Scala—重新格式化/分组列表/映射到映射的更好方法

我有一个问题，就是如何以功能性的方式对列表进行分组/格式化的最佳方式或正确方式进行编码

示例列表如下所示

 val list = List (
        Map(
          "name" -> "AAA",
          "id" -> "1",
          "category" -> "1",
          "sub_category" -> "1"
        ),

        Map(
          "name" -> "BBB",
          "id" -> "2",
          "category" -> "1",
          "sub_category" -> "2"
        ),

        Map(
          "name" -> "CCC",
          "id" -> "3",
          "category" -> "1",
          "sub_category" -> "2"
        ),

        Map(
          "name" -> "DDD",
          "id" -> "4",
          "category" -> "2",
          "sub_category" -> "1"
        ),

        Map(
          "name" -> "EEE",
          "id" -> "5",
          "category" -> "2",
          "sub_category" -> "2"
        )
      )

Map(
        2 -> Map(
          2 -> MutableList(
            Map(name -> EEE, id -> 5, category -> 2, sub_category -> 2)
          ),

          1 -> MutableList(
            Map(name -> DDD, id -> 4, category -> 2, sub_category -> 1)
          )
        ),

        1 -> Map(
          2 -> MutableList(
            Map(name -> BBB, id -> 2, category -> 1, sub_category -> 2),
            Map(name -> CCC, id -> 3, category -> 1, sub_category -> 2)
          ),

          1 -> MutableList(
            Map(name -> AAA, id -> 1, category -> 1, sub_category -> 1)
          )
        )
      )

我想按类别和子类别分组。预期的结果是这样的

 val list = List (
        Map(
          "name" -> "AAA",
          "id" -> "1",
          "category" -> "1",
          "sub_category" -> "1"
        ),

        Map(
          "name" -> "BBB",
          "id" -> "2",
          "category" -> "1",
          "sub_category" -> "2"
        ),

        Map(
          "name" -> "CCC",
          "id" -> "3",
          "category" -> "1",
          "sub_category" -> "2"
        ),

        Map(
          "name" -> "DDD",
          "id" -> "4",
          "category" -> "2",
          "sub_category" -> "1"
        ),

        Map(
          "name" -> "EEE",
          "id" -> "5",
          "category" -> "2",
          "sub_category" -> "2"
        )
      )

Map(
        2 -> Map(
          2 -> MutableList(
            Map(name -> EEE, id -> 5, category -> 2, sub_category -> 2)
          ),

          1 -> MutableList(
            Map(name -> DDD, id -> 4, category -> 2, sub_category -> 1)
          )
        ),

        1 -> Map(
          2 -> MutableList(
            Map(name -> BBB, id -> 2, category -> 1, sub_category -> 2),
            Map(name -> CCC, id -> 3, category -> 1, sub_category -> 2)
          ),

          1 -> MutableList(
            Map(name -> AAA, id -> 1, category -> 1, sub_category -> 1)
          )
        )
      )

预期输出可以包含List或MutableList，我已经完成了如下代码

val filtered:mutable.Map[Int,mutable.Map[Int,mutable.MutableList[Map[String,String]]]] =  mutable.Map()

      for(each <- list) {

        if(filtered.contains(each("category").toInt)) {

          if(filtered(each("category").toInt).contains(each("sub_category").toInt)) {

            filtered(each("category").toInt)(each("sub_category").toInt) += each

          } else {
            filtered(each("category").toInt) += (
              each("sub_category").toInt -> mutable.MutableList(each)
              )
          }
        } else {
          filtered += (
            each("category").toInt -> mutable.Map(each("sub_category").toInt -> mutable.MutableList(each))
            )
        }
      }

val筛选：mutable.Map[Int，mutable.Map[Int，mutable.MutableList[Map[String，String]]]]=mutable.Map（）
for（每个可变。可变列表（每个）
)
}
}否则{
过滤+=(
每个（“类别”）.toInt->mutable.Map（每个（“子类别”）.toInt->mutable.MutableList（每个））
)
}
}

---

我得到了结果，但这不是我想要的功能性方法。有人能帮我吗？

我想这得到了您想要的结果

list.groupBy(_("category")).mapValues(_.groupBy(_("sub_category")))

---

我不得不说，看起来您正在使用

Map

，而设计良好的

案例类

将是更好的方法。

我认为这会得到您想要的结果

list.groupBy(_("category")).mapValues(_.groupBy(_("sub_category")))

我不得不说，看起来您正在使用

Map

，而设计良好的

case类将是更好的方法