Question

我在play框架的scala模板中有一个字符串列表我希望一次迭代超过一半的列表，然后第二次迭代列表的另一半。
我不知道如何为此编写高效的迭代器。

我试过了 @for(i <- 0 until list.length/2 ) {list(i) }
然后进行第二次循环

@for(i <- list.length/2+1 until list.length ) 
 { list(i) }

这有效，但复杂性变高。

后来我做了

  @defining(list.size) { size => 
    @for(i <- 0 until size/2) 
    {list(i) }
  }

现在它似乎工作正常。

Answer 1

这是一种方式。

scala> List("a","b","c","d","e")
res0: List[String] = List(a, b, c, d, e)

scala> res0.splitAt(res0.size/2)
res1: (List[String], List[String]) = (List(a, b),List(c, d, e))

scala> res1._1.foreach(println(_))
a
b

scala> res1._2.foreach(println(_))
c
d
e

Answer 2

使用滑动创建迭代器，

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

scala> val step = (input.length + 1 ) / 2
step: Int = 2

scala> val sliding = input.sliding(step, step)
sliding: Iterator[List[Int]] = non-empty iterator

scala> val left = sliding.next()
left: List[Int] = List(1, 2)

scala> val right = sliding.next()
right: List[Int] = List(3)

scala> left.foreach(println)
1
2

scala> right.foreach(println)
3

或使用take＆amp;降，

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

scala> val step = (input.length + 1 ) / 2
step: Int = 2

scala> input.take(step).foreach(println)
1
2

scala> input.drop(step).foreach(println)
3

Answer 3

Iterator可能是个不错的选择。

scala> val list = List(1,2,3,4,5,6)
list: List[Int] = List(1, 2, 3, 4, 5, 6)

scala> val iterator = list.toIterator
iterator: Iterator[Int] = non-empty iterator

scala> val half = (0 until list.size / 2)
half: scala.collection.immutable.Range = Range(0, 1, 2)

scala> val left = (list.size / 2 until list.size)
left: scala.collection.immutable.Range = Range(3, 4, 5)

scala> half foreach { x => println(x + ":" + iterator.next) }
0:1
1:2
2:3

scala> left foreach { x => println(x + ":" + iterator.next) }
3:4
4:5
5:6

迭代scala列表

3 个答案: