3

例如，对于任何给定的列表或数组

val list = (1 to 3).toList
val array = (1 to 3).toArray

以及从集合类型映射到集合类型的给定函数，例如

def f(v: Int): Int = v + 10

如何应用于orf的第 i 个元素使得listarray

list.myApply(f, ith = 2)
res: List(1,12,3)

并且

array.myApply(f, ith = 2)
res: Array(1,12,3)
4

3 回答 3

7

tl;博士

import scala.collection.SeqLike
import scala.collection.generic.CanBuildFrom

implicit class Seq_[A, Repr, 
    S : ({type L[X] = X => SeqLike[A, Repr]})#L](seq: S) {

  def myApply[B >: A, That](f: A => B, ith: Int)
      (implicit bf: CanBuildFrom[Repr, B, That]): That =
    seq.updated(ith - 1, f(seq(ith - 1)))
}

讨论

一个简单的近似：

implicit class Seq_[A](seq: Seq[A]) {
  def myApply(f: A => A, ith: Int): Seq[A] =
    seq.updated(ith - 1, f(seq(ith - 1)))
}

示例用法：

scala> (1 to 3).toList.myApply(_ + 10, ith = 2)
res: Seq[Int] = List(1, 12, 3)

尝试的实际解决方案：

implicit class Seq_[A, Repr <: SeqLike[A, Repr]](seq: Repr) {
  def myApply[B >: A, That](f: A => B, ith: Int)
                           (implicit bf: CanBuildFrom[Repr, B, That]): That =
    seq.updated(ith - 1, f(seq(ith - 1)))
}

不幸的是，隐式不起作用。我不确定为什么。

scala> Seq_[Int, List[Int]]((1 to 3).toList).myApply(_ + 10, ith = 2)
res: List[Int] = List(1, 12, 3)

scala> Seq_[Int, List[Int]]((1 to 3).toList).myApply(_.toString + "*", ith = 2)
res: List[Any] = List(1, 2*, 3)

编辑：修复它！

implicit class Seq_[A, Repr](seq: SeqLike[A, Repr]) {
  def myApply[B >: A, That](f: A => B, ith: Int)
                           (implicit bf: CanBuildFrom[Repr, B, That]): That =
    seq.updated(ith - 1, f(seq(ith - 1)))
}

例子：

scala> (1 to 3).toList.myApply(_ + 10, ith = 2)
res: List[Int] = List(1, 12, 3)

scala> (1 to 3).toVector.myApply(Math.pow(2, _), ith = 3)
res: scala.collection.immutable.Vector[AnyVal] = Vector(1, 2, 8.0)

但我刚刚意识到你也希望它适用于Array，但事实并非SeqLike如此，所以让我再想一想......

啊，有一个从toPredef的隐式转换，它是 的子类型，所以我们只需要使用视图绑定。ArrayArrayOpsSeqLike

implicit class Seq_[A, Repr <% SeqLike[A, Repr]](seq: Repr) {
  def myApply[B >: A, That](f: A => B, ith: Int)
                           (implicit bf: CanBuildFrom[Repr, B, That]): That =
    seq.updated(ith - 1, f(seq(ith - 1)))
}

最后我们有正确的行为：

scala> (1 to 3).toList.myApply(_ + 10, ith = 2)
res: List[Int] = List(1, 12, 3)

scala> (1 to 3).toArray.myApply(Math.pow(2, _), ith = 3)
res: Array[AnyVal] = Array(1, 2, 8.0)

再次编辑 - samthebest 通知我视图边界已被弃用，因此使用本指南，我们可以将其替换为外观非常难看的上下文边界。

implicit class Seq_[A, Repr, 
    S : ({type L[X] = X => SeqLike[A, Repr]})#L](seq: S) {

  def myApply[B >: A, That](f: A => B, ith: Int)
      (implicit bf: CanBuildFrom[Repr, B, That]): That =
    seq.updated(ith - 1, f(seq(ith - 1)))
}
于 2014-09-12T05:07:07.713 回答
3

有人刚刚问过补丁，所以也许这是重复的：

scala> val list = (1 to 3).toList
list: List[Int] = List(1, 2, 3)

scala> def f(v: Int): Int = v + 10
f: (v: Int)Int

scala> def g(is: Seq[Int], f: Int => Int, i: Int) = is.patch(i, Seq(f(is(i))), 1)
g: (is: Seq[Int], f: Int => Int, i: Int)Seq[Int]

scala> g(list, f, 1)
res1: Seq[Int] = List(1, 12, 3)

概括一个 smidgen：

scala> def g(is: collection.Seq[Int], f: Int => Int, i: Int, count: Int = 1) = is.patch(i, is.slice(i, i + count) map f, count)
g: (is: Seq[Int], f: Int => Int, i: Int, count: Int)Seq[Int]

scala> g(list, f, 1)
res2: Seq[Int] = List(1, 12, 3)

scala> g(list, f, 1, 2)
res3: Seq[Int] = List(1, 12, 13)

这是我的第一次尝试，正如 Chris 提示的那样：

scala> def g(is: collection.mutable.Seq[Int], f: Int => Int, i: Int) = is(i) = f(is(i))
g: (is: scala.collection.mutable.Seq[Int], f: Int => Int, i: Int)Unit

scala> val as = (1 to 10).toArray
as: Array[Int] = Array(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)

scala> g(as, f, 1)

scala> as
res7: Array[Int] = Array(1, 12, 3, 4, 5, 6, 7, 8, 9, 10)

正如克里斯所说：

scala> def g(is: collection.Seq[Int], f: Int => Int, i: Int) = is.updated(i, f(is(i)))
g: (is: Seq[Int], f: Int => Int, i: Int)Seq[Int]

晚安，格雷西。

于 2014-09-12T05:16:21.517 回答
0

使用隐式向现有集合添加其他方法非常复杂。如果使用外部方法OK，这是一个解决方案。几乎在那里，但不完全。Scala IDE 工作表中的示例

object SeqOps {
  def applyToith(col: Seq[Int], f: Int => Int, ith: Int): Seq[Int] = {
    val indexCol = col.zipWithIndex
    indexCol.map {
      a => if (a._2 == ith) f(a._1) else a._1
    }
  }                                       //> applyToith: (col: Seq[Int], f: Int => Int, ith: Int)Seq[Int]

  def f(i: Int) = i + 10                  //> f: (i: Int)Int
  val l = List(1, 2, 3)                   //> l  : List[Int] = List(1, 2, 3)
  applyToith(l, f _, 0)                   //> res0: Seq[Int] = List(11, 2, 3)

  val a = Array(1, 2, 3)                  //> a  : Array[Int] = Array(1, 2, 3)
  applyToith(a, f _, 1)                   //> res1: Seq[Int] = ArrayBuffer(1, 12, 3)

  val v = Vector(1, 2, 3)                 //> v  : scala.collection.immutable.Vector[Int] = Vector(1, 2, 3)
  applyToith(v, f _, 2)                   //> res2: Seq[Int] = Vector(1, 2, 13)

}

在数组的情况下，它返回一个 ArrayBuffer 而不是 Array。对于所有其他 Seq 类型都可以正常工作。我尝试了许多其他组合，但没有解决这个问题。

val a : Seq[Int] = Array(1, 2)
a.zipWithIndex

此 zipWithIndex 返回一个 ArrayBuffer，但如果val a: Array[Int]使用，则zipWithIndex返回一个 Array。

我猜想将 Java 数组改装成集合的变幻莫测。

于 2014-09-12T05:10:44.087 回答