惯用Scala懒惰地产生组合的方式

发布时间：2020-12-16 19:20:43 所属栏目：安全来源：网络整理

导读：我想生成一些值的组合,如下面的代码所示： object ContinueGenerate { val foods = List("A","B","C") val places = List("P1","P2","P3") val communities = List("C1","C2","C3","C4") case class Combination(food: String,place: String,community: Stri

我想生成一些值的组合,如下面的代码所示：

object ContinueGenerate {

  val foods = List("A","B","C")
  val places = List("P1","P2","P3")
  val communities = List("C1","C2","C3","C4")

  case class Combination(food: String,place: String,community: String)

  def allCombinations() = {
    for {
      food <- foods; place <- places; community <- communities
    } yield Combination(food,place,community)
  }

  def main(args: Array[String]) {
    allCombinations foreach println
  }

}

然而,这种方法的问题在于,所有数据都是立即生成的.当食物,地方和社区的规模变得非常大时,这是一个大问题.除了这三个之外,还可能有其他参数.

因此,我希望能够以连续样式生成组合,以便仅在请求时生成组合.

什么是惯用的Scala方式呢？

解决方法

您可以通过在每个列表上使用View来执行此操作.在下面的代码中,我添加了一个副作用,因此在为每个元素调用yield时它是可见的.

val foods = List("A","C")
val places = List("P1","P3")
val communities = List("C1","C4")

case class Combination(food: String,community: String)

def allCombinations() =
  for {
    food <- foods; place <- places; community <- communities
  } yield {
    val comb = Combination(food,community)
    println(comb)
    comb
  }

//Prints all items
val combinations = allCombinations()

def allCombinationsView() =
  for {
    //Use a view of each list
    food <- foods.view; place <- places.view; community <- communities.view
  } yield {
    val comb = Combination(food,community)
    println(comb)
    comb
  }
//Prints nothing
val combinationsView = allCombinationsView()

//Prints 5 items
val five = combinationsView.take(5).toList

（编辑：李大同）

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