迭代器元素的条件连接 - A Scala 惯用解决方案

Question

我有一个字符串迭代器，并希望连接一个与谓词匹配的元素前面的每个元素，例如对于迭代器的Iterator("a", "b", "c break", "d break", "e")和.line.endsWith("break")的谓词我想打印出来

(Group: 0): a-b-c break
(Group: 1): d break
(Group: 2): e

（无需一次多组持有memory）

我知道我可以使用下面这样的迭代器来实现这一点，但是必须有一种更“Scala”的方式来编写它，对吧？

import scala.collection.mutable.ListBuffer

object IteratingAndAccumulating extends App {
  class AccumulatingIterator(lines: Iterator[String])extends Iterator[ListBuffer[String]] {
    override def hasNext: Boolean = lines.hasNext

    override def next(): ListBuffer[String] = getNextLine(lines, new ListBuffer[String])

    def getNextLine(lines: Iterator[String], accumulator: ListBuffer[String]): ListBuffer[String] = {
      val line = lines.next
      accumulator += line
      if (line.endsWith("break") || !lines.hasNext) accumulator
      else getNextLine(lines, accumulator)
    }
  }

  new AccumulatingIterator(Iterator("a", "b", "c break", "d break", "e"))
    .map(_.mkString("-")).zipWithIndex.foreach{
    case (conc, i) =>
      println(s"(Group: $i): $conc")
  }
}

非常感谢，

菲尔

Answer 1

如果您不介意一次将全部内容加载到 memory 中，这是一个简单的解决方案：

  val lines: List[List[String]] = it.foldLeft(List(List.empty[String])) { 
     case (head::tail, x) if predicate(x) => Nil :: (x::head) :: tail
     case (head::tail, x) => (x::head ) :: tail
  }.dropWhile(_.empty).map(_.reverse).reverse

如果您宁愿逐个遍历字符串和组，它会涉及更多一点：

// first "instrument" the iterator, by "demarkating" group boundaries with None:
val instrumented: Iterator[Option[String]] = it.flatMap { 
  case x if predicate(x) => Seq(Some(x), None)
  case x => Seq(Some(x))
}

// And now, wrap it around into another iterator, constructing groups:
val lines: Iterator[Iterator[String]] = Iterator.continually { 
    instrumented.takeWhile(_.nonEmpty).flatten
}.takeWhile(_.nonEmpty)