如何在 Scala 中调用一个方法 n 次? [英] How to call a method n times in Scala?

问题描述

我有一个案例，我想调用一个方法 n 次，其中 n 是一个 Int.在 Scala 中是否有一种以函数式"方式执行此操作的好方法?

I have a case where I want to call a method n times, where n is an Int. Is there a good way to do this in a "functional" way in Scala?

case class Event(name: String, quantity: Int, value: Option[BigDecimal])

// a list of events
val lst = List(
    Event("supply", 3, Some(new java.math.BigDecimal("39.00"))),
    Event("sale", 1, None),
    Event("supply", 1, Some(new java.math.BigDecimal("41.00")))
    )

// a mutable queue
val queue = new scala.collection.mutable.Queue[BigDecimal]

lst.map { event =>
    event.name match {
        case "supply" => // call queue.enqueue(event.value) event.quantity times
        case "sale" =>   // call queue.dequeue() event.quantity times
    }
}

我认为闭包是一个很好的解决方案，但我无法让它工作.我也尝试过 for 循环，但这不是一个漂亮的函数式解决方案.

I think a closure is a good solution for this, but I can't get it working. I have also tried with a for-loop, but it's not a beautiful functional solution.

推荐答案

一个更实用的解决方案是使用具有不可变队列和 Queue 的 fill 的折叠和 drop 方法:

A more functional solution would be to use a fold with an immutable queue and Queue's fill and drop methods:

 val queue = lst.foldLeft(Queue.empty[Option[BigDecimal]]) { (q, e) =>
   e.name match {
     case "supply" => q ++ Queue.fill(e.quantity)(e.value)
     case "sale"   => q.drop(e.quantity)
   }
 }

或者更好的是，在 Event 的子类中捕捉你的 "supply"/"sale" 区别，避免尴尬的 Option[BigDecimal] 业务:

Or even better, capture your "supply"/"sale" distinction in subclasses of Event and avoid the awkward Option[BigDecimal] business:

sealed trait Event { def quantity: Int }
case class Supply(quantity: Int, value: BigDecimal) extends Event
case class Sale(quantity: Int) extends Event

val lst = List(
  Supply(3, BigDecimal("39.00")),
  Sale(1),
  Supply(1, BigDecimal("41.00"))
)

val queue = lst.foldLeft(Queue.empty[BigDecimal]) { (q, e) => e match {
  case Sale(quantity)          => q.drop(quantity)
  case Supply(quantity, value) => q ++ Queue.fill(quantity)(value)
}}

这并没有直接回答您的问题(如何以指定的次数调用函数)，但它绝对更符合习惯.

This doesn't directly answer your question (how to call a function a specified number of times), but it's definitely more idiomatic.

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