在不同的案例类scala的相同字段上按列表分组

Question

如何对包含不同案例类的两个列表进行分组,但两个类都有相同的字段:

case class X(a:Long, b:Long, c:Long, d:Long)
case class Y(a:Long, e:Long, f :Long)

val i = List(X(10,10,8,8))
val j = List(Y(10,10,8))

val joined = a++b
joined.groupBy(_.a)

错误:值a不是具有Seri​​alizable的Product的成员

谢谢

Answer 1

不要害怕模式匹配:

scala> val joined = (i++j) groupBy { case x: X => x.a case y: Y => y.a }
joined: scala.collection.immutable.Map[Long,List[Product with Serializable]] = Map(10 -> List(X(10,10,8,8), Y(10,10,8)))

粗暴但按要求:

scala> val joined = ((i++j).asInstanceOf[List[{ def a: Long }]]) groupBy(_.a)
warning: there were 1 feature warning(s); re-run with -feature for details
joined: scala.collection.immutable.Map[Long,List[AnyRef{def a: Long}]] = Map(10 -> List(X(10,10,8,8), Y(10,10,8)))

要么

scala> val joined = (i++j) groupBy { case x => x.asInstanceOf[{ def a: Long }].a }
warning: there were 1 feature warning(s); re-run with -feature for details
joined: scala.collection.immutable.Map[Long,List[Product with Serializable]] = Map(10 -> List(X(10,10,8,8), Y(10,10,8)))

产生的集合特别好吗？

更新:

(i++j) groupBy (_.asInstanceOf[{ def a: Long }].a)

我想知道IDE是否有重构,"转换为下划线"？