Scala 实现 Haskell's groupBy

Question

我正在寻找Haskell的groupBy的Scala实现。

行为应该是这样的:

isD :: Char -> Bool
isD c = elem c "123456789-_ "
groupBy (a b -> isD a == isD b) "this is a line with 0123344334343434343434-343 3345"
["this"," ","is"," ","a"," ","line"," ","with"," 0123344334343434343434-343 3345"]

我尝试了Scala的groupBy函数，但是它只需要一个参数的函数，而不是Haskell的2。我还研究了分区，但是它只返回一个元组。

我正在寻找的函数应该将匹配谓词的每个连续元素分组。

Answer 1

这样的问题似乎经常出现，在我看来，这是一个很好的迹象，Rex Kerr的groupedWhile方法应该包含在标准集合库中。但是，如果您不想将其复制/粘贴到您的项目中…

我喜欢你的递归解决方案，但它实际上并没有输出正确的东西(即字符串)，所以这里是我如何改变它:

def groupBy(s: String)(f: (Char, Char) => Boolean): List[String] = s match {
  case "" => Nil
  case x => 
    val (same, rest) = x span (i => f(x.head, i))
    same :: groupBy(rest)(f)
}

然后，使用您的函数并在REPL中尝试:

val isD = (x: Char) => "123456789-_ " contains x
groupBy("this is a line with 0123344334343434343434-343 3345")(isD(_) == isD(_))

结果是List[String]，这可能是您真正想要的。

Answer 2

现在使用这个，感谢答案:

def groupByS(eq: (Char,Char) => Boolean, list: List[Char]): List[List[Char]] = {
    list match {
    case head :: tail => {
      val newHead = head :: tail.takeWhile(eq(head,_))
      newHead :: groupByS(eq, tail.dropWhile(eq(head,_)))
    }
    case nil => List.empty
  }
}

这个可以改进;)

Answer 3

将Haskell版本转换为Scala肯定不会太难。这是Haskell对groupBy的定义。它使用span;我不知道Scala中是否有等价的span，或者你是否需要翻译Haskell对span的定义

Answer 4

我的版本，只是瞎搞——不太确定。我知道Haskell比Scala好，但我想学习Scala:

object GroupByTest extends App {    
  val ds = Set('0', '1', '2', '3', '4', '5', '6', '7', '8', '9', '-', '_', ' ')
  def isD(d: Char) = ds contains d
  def hgroupBy[A](op: A => (A => Boolean), a: List[A]): List[List[A]] = 
    a match {
      case Nil => List.empty
      case x :: xs =>
        val t = xs span op(x)         
        (x :: t._1) :: hgroupBy(op, t._2)        
    }
  val lambda: Char => Char => Boolean = x => y => isD(x) == isD(y)
  println(hgroupBy(lambda, "this is a line with 0123344334343434343434-343 3345".toList))
}

Answer 5

def partitionBy[T, U](list: List[T])(f: T => U ): List[List[T]] = {
  def partitionList(acc: List[List[T]], list: List[T]): List[List[T]] = {
    list match {
      case Nil => acc
      case head :: tail if f(acc.last.head) == f(head) => partitionList(acc.updated(acc.length - 1, head :: acc.last), tail)
      case head :: tail => partitionList(acc ::: List(head) :: Nil, tail)
    }
  }
  if (list.isEmpty) List.empty
  else partitionList(List(List(list.head)), list.tail)
}
partitionBy("112211".toList)(identity)
//res: List[List[Char]] = List(List(1, 1), List(2, 2), List(1, 1))

val l = List("mario", "adam", "greg", "ala", "ola")
partitionBy(l)(_.length)
//res: List[List[String]] = List(List(mario), List(greg, adam), List(ola, ala))