我正在从 Apache Spark 上的数据库构建家谱,使用递归搜索为数据库中的每个人找到最终的 parent (即位于家谱顶部的人)。
假设搜索他们的id时返回的第一个人是正确的 parent
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))
它给出了以下错误
"Caused by: org.apache.spark.SparkException: RDD transformations and actions can only be invoked by the driver, not inside of other transformations; for example,
rdd1.map(x => rdd2.values.count() * x)
is invalid because the values transformation and count action cannot be performed inside of therdd1.map
transformation. For more information, see SPARK-5063."
我从阅读其他类似问题中了解到问题在于我正在调用
findUltimateParentId
在 foreach 循环中,如果我从 shell 中使用一个人的 id 调用该方法,它会返回正确的最终 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?
最佳答案
如果我理解正确的话——这里有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“最深的家庭”(从祖先到 child 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,以更好地了解其工作原理:
/**
* 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
Person
应该很容易对象,并映射输出 WithAncestor
对象变成你需要的任何东西。请注意,此代码假定如果任何人具有 parentId X - 输入中实际存在具有该 id 的另一个人
关于scala - Apache Spark 中的递归方法调用,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/35464374/