[英]could not find implicit value for evidence parameter of type Ordered[T]
我實際上已經被封鎖了大約4個小時了。 我想得到一個按其int值排序的對象列表[String,Int]。 函數partiotion工作正常,所以應該是bestN,但是當把它加載到我的解釋器中時,我得到:
<console>:15: error: could not find implicit value for evidence parameter of type Ordered[T]
在我的謂詞上。 有人看到問題是什么嗎? 我此刻真的很絕望......
這是代碼:
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)
}
編輯:第一個函數通過將謂詞應用於每個成員將List分為兩個,bestN函數應返回列表列表中最低n個成員的List。 這個班級可以使Pairs具有可比性,在這種情況下我想做的是:
val z = List(Pair("alfred",1),Pair("peter",4),Pair("Xaver",1),Pair("Ulf",2),Pair("Alfons",6),Pair("Gulliver",3))
有了這個給定的List我想得到例如:
bestN(z, 3)
結果:
(("alfred",1), ("Xaver",1), ("Ulf",2))
看起來您的分區函數不需要Ordered T,因為它只調用謂詞函數。
以下不起作用(大概)但僅僅編譯。 代碼審查的其他事項將是額外的括號和類似的東西。
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))
}
我發現分區功能難以閱讀; 你需要一個功能來分割所有的parens。 這里有幾個公式,它們也使用過濾器接受的結果左邊的約定,拒絕正確。
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)
}
這可能更接近原始代碼的意圖:
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
}
}
箭頭符號更常見:
val z = List(
"alfred" -> 1,
"peter" -> 4,
"Xaver" -> 1,
"Ulf" -> 2,
"Alfons" -> 6,
"Gulliver" -> 3
)
我懷疑我錯過了什么,但無論如何我會發布一些代碼。
對於bestN
,你知道你可以這樣做嗎?
val listOfPairs = List(Pair("alfred",1),Pair("peter",4),Pair("Xaver",1),Pair("Ulf",2),Pair("Alfons",6),Pair("Gulliver",3))
val bottomThree = listOfPairs.sortBy(_._2).take(3)
哪個給你:
List((alfred,1), (Xaver,1), (Ulf,2))
對於partition
功能,您可以這樣做(假設您希望所有對都低於4):
val partitioned = listOfPairs.partition(_._2 < 4)
給出(左邊全部低於4,右邊全部更高):
(List((alfred,1), (Xaver,1), (Ulf,2), (Gulliver,3)),List((peter,4), (Alfons,6)))
只是與您分享:這有效! 非常感謝所有幫助過我的人,你們都很棒!
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 => {
def cmp(a: (U,V), b: (U,V)) = (a: OrderedPair[U,V]) <= (b: OrderedPair[U,V])
val (smaller,bigger) = partition(((x:(U,V)) => cmp(pivot, x)), list)
val s = smaller.size
//println(n + " :" + s)
//println("size:" + smaller.size + "Pivot: " + pivot + " Smaller part: " + smaller + " bigger: " + bigger)
if (s == n) smaller
else if (s+1 == n) pivot::smaller
else if (s < n) bestN(bigger, n-s)
else bestN(smaller, n)
}
case Nil => Nil
}
}
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)
}
implicit final def OrderedPair[T, V <% Ordered[V]](p : Pair[T, V]) : OrderedPair[T,V] = new OrderedPair(p)
val z = List(Pair("alfred",1),Pair("peter",1),Pair("Xaver",1),Pair("Ulf",2),Pair("Alfons",6),Pair("Gulliver",3))
println(bestN(z, 3))
println(bestN(z, 4))
println(bestN(z, 1))
}
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