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如何在Scala中执行嵌套groupBy

nested group-by collections scala json

Max Heiber · 技术社区 · 10 年前

我有一个案例对象列表,如下所示:

AppInfo(client = "client1", project = "project1", version = "version1")
AppInfo(client = "client1", project = "project1", version = "version2")
AppInfo(client = "client2", project = "project3", version = "version1")
AppInfo(client = "client2", project = "project4", version = "version1")

需要做一个这样的嵌套结构:

Map(
  "clients" -> List(
    Map(
      "name" -> "client1",
      "projects" -> List(
        Map(
          "name" -> "project1",
          "versions" -> List(
            "version1",
            "version2"
          )
        )
      )
    ),
    Map(
      "name" -> "client2",
      "projects" -> List(
        Map(
          "name" -> "project3",
          "versions" -> List(
            "version1"
          )
        ),
        Map(
          "name" -> "project4",
          "versions" -> List(
            "version1"
          )
        )
      )
    )
  )
)

这看起来很糟糕,但它将序列化为这个非常简单的JSON:

{
  "clients": [
    {
      "name": "client1",
      "projects": [
        {
          "name": "project1",
          "versions": [
            "version1",
            "version2"
          ]
        }
      ]
    },
    {
      "name": "client2",
      "projects": [
        {
          "name": "project3",
          "versions": [
            "version1"
          ]
        },
        {
          "name": "project4",
          "versions": [
            "version1"
          ]
        }
      ]
    }
  ]
}

有什么合理的方法可以做到这一点吗?现在我在列表中的groupBys中的maps中有地图。

编辑

类似于 Specter Clojure图书馆可能在这里有所帮助。

3 回复 | 直到 10 年前

Dani 10 年前

怎么样:

for {
    (clientName, clientInfos) <- infoList.groupBy(_.client)
} yield {
    val clientProjects = clientInfos.groupBy(_.project)
    val projectSection = clientProjects.map { case(name, infos) => Map("name" -> name, "versions" -> infos.map(_.version)) }

    Map("name" -> clientName, "projects" -> projectSection)
}

它不会减少 map 或 groupBy 调用,但这是我能够组织代码的最方便的方式。

Max Heiber 10 年前

给定一个列表 AppInfo case对象已调用 appInfos 做 byClient(appInfos) :

def byVersion(appInfos: List[AppInfo]) = appInfos
                                          .map(_.version)

def byProject(appInfos: List[AppInfo]) = appInfos
                                          .groupBy(_.project)
                                          .map(tuple => Map("name" -> tuple._1, "versions" -> byVersion(tuple._2)))

def byClient(appInfos: List[AppInfo]) = appInfos
                                          .groupBy(_.client)
                                          .map(tuple => Map("name" -> tuple._1, "projects" -> byProject(tuple._2)))

这是重复的,但能完成工作。

Max Heiber 10 年前

案例类将为您提供比地图更好的类型安全性,但更丑陋:

给定一个列表 AppInfo case对象已调用 appInfos 做 byClient2(appInfos) :

case class ProjectApps(name: String, versions: List[String])

object ProjectApps{
  def apply(tuple: (String, List[AppInfo])): ProjectApps =
                ProjectApps(name=tuple._1, versions=tuple._2.map(_.version))
}


case class ClientApps(name: String, projects: List[ProjectApps])

object ClientApps {
  def apply(tuple: (String, List[AppInfo])): ClientApps = {
    val name = tuple._1
    val projects = tuple._2.groupBy(_.project).map(ProjectApps.apply).toList
    ClientApps(name, projects)
  }
}


def byClient2(appInfos: List[AppInfo]): List[ClientApps] = 
                     appInfos.groupBy(_.client).map(ClientApps.apply).toList