scala - 用于嵌套集合的 zipWithIndex(没有可变状态的发布)

Question

有一些嵌套的集合:

val xs = List(
           List("a","b"), 
           List("c"), 
           List("d", "e", "f"))

我想为嵌套列表的元素创建唯一索引:

val result = List(
         List(("a", 0), ("b", 1)), 
         List(("c", 2)),
         List(("d", 3), ("e", 4), ("f", 5)))

这是一个糟糕的解决方案(使用可变状态):

val indexes:List[Int] = xs.flatten.zipWithIndex.map(_._2)
var idx = 0
val result = xs.map(_.map{ el =>
   val copy = (el, indexes(idx))
   idx = idx + 1
   copy
})

如何在没有可变状态的情况下释放此任务？

Answer 1

解决方案 1(使用 fold):

scala> xs.foldLeft((List[List[(String, Int)]](), 0)){ 
         case ((r, i), x) => (r:+x.zip(Stream.from(i)), i+x.size) 
       }._1
res1: List[List[(String, Int)]] = List(List((a,0), (b,1)), List((c,2)), List((d,3), (e,4), (f,5)))

解决方案2(使用递归)

scala> def deepZip[A](ls: List[List[A]], i: Int = 0): List[List[(A, Int)]] = ls match {
     |   case Nil => Nil
     |   case x::xs => x.zip(Stream.from(i)) :: deepZip(xs, i+x.size)
     | }
deepZip: [A](ls: List[List[A]], i: Int)List[List[(A, Int)]]

scala> deepZip(xs)
res2: List[List[(String, Int)]] = List(List((a,0), (b,1)), List((c,2)), List((d,3), (e,4), (f,5)))

解决方案3:

scala> (xs, xs.map(_.size).scanLeft(0){ _+_ }).zipped map { (a, b) => a.zip(Stream.from(b)) }
res3: List[List[(String, Int)]] = List(List((a,0), (b,1)), List((c,2)), List((d,3), (e,4), (f,5)))