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Serializable的产品的成员

感谢

Answer 1

不要模式匹配：

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)))

粗略但按要求：

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)))

或

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)))

生成的集合特别好吗？

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”？

Answer 2

您还可以使用Product特征提取第一项：

scala> val joined= (i++j).groupBy(_.productIterator.toList.head)
joined: scala.collection.immutable.Map[Any,List[Product with Serializable]] = Map(10 -> List(X(10,10,8,8), Y(10,10,8)))

Answer 3

我用这个解决了它：

trait genericClass{
    def a:Long

}
case class X(override val a:Long, b:Long, c:Long, d:Long) extends genericClass
case class Y(override val a:Long, e :Long, f :Long) extends genericClass

然后：

val c = (X++Y).groupBy(_.a)

感谢

在不同的案例类scala的相同字段上按列表分组

3 个答案: