Question

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foods，places和communities的大小变得非常大时，这是一个大问题。除了这三个之外，还可能有其他参数。

所以我希望能够以连续样式生成组合，这样只有在请求时才会生成组合。

Scala的惯用方法是什么？

Answer 1

您使用流：

object ContinueGenerate {

  val foods = Stream("A", "B", "C")
  val places = Stream("P1", "P2", "P3")
  val communities = Stream("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
  }

}

A Stream缓存所有数据。如果您只想迭代一次，请改用Iterator，这应该是垃圾收集已遍历的元素。

Answer 2

您可以在每个列表上使用View来执行此操作。在下面的代码中，我添加了一个副作用，以便在为每个元素调用yield时可见。

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 {
    val comb = Combination(food, place, 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, place, community)
    println(comb)
    comb
  }
//Prints nothing
val combinationsView = allCombinationsView()

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

惯用Scala懒惰地产生组合的方式

2 个答案: