流式传输两个集合并根据公共属性收集地图

Question

我环顾四周，但没有找到优雅的解决方案。我要做的是有一个很好的简化（双关语意图）解决方案，以从具有1个共享属性的两个集合（ID）中创建一个地图。地图应为匹配对象的键值。

我暂时完成的工作是集合1上的一个foreach，在集合2上的foreach中以获取匹配模型，然后对其进行操作。我想创建地图，然后单独对地图中的每个对进行操作。我试图提出一个简单的例子，以使其更加清楚。

data class FirstNameModel(val idNumber: String, val firstName: String)
data class LastNameModel(val idNumber: String, val lastName: String)

val randomFirstNameList = listOf(
   FirstNameModel("5631ab", "Bob"),
   FirstNameModel("ca790a", "George"),
   FirstNameModel("j8f1sa", "Alice")
)

val randomLastNameList = listOf(
   LastNameModel("j8f1sa", "Smith"),
   LastNameModel("5631ab", "Johnson"),
   LastNameModel("ca790a", "Takai")
)

// stream function to correctly create map (not just a null one like below).

val map: Map<FirstNameModel, LastNameModel>? = null

fun printIt() {
   map?.forEach {
       println("Name for id ${it.key.idNumber} is ${it.key.firstName} ${it.value.lastName}")
   }
//    should print something like:
//            Name for id 5631ab is Bob Johnson
//            Name for id ca790a is George Takai
//            Name for id j8f1sa is Alice Smith
}

我目前一直在Kotlin尝试过此操作，但这是我有时在Java中的用例，所以对这两者都感到好奇。

Answer 1

这将起作用，包括 @扫地机（谢谢！）中提到的案例下面的评论：

val map: Map<FirstNameModel, LastNameModel> = randomFirstNameList.map { it.idNumber }
  .plus(randomLastNameList.map { it.idNumber })
  .distinct()
  .associate { idNumber ->
    (randomFirstNameList.firstOrNull { it.idNumber == idNumber } ?: FirstNameModel(idNumber, "")) to
    (randomLastNameList.firstOrNull { it.idNumber == idNumber } ?: LastNameModel(idNumber, ""))
  }

但是，引入数据类以保持结果很有意义：

data class NameModel(val idNumber: String, val firstName: String, val lastName: String)

val map: List<NameModel> = randomFirstNameList.map { it.idNumber }
  .plus(randomLastNameList.map { it.idNumber })
  .distinct()
  .map { idNumber ->
    NameModel(
      idNumber,
      randomFirstNameList.firstOrNull { it.idNumber == idNumber }?.firstName ?: "",
      randomLastNameList.firstOrNull { it.idNumber == idNumber }?.lastName ?: ""
    )
  }

输出（包括一个姓氏输入，没有相应的名字输入）：

NameModel(idNumber=5631ab, firstName=Bob, lastName=Johnson)
NameModel(idNumber=ca790a, firstName=George, lastName=Takai)
NameModel(idNumber=j8f1sa, firstName=Alice, lastName=Smith)
NameModel(idNumber=999999, firstName=, lastName=NoFirstName)

附加说明：

如果两个列表非常大，则重复的 firstornull 调用可能会增加时间消耗。在这种情况下，为第一个和姓氏创建查找图是很有意义的：

val firstNameMap = randomFirstNameList.associate { it.idNumber to it.firstName }
val lastNameMap = randomLastNameList.associate { it.idNumber to it.lastName }

val map: List<NameModel> = randomFirstNameList.map { it.idNumber }
  .plus(randomLastNameList.map { it.idNumber })
  .distinct()
  .map { idNumber ->
    NameModel(
      idNumber,
      firstNameMap.getOrElse(idNumber) { "" },
      lastNameMap.getOrElse(idNumber) { "" }
    )
  }