Scala unFlatMap

Question

我想执行类似 unFlatMap 的操作。假设我有流:

Stream("|","A","d","a","m","|","J","o","h","n", .. .)

流可以是无限的。我想将其转换为:

Stream("亚当", "约翰", ...)

当然这只是例子。一般来说，我想对由分隔符分隔的元素执行一些操作，通用签名将是:

def unFlatMap[B](isSeparator:A => Boolean)(group:Seq[A] => B):TraversableOnce[B]

如何以干净且内存高效的方式做到这一点？

Answer 1

你可以这样做:

def groupStream[A, B](s: Stream[A])(isSeparator: A => Boolean)(group: Seq[A] => B): Stream[B] = 
  group(s.takeWhile(!isSeparator(_)).toList) #:: groupStream(s.dropWhile(!isSeparator(_)).drop(1))(isSeparator)(group)

或者如果你想要一个更容易阅读但更冗长的版本:

def groupStream[A, B](s: Stream[A])(isSeparator: A => Boolean)(group: Seq[A] => B): Stream[B] = {
  def isNotSeparator(i: A): Boolean = ! isSeparator(i)

  def doGroupStream(s: Stream[A]): Stream[B] = 
    group(s.takeWhile(isNotSeparator).toList) #:: doGroupStream(s.dropWhile(isNotSeparator).drop(1))

  doGroupStream(s)
}

如果你想在 Stream 上使用隐式方法，你也可以这样做

implicit class ImprovedStream[A](val s: Stream[A]) extends AnyVal {
    def groupStream[B](isSeparator: A => Boolean)(group: Seq[A] => B): Stream[B] = {
      def isNotSeparator(i: A): Boolean = ! isSeparator(i)

      def doGroupStream(st: Stream[A]): Stream[B] = 
        group(st.takeWhile(isNotSeparator).toList) #:: doGroupStream(st.dropWhile(isNotSeparator).drop(1))

      doGroupStream(s)
    }
}

现在，使用您的示例:

val a = Stream("|" ,"A","d","a","m","|","J","o","h","n", "|", "M", "a", "r", "y", "|", "J", "o", "e")

val c = groupStream(a)(_ == "|")(_.mkString)

c.take(10).toList
//  List[String] = List("", Adam, John, Mary, Joe, "", "", "", "", "")

使用隐式版本:

val c = groupStream(a)(_ == "|")(_.mkString)