如何在 Scala 中对可变长度的重复序列进行分组 [英] How to group a variable-length, repeating sequence in Scala

问题描述

我有一组以某种模式重复的整数:

I have a collection of ints that repeat themselves in a pattern:

val repeatingSequence = List(1,2,3,1,2,3,4,1,2,1,2,3,4,5)

当模式重复时，我想将列表分割；在这种情况下，当序列回到 1 时:

I'd like to section that List up when the pattern repeats itself; in this case, when the sequence goes back to 1:

val groupedBySequence = List(List(1,2,3), List(1,2,3,4), List(1,2), List(1,2,3,4,5))

请注意，当序列跳回 1 时我正在分组，但序列可以是任意长度.我和我的同事通过添加一个名为groupWhen"的额外方法解决了这个问题

Notice that I'm grouping when the sequence jumps back to 1, but that the sequence can be of arbitrary length. My colleague and I have solved it by adding an additional method called 'groupWhen'

class IteratorW[A](itr: Iterator[A]) {
  def groupWhen(fn: A => Boolean): Iterator[Seq[A]] = {
    val bitr = itr.buffered
    new Iterator[Seq[A]] {
      override def hasNext = bitr.hasNext
      override def next = {
        val xs = collection.mutable.ListBuffer(bitr.next)
        while (bitr.hasNext && !fn(bitr.head)) xs += bitr.next
        xs.toSeq
      }
    }
  }
}
implicit def ToIteratorW[A](itr: Iterator[A]): IteratorW[A] = new IteratorW(itr)

> repeatingSequence.iterator.groupWhen(_ == 1).toSeq
List(List(1,2,3), List(1,2,3,4), List(1,2), List(1,2,3,4,5))

然而，我们都觉得收藏库中潜藏着更优雅的解决方案.

However, we both feel like there's a more elegant solution lurking in the collection library.

推荐答案

给定一个迭代器 itr，这可以解决问题:

Given an iterator itr, this will do the trick:

val head = iter.next()
val out = (
  Iterator continually {iter takeWhile (_ != head)}
  takeWhile {!_.isEmpty}
  map {head :: _.toList}
).toList

这篇关于如何在 Scala 中对可变长度的重复序列进行分组的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！