Question

我正在努力寻找列表的特殊映射。如果我直接向您展示具体实例，您可以更容易地理解我的问题：

如何转换

(cat, List((0, 45.42), (1, 12.45), (2, 91.45))
(dog, List((0, 23.31), (1, 10.23), (2, 52.25))

到

(0, List((cat, 45.42)), List((dog, 23.31))
(1, List((cat, 12.45)), List((dog, 10.23))
(2, List((cat, 91.45)), List((dog, 52.25)))

类型基本上是：

[(String, List[(Int, Double)])]

到

[(Int, List[(String, Double)])]

是否可以使用Scala的链式函数编程函数执行此类操作？

Answer 1

我不确定它是最美丽的惯用解决方案，但它有效

  val original: Map[String, List[(Int, Double)]] = Map(
    "cat" -> List((0, 45.42), (1, 12.45), (2, 91.45)),
    "dog" -> List((0, 23.31), (1, 10.23), (2, 52.25))
  )

  val flatten = for {
    (s, v) <- original
    (i, d) <- v
  } yield (i, s, d)

  implicit class RichTuple2[A, B, C](t: (A, B, C)) {
    def tail: (B, C) = (t._2, t._3)
  }

  val converted = flatten
    .groupBy(_._1)
    .mapValues(_.map(_.tail))

  println(converted)

Answer 2

这是：

scala> val ori: Map[String, List[(Int, Double)]] = Map(
     |   "cat" -> List((0, 45.42), (1, 12.45), (2, 91.45)),
     |   "dog" -> List((0, 23.31), (1, 10.23), (2, 52.25))
     | )
ori: Map[String,List[(Int, Double)]] = Map(cat -> List((0,45.42), (1,12.45), (2,91.45)), dog -> List((0,23.31), (1,10.23), (2,52.25)))

scala> ori.foldLeft(Map[Int, List[(String, Double)]]()){ case (m, (k, v)) =>
     |   v.foldLeft(m){ case(r, (i, d)) => r.updated(i, r.getOrElse(i, Nil) :+ (k, d)) }
     | }
res1: scala.collection.immutable.Map[Int,List[(String, Double)]] = Map(
       0 -> List((cat,45.42), (dog,23.31)), 
       1 -> List((cat,12.45), (dog,10.23)), 
       2 -> List((cat,91.45), (dog,52.25)))

Scala将map [String，List [Int，Double]]转换为[Int，List [String，Double]]

2 个答案: