = TreeSet(1 is a number, 
                                                                10 is a number, 
                                                                2 is a number, 
                                                                3 is a number, 
                                                                4 is a number, 
                                                                5 is a number, 
                                                                6 is a number, 
                                                                7 is a number, 
                                                                8 is a number, 
                                                                9 is a number) 
// The default Ordering[String] is an "asciibetical" sort, 
// so 10 comes between 1 and 2. :)

但是,当您的映射器函数返回一个未知Ordering的类型时,隐式on SortedSet不匹配(具体而言,没有找到其隐式参数的值),因此编译器看起来"向上"以获得兼容CanBuildFrom并从中找到通用的Set.

scala> case class Foo(i: Int)
defined class Foo

scala> val result3 = ss.map(Foo(_))
result3: scala.collection.immutable.Set[Foo] = Set(Foo(10), Foo(4), Foo(6), Foo(7), Foo(1), Foo(3), Foo(5), Foo(8), Foo(9), Foo(2))

// The default Set is a hash set, therefore ordering is not preserved

当然,你可以通过简单地提供一个实例来解决这个问题Ordering[Foo]:

scala> implicit val fooIsOrdered: Ordering[Foo] = Ordering.by(_.i)
fooIsOrdered: Ordering[Foo] = scala.math.Ordering$$anon$9@7512dbf2

scala> val result4 = ss.map(Foo(_))
result4: scala.collection.immutable.SortedSet[Foo] = TreeSet(Foo(1), Foo(2), 
                                                       Foo(3), Foo(4), Foo(5), 
                                                       Foo(6), Foo(7), Foo(8), 
                                                       Foo(9), Foo(10))
  // And we're back!

最后,请注意,玩具示例通常不会出现问题,因为Scala集合库具有针对小(n <= 6)集和映射的特殊实现.