给定一个List[CaseClass],我想通过两个类属性对元素进行分组,并获得嵌套的Maps
case class CaseClass(
a String,
b String,
c String
)
val collection: List[CaseClass] = List(
CaseClass("a1","b1","c1"),
CaseClass("a1","b1","c2"),
CaseClass("a1","b2","c3"),
CaseClass("a2","b2","c4"),
CaseClass("a2","b2","c5"),
)
val res = collection.groupBy(_.a).map{ case (k,v) => v.groupBy(_.b)}
我得到的是按"b"属性分组的Map[String,CaseClass]对象。但我想得到按"a"(外部映射)和"b"(内部映射)分组的Map[String,Map[String,CaseClass]],如下所示:
Map("a1" ->
Map("b1" -> List( CaseClass("a1","b1","c1"),
CaseClass("a1","b1","c2"))),
Map("b2" -> List( CaseClass("a1","b2","c3"))),
"a2" ->
Map("b2" -> List( CaseClass("a2","b2","c4"),
CaseClass("a2","b2","c5"))))
我应该如何更改代码?
问题是(k,v) => v.groupBy(_.b)
需要将(key, value)
映射到(key, value)
,但这是将(key, value)
映射到value
。钥匙被丢弃了。
所以你只需要保留密钥:
val res = collection.groupBy(_.a).map{ case (k,v) => (k, v.groupBy(_.b)) }