Question

问题1：

val a: List[String] = List("0","aa","1","2","3","aa","4","5","6","7")

如何在＆＃34; aa＆＃34;之后获得接下来的两个元素然后将这些元素放入每个独立的列表中：

期望：

List(List("1","2"),List("4","5"))

问题2：

val b: List[String] = List("0","aa","1","2","3","#","aa","4","5","#","6","7")

如何获得＆＃34; aa＆＃34;之间的元素。和＆＃34;＃＆＃34; （排除＆＃34; aa＆＃34;和＆＃34;＃＆＃34;）然后将这些元素放入每个独立的列表中：

期望：

List(List("1","2","3"),List("4","5"))

Answer 1

您可以使用var request = { Ids: ["id1", "id2"], Types = ["type1", "type2"], StoreId: "123456" } $http({method: 'POST',header:{Content-Type: application/json},url: targetUri,data: request}).then(function(response) { console.log(response) })和sliding来获取tails个相关的子列表，然后使用Iterator来过滤掉您需要的collect个并在一次操作中将它们转换为所需的格式：

对于第一个问题：

List

或者：

scala> val a: List[String] = List("0","aa","1","2","3","aa","4","5","6","7")
a: List[String] = List(0, aa, 1, 2, 3, aa, 4, 5, 6, 7)

scala> a.sliding(3).collect {
  case "aa" :: rest => rest
}.toList
res1: List[List[String]] = List(List(1, 2), List(4, 5))

关于第二个问题：

scala> a.tails.collect {
  case "aa" :: rest => rest.take(2)
}.toList
res2: List[List[String]] = List(List(1, 2), List(4, 5))

Answer 2

以下是第一个问题的答案：

val a: List[String] = List("0","aa","1","2","3","aa","4","5","6","7")

def arraySplit(in : List[String], accA: List[List[String]], accB: List[String]) : List[List[String]] = in match {
  case h::tail => if (h == "aa" ) arraySplit(tail, accB::accA, List()) else arraySplit(tail, accA, h::accB )
  case List() => accB::accA
}

val r = arraySplit(a, List(), List())

r.reverse.tail.map(x => x.reverse.take(2) )

但这不是最佳解决方案。

Answer 3

对于问题1，使用tails会很有趣

val a: List[String] = List("0","aa","1","2","3","aa","4","5","6","7")
a.tails.toList.filter(_.headOption == Some("aa")).map(_.drop(1).take(2)) 

>> List[Seq[String]] = List(List("1", "2"), List("4", "5"))

如何在指定元素之后获取一些元素，然后将这些元素放入每个独立的List中

3 个答案: