Question

我想在Scala中进行总结。

我有一个案例类实例列表。

case class Structure(structureFirst:String,structureSecond:String,structureThird:String,count:Int)

val structures = List(
  Structure("first001","second001","third001",1),

  Structure("first002","second002","third002",2),
  Structure("first002","second002","third002",2),

  Structure("first003","second003","third003",3),
  Structure("first003","second003","third003",3),
  Structure("first003","second003","third003",3),

  Structure("first004","second004","third004",4),
  Structure("first004","second004","third004",4),
  Structure("first004","second004","third004",4),
  Structure("first004","second004","third004",4)
)

然后我想像这个SQL一样总结。

SELECT SUM(Structure.count) FROM structures GROUP BY Structure.structureFirst,Structure.structureSecond,Structure.structureThird;

这是输出。

val result = List(
  Structure("first001","second001","third001",1),
  Structure("first002","second002","third002",4),
  Structure("first003","second003","third003",9),
  Structure("first004","second004","third004",16)
)

我如何总结这个？

感谢。

Answer 1

我绝不是世界上最好的Scala开发者，但我的新手尝试屈服这个

val result = for (structs <- structures.groupBy(_.structureFirst).values.toList)
  yield { 
    structs.reduce((item, cur) => item.copy(count=item.count + cur.count))
  }

result: List[Structure] = List(
 Structure(first002,second002,third002,4), 
 Structure(first004,second004,third004,16), 
 Structure(first001,second001,third001,1), 
 Structure(first003,second003,third003,9)
)

Answer 2

使用groupBy，在这几乎是一行中：

structures.groupBy(s => (s.structureFirst, s.structureSecond, s.structureThird)).map {
   case (key, ss) => 
      ss.reduce((a,b) => a.copy(count = a.count + b.count) ) 
}.toList.
sortBy (_.count)

如何通过SQL在Scala中汇总

2 个答案: