我正在使用Apache Spark上的数据库构建一个家族树,使用递归搜索来查找数据库中每个人的最终父级(即家族树顶部的人)。 出于这个目的,假设在搜索他们的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))
这会产生以下错误“由以下引起:org.apache.spark.SparkException:RDD转换和操作只能由驱动程序调用,而不能在其他转换内部调用;例如,rdd1.map(x => rdd2.values.count()* x)无效,因为无法在rdd1.map转换中执行值转换和计数操作。有关更多信息,请参阅SPARK-5063。“
我从阅读其他类似的问题中了解到,问题是我在foreach循环中调用了findUltimateParentId,如果我从shell调用带有person id的方法,则返回正确的最终父id
但是,没有其他建议的解决方案对我有用,或者至少我看不到如何在我的程序中实现它们,任何人都可以帮忙吗?
答案 0 :(得分:5)
如果我理解正确 - 这是一个适用于任何大小的输入的解决方案(虽然性能可能不是很好) - 它在RDD上执行N次迭代,其中N是“最深的家族”(从祖先到最大的距离)孩子)在输入中:
String
这是一个更好地理解其工作原理的测试:
// 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))))
}
}
将输入映射到这些/**
* 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))
)
}
对象并将输出Person
对象映射到您需要的任何对象应该很容易。请注意,此代码假定如果任何人具有parentId X,则具有该id的另一个人实际存在于输入
答案 1 :(得分:0)
使用SparkContext.broadcast修复此问题:
val peopleById = peopleRDD.keyBy(f => f.id)
val broadcastedPeople = sc.broadcast(peopleById.collectAsMap())
def findUltimateParentId(personId: String) : String = {
if((personId == null) || (personId.length() == 0))
return "-1"
val personOption = broadcastedPeople.value.get(personId)
if(personOption.isEmpty) {
return "0";
}
val person = personOption.get
if(person.personId == 0 || person.orgId == person.personId) {
return person.id
}
else {
return findUltimateParentId(person.parentId)
}
}
val ultimateParentIds = peopleRDD.foreach(f => f.findUltimateParentId(f.parentId))
现在工作得很好!