Apache Spark 中的递归方法调用

我正在从 Apache Spark 上的数据库构建一个家谱，使用递归搜索来查找数据库中每个人的最终父级（即家谱顶部的人）。

假设搜索 id 时返回的第一个人是正确的父母

val peopleById = peopleRDD.keyBy(f => f.id)
def findUltimateParentId(personId: String) : String = {

    if((personId == null) || (personId.length() == 0))
        return "-1"

    val personSeq = peopleById.lookup(personId)
    val person = personSeq(0)
    if(person.personId == "0 "|| person.id == person.parentId) {

        return person.id

    }
    else {

        return findUltimateParentId(person.parentId)

    }

}

val ultimateParentIds = peopleRDD.foreach(f => f.findUltimateParentId(f.parentId))

它给出以下错误

“由 org.apache.spark.SparkException 引起：RDD 转换和操作只能由驱动程序调用，不能在其他转换内部调用；例如，rdd1.map(x => rdd2.values.count() * x)无效，因为值转换和计数操作无法在rdd1.map转型。有关更多信息，请参阅 SPARK-5063。”

我通过阅读其他类似的问题了解到问题是我正在打电话findUltimateParentId从 foreach 循环中，如果我使用一个人的 id 从 shell 调用该方法，它会返回正确的最终结果parent id

但是，其他建议的解决方案都不适合我，或者至少我看不到如何在我的程序中实现它们，有人可以帮忙吗？

如果我理解正确的话 - 这是一个适用于任何大小输入的解决方案（尽管性能可能不是很好） - 它在 RDD 上执行 N 次迭代，其中 N 是“最深的家族”（从祖先到孩子的最大距离）输入：

// representation of input: each person has an ID and an optional parent ID
case class Person(id: Int, parentId: Option[Int])

// representation of result: each person is optionally attached its "ultimate" ancestor,
// or none if it had no parent id in the first place
case class WithAncestor(person: Person, ancestor: Option[Person]) {
  def hasGrandparent: Boolean = ancestor.exists(_.parentId.isDefined)
}

object RecursiveParentLookup {
  // requested method
  def findUltimateParent(rdd: RDD[Person]): RDD[WithAncestor] = {

    // all persons keyed by id
    def byId = rdd.keyBy(_.id).cache()

    // recursive function that "climbs" one generation at each iteration
    def climbOneGeneration(persons: RDD[WithAncestor]): RDD[WithAncestor] = {
      val cached = persons.cache()
      // find which persons can climb further up family tree
      val haveGrandparents = cached.filter(_.hasGrandparent)

      if (haveGrandparents.isEmpty()) {
        cached // we're done, return result
      } else {
        val done = cached.filter(!_.hasGrandparent) // these are done, we'll return them as-is
        // for those who can - join with persons to find the grandparent and attach it instead of parent
        val withGrandparents = haveGrandparents
          .keyBy(_.ancestor.get.parentId.get) // grandparent id
          .join(byId)
          .values
          .map({ case (withAncestor, grandparent) => WithAncestor(withAncestor.person, Some(grandparent)) })
        // call this method recursively on the result
        done ++ climbOneGeneration(withGrandparents)
      }
    }

    // call recursive method - start by assuming each person is its own parent, if it has one:
    climbOneGeneration(rdd.map(p => WithAncestor(p, p.parentId.map(i => p))))
  }

}

这是一个测试，可以更好地理解它是如何工作的：

/**
  *     Example input tree:
  *
  *            1             5
  *            |             |
  *      ----- 2 -----       6
  *      |           |
  *      3           4
  *
  */

val person1 = Person(1, None)
val person2 = Person(2, Some(1))
val person3 = Person(3, Some(2))
val person4 = Person(4, Some(2))
val person5 = Person(5, None)
val person6 = Person(6, Some(5))

test("find ultimate parent") {
  val input = sc.parallelize(Seq(person1, person2, person3, person4, person5, person6))
  val result = RecursiveParentLookup.findUltimateParent(input).collect()
  result should contain theSameElementsAs Seq(
    WithAncestor(person1, None),
    WithAncestor(person2, Some(person1)),
    WithAncestor(person3, Some(person1)),
    WithAncestor(person4, Some(person1)),
    WithAncestor(person5, None),
    WithAncestor(person6, Some(person5))
  )
}

将您的输入映射到这些应该很容易Person对象，并映射输出WithAncestor对象变成你需要的任何东西。请注意，此代码假设如果任何人有parentId X - 输入中实际上存在另一个具有该id的人