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迭代循环方式

发布于 2024-09-10 11:39:53 字数 68 浏览 9 评论 0原文

我需要通过循环方式迭代列表。我还需要向列表中添加新元素并迭代所有元素（旧元素和新元素），我该怎么做？他们有什么数据结构吗？

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落花随流水 2024-09-17 11:39:53

一种选择是使用 Stream 类创建惰性、循环、无限序列：

scala> val values = List(1, 2, 3)
values: List[Int] = List(1, 2, 3)

scala> Stream.continually(values.toStream).flatten.take(9).toList
res2: List[Int] = List(1, 2, 3, 1, 2, 3, 1, 2, 3)

或者这样：

val values = List(1, 2, 3)

def circularStream(values: List[Int],
                   remaining: List[Int] = List()): Stream[Int] = {

  if (remaining.isEmpty)
    circularStream(values,values)
  else
    Stream.cons(remaining.head, circularStream(values, remaining.drop(1)))
}

circularStream(values).take(9).toList //Same result as example #1

One option is to use the Stream class to create a lazy, circular, infinite sequence:

scala> val values = List(1, 2, 3)
values: List[Int] = List(1, 2, 3)

scala> Stream.continually(values.toStream).flatten.take(9).toList
res2: List[Int] = List(1, 2, 3, 1, 2, 3, 1, 2, 3)

or this way:

val values = List(1, 2, 3)

def circularStream(values: List[Int],
                   remaining: List[Int] = List()): Stream[Int] = {

  if (remaining.isEmpty)
    circularStream(values,values)
  else
    Stream.cons(remaining.head, circularStream(values, remaining.drop(1)))
}

circularStream(values).take(9).toList //Same result as example #1

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初相遇 2024-09-17 11:39:53

def forever:Stream[Int] = Stream(1,2,3) append forever

def forever:Stream[Int] = Stream(1,2,3) append forever

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与他有关 2024-09-17 11:39:53

这种东西确实应该出现在标准流库中，但似乎并非如此。 dbryne 的流答案效果很好，或者如果您更喜欢它的理解形式，

val listToRepeat:List[Foo]
val forever:Stream[Foo] = for(x<-Stream.continually(1); y<-listToRepeat) yield y

即使您忽略了该值，第一个流生成器也能让事情永远进行下去。第二个生成器隐式地扁平化为您想要的无限流。

This sort of thing really deserves to be in standard stream library, but doesn't appear to be. dbryne's answer with a stream works well, or if you prefer it in for-comprehension form

val listToRepeat:List[Foo]
val forever:Stream[Foo] = for(x<-Stream.continually(1); y<-listToRepeat) yield y

The first stream generator keeps things going forever even though you are ignoring the value. The second generator gets implicitly flattened into the infinite stream you want.

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浊酒尽余欢 2024-09-17 11:39:53

我想也许这就是你想要的；即使您正在迭代列表，也能够将新元素添加到列表中。代码很丑陋，但似乎可以工作。

import scala.collection.mutable.Queue

class Circular[A](list: Seq[A]) extends Iterator[A]{

  val elements = new Queue[A] ++= list
  var pos = 0

  def next = {
    if (pos == elements.length) 
      pos = 0
    val value = elements(pos)
    pos = pos + 1
    value
  }

  def hasNext = !elements.isEmpty
  def add(a: A): Unit = { elements += a }
  override def toString = elements.toString

}

你可以这样使用它：

scala> var circ = new Circular(List(1,2))
res26: Circular[Int] = Queue(1,2)
scala> circ.next
res27: Int = 1
scala> circ.next
res28: Int = 2
scala> circ.next
res29: Int = 1
scala> circ.add(5)
scala> circ.next
res30: Int = 2
scala> circ.next
res31: Int = 5
scala> circ
res32: Circular[Int] = Queue(1,2,5)

I think maybe this is what you want; the ability to add new elements to your list even as you are iterating it. The code is ugly but it seems to work.

import scala.collection.mutable.Queue

class Circular[A](list: Seq[A]) extends Iterator[A]{

  val elements = new Queue[A] ++= list
  var pos = 0

  def next = {
    if (pos == elements.length) 
      pos = 0
    val value = elements(pos)
    pos = pos + 1
    value
  }

  def hasNext = !elements.isEmpty
  def add(a: A): Unit = { elements += a }
  override def toString = elements.toString

}

You can use it like this:

scala> var circ = new Circular(List(1,2))
res26: Circular[Int] = Queue(1,2)
scala> circ.next
res27: Int = 1
scala> circ.next
res28: Int = 2
scala> circ.next
res29: Int = 1
scala> circ.add(5)
scala> circ.next
res30: Int = 2
scala> circ.next
res31: Int = 5
scala> circ
res32: Circular[Int] = Queue(1,2,5)

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