什么是Scala通过分隔符拆分List的惯用方法?

时间:2013-01-30 21:19:11

标签: java scala

如果我有一个String类型的列表,

scala> val items = List("Apple","Banana","Orange","Tomato","Grapes","BREAK","Salt","Pepper","BREAK","Fish","Chicken","Beef")
items: List[java.lang.String] = List(Apple, Banana, Orange, Tomato, Grapes, BREAK, Salt, Pepper, BREAK, Fish, Chicken, Beef)

如何根据特定的字符串/模式(n,在这种情况下)将其拆分为"BREAK"个单独的列表。

我已经考虑过使用"BREAK"找到indexOf的位置,然后按照这个方式拆分列表,或者使用与takeWhile (i => i != "BREAK")类似的方法,但我想知道是否有更好的方法?

如果有帮助,我知道items列表中只有3组项目(因此有2个"BREAK"标记)。

6 个答案:

答案 0 :(得分:8)

def splitBySeparator[T]( l: List[T], sep: T ): List[List[T]] = {
  l.span( _ != sep ) match {
    case (hd, _ :: tl) => hd :: splitBySeparator( tl, sep )
    case (hd, _) => List(hd)
  }
}

val items = List("Apple","Banana","Orange","Tomato","Grapes","BREAK","Salt","Pepper","BREAK","Fish","Chicken","Beef")
splitBySeparator(items, "BREAK")

结果:

res1: List[List[String]] = List(List(Apple, Banana, Orange, Tomato, Grapes), List(Salt, Pepper), List(Fish, Chicken, Beef))

更新:上述版本虽然简洁有效,但有两个问题:它不能很好地处理边缘情况(如List("BREAK")List("BREAK", "Apple", "BREAK"),而不是tail recursive。所以这是另一个(命令性)版本修复它:

import collection.mutable.ListBuffer
def splitBySeparator[T]( l: Seq[T], sep: T ): Seq[Seq[T]] = {
  val b = ListBuffer(ListBuffer[T]())
  l foreach { e =>
    if ( e == sep ) {
      if  ( !b.last.isEmpty ) b += ListBuffer[T]()
    }
    else b.last += e
  }
  b.map(_.toSeq)
}

它内部使用ListBuffer,就像我在List.span的第一个版本中使用的splitBySeparator的实现一样。

答案 1 :(得分:5)

另一种选择:

val l = Seq(1, 2, 3, 4, 5, 9, 1, 2, 3, 4, 5, 9, 1, 2, 3, 4, 5, 9, 1, 2, 3, 4, 5)

l.foldLeft(Seq(Seq.empty[Int])) {
  (acc, i) =>
    if (i == 9) acc :+ Seq.empty
    else acc.init :+ (acc.last :+ i)
}

// produces:
List(List(1, 2, 3, 4, 5), List(1, 2, 3, 4, 5), List(1, 2, 3, 4, 5), List(1, 2, 3, 4, 5))

答案 2 :(得分:0)

这个怎么样:使用scan来确定列表中每个元素属于哪个部分。

val l = List("Apple","Banana","Orange","Tomato","Grapes","BREAK","Salt","Pepper","BREAK","Fish","Chicken","Beef")
val count = l.scanLeft(0) { (n, s) => if (s=="BREAK") n+1 else n } drop(1)
val paired = l zip count
(0 to count.last) map { sec => 
  paired flatMap { case (x, c) => if (c==sec && x!="BREAK") Some(x) else None }  
}
// Vector(List(Apple, Banana, Orange, Tomato, Grapes), List(Salt, Pepper), List(Fish, Chicken, Beef))

答案 3 :(得分:0)

这也不是尾递归,但它适用于边缘情况:

def splitsies[T](l:List[T], sep:T) : List[List[T]] = l match {
  case head :: tail =>
    if (head != sep)
      splitsies(tail,sep) match {
        case h :: t => (head :: h) :: t
        case Nil => List(List(head))
      }
    else
      List() :: splitsies(tail, sep)
  case Nil => List()
}

唯一令人讨厌的事情:

scala> splitsies(List("BREAK","Tiger"),"BREAK")
res6: List[List[String]] = List(List(), List(Tiger))

如果你想更好地处理分隔符启动的情况,请查看与Martin answer中使用span不同的东西(稍微不同的问题)。

答案 4 :(得分:0)

使用List.unfold(Scala 2.13及更高版本):

val p: String => Boolean = _ != "BREAK"

val result: List[List[String]] = List.unfold(items) {
  case Nil =>
    None
  case l if p(l.head) =>
    Some(l.span(p))
  case _ :: tail =>
    Some(tail.span(p))
}

代码在Scastie处运行。

使用reverse + foldLeft

def splitAtElement[T](list: List[T], element: T): List[List[T]] = {
  list.reverse.foldLeft(List(List[T]()))((l, currentElement) => {
    if (currentElement == element) {
      List() :: l
    } else {
      (currentElement :: l.head) :: l.tail
    }
  })
}

代码在Scastie处运行。

使用foldRight

def splitBySeparator[T](list: List[T], sep: T): List[List[T]] = {
  list.foldRight(List(List[T]()))((s, l) => {
    if (sep == s) {
      List() :: l
    } else {
      (s :: l.head) :: l.tail
    }
  }).filter(_.nonEmpty)
}

代码在Scastie处运行。

答案 5 :(得分:-1)

val q = items.mkString(",").split("BREAK").map("(^,|,$)".r.replaceAllIn(_, "")).map(_.split(","))

这里","是一个唯一的分隔符,不会出现在项目列表中的任何字符串中。如果需要,我们可以选择不同的分隔符。

items.mkString(",")将所有内容组合成一个字符串

.split("BREAK") // which we then split using "BREAK" as delimiter to get a list

.map("(^,|,$)".r.replaceAllIn(_, "")) // removes the leading/trailing commas of each element of the list in previous step

.map(_.split(",")) // splits each element using comma as seperator to give a list of lists


scala> val q = items.mkString(",").split("BREAK").map("(^,|,$)".r.replaceAllIn(_, "")).map(_.split(","))
q: Array[Array[String]] = Array(Array(Apple, Banana, Orange, Tomato, Grapes), Array(Salt, Pepper), Array(Fish, Chicken, Beef))

scala> q(0)
res21: Array[String] = Array(Apple, Banana, Orange, Tomato, Grapes)

scala> q(1)
res22: Array[String] = Array(Salt, Pepper)

scala> q(2)
res23: Array[String] = Array(Fish, Chicken, Beef)