Scala如何将两个列表匹配到Map或Tuple

Question

val listA = List("one", "two", "three")

val listB = List(1, 3, 4, 9, 2, 6)

Result:

val m: Map[String, Int] = Map(one -> 1, two -> 2, three -> 3)

I want to pair listA and listB to a Map with corresponding key/value. I have tried to use the zip method but it only support sequential merging. How can I achieve the above result?

Answer 1

如果f(x)是可以返回给定字母值的相应数值的函数，则只需将listA中的每个x listA到一个对(x, f(x)) ，然后将结果对的列表转换为a使用.toMap地图。

Answer 2

带有已sorted listB的zip ：

listA.zip(listB.sorted).toMap

Answer 3

Do you intend:

scala> val listA = List("one", "two", "three", "four")
listA: List[String] = List(one, two, three, four)

scala> val listB = List(1, 3, 4, 9, 2, 6)
listB: List[Int] = List(1, 3, 4, 9, 2, 6)

scala> val m = Map("one"->1, "two"->20, "three"->3)
m: scala.collection.immutable.Map[String,Int] = Map(one -> 1, two -> 20, three -> 3)

scala> listA flatMap { case k if (m contains k) && (listB contains m(k)) => Some(k, m(k)) ; case _ => None }
res4: List[(String, Int)] = List((one,1), (three,3))