如何在 Scala 中调用一个方法 n 次？

How to call a method n times in Scala?

提问by Jonas

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?

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“功能”方式执行此操作的好方法？

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
    }
}

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.

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功能解决方案。

采纳答案by Travis Brown

A more functional solution would be to use a fold with an immutable queue and Queue's filland dropmethods:

一个功能更强大的解决方案将是使用一个倍的不可变队列和Queue的fill和drop的方法：

 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)
   }
 }

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

或者更好的是，在子类中捕获您的"supply"/"sale"区别Event并避免尴尬的Option[BigDecimal]业务：

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)
}}

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

这并没有直接回答您的问题（如何以指定次数调用函数），但它绝对更惯用。

回答by tenshi

The simplest solution is to use range, I think:

最简单的解决方案是使用范围，我认为：

(1 to n) foreach (x => /* do something */)

But you can also create this small helper function:

但是你也可以创建这个小辅助函数：

implicit def intTimes(i: Int) = new {
    def times(fn: => Unit) = (1 to i) foreach (x => fn)
}

10 times println("hello")

this code will print "hello" 10 times. Implicit conversion intTimesmakes method timesavailable on all ints. So in your case it should look like this:

此代码将打印“你好”10 次。隐式转换intTimes使方法times可用于所有整数。所以在你的情况下它应该是这样的：

event.quantity times queue.enqueue(event.value) 
event.quantity times queue.dequeue() 

回答by oxbow_lakes

Not quite an answer to your question, but if you had an endomorphism (i.e. a transformation A => A), then using scalazyou could use the natural monoid for Endo[A]

不是你问题的答案，但如果你有一个内同态（即转换A => A），那么使用scalaz你可以使用自然幺半群Endo[A]

N times func apply target

So that:

以便：

scala> import scalaz._; import Scalaz._
import scalaz._
import Scalaz._

scala> Endo((_:Int) * 2).multiply(5)
res3: scalaz.Endo[Int] = Endo(<function1>)

scala> res1(3)
res4: Int = 96

回答by Luigi Plinge

import List._

fill(10) { println("hello") }

Simple, built-in, and you get a List of Units as a souvenier!

简单，内置，你会得到一个单位列表作为纪念品！

But you'll never need to call a function multiple times if you're programming functionally.

但是，如果您进行函数式编程，则永远不需要多次调用一个函数。

回答by Alexey Romanov

With recursion:

递归：

def repeat(n: Int)(f: => Unit) { 
  if (n > 0) {
    f
    repeat(n-1)(f)
  }
}

repeat(event.quantity) { queue.enqueue(event.value) }