在 Scala 中加入两个列表时如何处理不匹配的键

Question

我是 scala 的新手，在按键值合并或加入（不知道 scala 中的确切术语）两个列表时遇到问题。在一个案例中，键值不匹配，我得到了 java.util.NoSuchElementException: key not found 错误。但我只想匹配案例。我的示例代码片段如下：

val users = List(
  (1,1,4.0),
  (1,3,4.0),
  (1,6,4.0)
)

val cars = List(
  (1,1,"ww"),
  (1,3,"mds"),
  (2,6,"pgt")
)

val usersmapped = users.map(i => ((i._1,i._2),i)).toMap
val carsmapped = cars.map(i => ((i._1,i._2),i)).toMap
val result  = carsmapped.map(ar => (ar._2,usersmapped(ar._1)._3))

以上代码的错误信息是：java.util.NoSuchElementException: key not found: (2,6)

我的目标是拥有这样的东西：

List(
(1,1,ww) -> 4.0, 
(1,3,mds)-> 4.0
)

或

List(
(1,1,ww) -> 4.0, 
(1,3,mds) -> 4.0,
(1,6,null) -> 4.0  //null means no match with other list 
)

尤其是第一个可以吗？我该怎么做？

我试过了，但它返回了 none:

  val result  = Try(carsmapped.map(ar => (ar._2,usersmapped(ar._1)._3)))
  ).toOption

如有帮助，将不胜感激

Answer 1

使用get method instead of "()". get method returning Option:

val users = List(
  (1,1,4.0),
  (1,3,4.0),
  (1,6,4.0)
)

val cars = List(
  (1,1,"ww"),
  (1,3,"mds"),
  (2,6,"pgt")
)

val usersmapped = users.map(i => ((i._1,i._2),i)).toMap
val carsmapped = cars.map(i => ((i._1,i._2),i)).toMap

carsmapped
  .map(ar => (ar._2, usersmapped.get(ar._1).map(_._3)))
  .foreach(println)

在 Scala 中加入两个列表时如何处理不匹配的键

How to handle not matching keys when joining two lists in scala

collections

functional-programming

scala