scala - 找不到 Ordered[T] 类型的证据参数的隐式值

Question

我现在实际上被阻止了大约 4 个小时。我想获得按其 int 值排序的 Pairs[String, Int] 列表。函数分区工作正常，bestN 也应该如此，但是当将它加载到我的解释器中时，我得到:

<console>:15: error: could not find implicit value for evidence parameter of type Ordered[T]

在我的谓词上。有人看到问题是什么吗？这一刻我真的很绝望……

这是代码:

def partition[T : Ordered](pred: (T)=>Boolean, list:List[T]): Pair[List[T],List[T]] = {
    list.foldLeft(Pair(List[T](),List[T]()))((pair,x) => if(pred(x))(pair._1, x::pair._2) else (x::pair._1, pair._2))
}

def bestN[T <% Ordered[T]](list:List[T], n:Int): List[T] = {
    list match {
        case pivot::other => {
            println("pivot: " + pivot)
            val (smaller,bigger) = partition(pivot <, list)
            val s = smaller.size
            println(smaller)
            if (s == n) smaller 
            else if (s+1 == n) pivot::smaller
            else if (s < n) bestN(bigger, n-s-1) 
            else bestN(smaller, n)
        }
        case Nil => Nil
    }
}

class OrderedPair[T, V <% Ordered[V]] (t:T, v:V) extends Pair[T,V](t,v) with Ordered[OrderedPair[T,V]] {
    def this(p:Pair[T,V]) = this(p._1, p._2)
    override def compare(that:OrderedPair[T,V]) : Int = this._2.compare(that._2)
}

编辑:第一个函数通过将谓词应用于每个成员将列表分成两个，bestN 函数应该返回列表列表中最低 n 个成员的列表。这个类(class)是为了让 Pairs 具有可比性，在这种情况下，我想做的是:

val z = List(Pair("alfred",1),Pair("peter",4),Pair("Xaver",1),Pair("Ulf",2),Pair("Alfons",6),Pair("Gulliver",3))

使用这个给定的列表，我想获得例如:

bestN(z, 3)

结果:

(("alfred",1), ("Xaver",1), ("Ulf",2))

Answer 1

看起来您的分区函数不需要 Ordered T，因为它只是调用谓词函数。

以下不起作用(大概)，而只是编译。代码审查的其他问题是额外的大括号和类似的东西。

package evident

object Test extends App {

  def partition[T](pred: (T)=>Boolean, list:List[T]): Pair[List[T],List[T]] = {
    list.foldLeft(Pair(List[T](),List[T]()))((pair,x) => if(pred(x))(pair._1, x::pair._2) else (x::pair._1, pair._2))
  }

  def bestN[U,V<%Ordered[V]](list:List[(U,V)], n:Int): List[(U,V)] = {
    list match {
      case pivot::other => {
        println(s"pivot: $pivot and rest ${other mkString ","}")
        def cmp(a: (U,V), b: (U,V)) = (a: OrderedPair[U,V]) < (b: OrderedPair[U,V])
        val (smaller,bigger) = partition(((x:(U,V)) => cmp(x, pivot)), list)
        //val (smaller,bigger) = list partition ((x:(U,V)) => cmp(x, pivot))
        println(s"smaller: ${smaller mkString ","} and bigger ${bigger mkString ","}")
        val s = smaller.size
        if (s == n) smaller
        else if (s+1 == n) pivot::smaller
        else if (s < n) bestN(bigger, n-s-1)
        else bestN(smaller, n)
      }
      case Nil => Nil
    }
  }

  implicit class OrderedPair[T, V <% Ordered[V]](tv: (T,V)) extends Pair(tv._1, tv._2) with Ordered[OrderedPair[T,V]] {
    override def compare(that:OrderedPair[T,V]) : Int = this._2.compare(that._2)
  }

  val z = List(Pair("alfred",1),Pair("peter",4),Pair("Xaver",1),Pair("Ulf",2),Pair("Alfons",6),Pair("Gulliver",3))
  println(bestN(z, 3))
}

我发现分区函数难以阅读；你需要一个函数来分割所有的括号。这里有几个公式，它们也使用过滤器接受的结果向左走，拒绝向右走的惯例。

def partition[T](p: T => Boolean, list: List[T]) = 
  ((List.empty[T], List.empty[T]) /: list) { (s, t) =>
    if (p(t)) (t :: s._1, s._2) else (s._1, t :: s._2)
  }
def partition2[T](p: T => Boolean, list: List[T]) =
  ((List.empty[T], List.empty[T]) /: list) {
    case ((is, not), t) if p(t) => (t :: is, not)
    case ((is, not), t)         => (is, t :: not)
  }
// like List.partition
def partition3[T](p: T => Boolean, list: List[T]) = {
  import collection.mutable.ListBuffer
  val is, not = new ListBuffer[T]
  for (t <- list) (if (p(t)) is else not) += t
  (is.toList, not.toList)
}

这可能更接近原始代码的意图:

def bestN[U, V <% Ordered[V]](list: List[(U,V)], n: Int): List[(U,V)] = {
  require(n >= 0)
  require(n <= list.length)
  if (n == 0) Nil
  else if (n == list.length) list
  else list match {
    case pivot :: other =>
      println(s"pivot: $pivot and rest ${other mkString ","}")
      def cmp(x: (U,V)) = x._2 < pivot._2
      val (smaller, bigger) = partition(cmp, other)     // other partition cmp
      println(s"smaller: ${smaller mkString ","} and bigger ${bigger mkString ","}")
      val s = smaller.size
      if (s == n) smaller
      else if (s == 0) pivot :: bestN(bigger, n - 1)
      else if (s < n) smaller ::: bestN(pivot :: bigger, n - s)
      else bestN(smaller, n)
    case Nil => Nil
  }
}

箭头符号更常见:

  val z = List(
    "alfred" -> 1,
    "peter" -> 4,
    "Xaver" -> 1,
    "Ulf" -> 2,
    "Alfons" -> 6,
    "Gulliver" -> 3
  )