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我正在使用 Spark 进行员工记录累积，为此我使用 Spark 的累加器。我使用 Map[empId, emp] 作为 accumulableCollection 以便我可以通过他们的 ID 搜索员工。我已经尝试了一切，但它不起作用。如果不支持我使用 accumulableCollection 或 Map 的方式存在任何逻辑问题，有人可以指出。以下是我的代码

package demo

import org.apache.spark.{SparkContext, SparkConf, Logging}

import org.apache.spark.SparkContext._
import scala.collection.mutable


object MapAccuApp extends App with Logging {
  case class Employee(id:String, name:String, dept:String)

  val conf = new SparkConf().setAppName("Employees") setMaster ("local[4]")
  val sc = new SparkContext(conf)

  implicit def empMapToSet(empIdToEmp: mutable.Map[String, Employee]): mutable.MutableList[Employee] = {
    empIdToEmp.foldLeft(mutable.MutableList[Employee]()) { (l, e) => l += e._2}
  }

  val empAccu = sc.accumulableCollection[mutable.Map[String, Employee], Employee](mutable.Map[String,Employee]())

  val employees = List(
    Employee("10001", "Tom", "Eng"),
    Employee("10002", "Roger", "Sales"),
    Employee("10003", "Rafael", "Sales"),
    Employee("10004", "David", "Sales"),
    Employee("10005", "Moore", "Sales"),
    Employee("10006", "Dawn", "Sales"),
    Employee("10007", "Stud", "Marketing"),
    Employee("10008", "Brown", "QA")
  )

  System.out.println("employee count " + employees.size)


  sc.parallelize(employees).foreach(e => {
    empAccu += e
  })

  System.out.println("empAccumulator size " + empAccu.value.size)
}
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1 回答 1

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对于您的问题，使用accumulableCollection似乎有点矫枉过正，如下所示：

import org.apache.spark.{AccumulableParam, Accumulable, SparkContext, SparkConf}

import scala.collection.mutable

case class Employee(id:String, name:String, dept:String)

val conf = new SparkConf().setAppName("Employees") setMaster ("local[4]")
val sc = new SparkContext(conf)

implicit def mapAccum =
    new AccumulableParam[mutable.Map[String,Employee], Employee]
{
  def addInPlace(t1: mutable.Map[String,Employee],
                 t2: mutable.Map[String,Employee])
      : mutable.Map[String,Employee] = {
    t1 ++= t2
    t1
  }
  def addAccumulator(t1: mutable.Map[String,Employee], e: Employee)
      : mutable.Map[String,Employee] = {
    t1 += (e.id -> e)
    t1
  }
  def zero(t: mutable.Map[String,Employee])
      : mutable.Map[String,Employee] = {
    mutable.Map[String,Employee]()
  }
}

val empAccu = sc.accumulable(mutable.Map[String,Employee]())

val employees = List(
  Employee("10001", "Tom", "Eng"),
  Employee("10002", "Roger", "Sales"),
  Employee("10003", "Rafael", "Sales"),
  Employee("10004", "David", "Sales"),
  Employee("10005", "Moore", "Sales"),
  Employee("10006", "Dawn", "Sales"),
  Employee("10007", "Stud", "Marketing"),
  Employee("10008", "Brown", "QA")
)

System.out.println("employee count " + employees.size)

sc.parallelize(employees).foreach(e => {
  empAccu += e
})

println("empAccumulator size " + empAccu.value.size)
empAccu.value.foreach(entry =>
  println("emp id = " + entry._1 + " name = " + entry._2.name))

虽然目前这方面的记录很差，但 Spark 代码库中的相关测试非常有启发性。

编辑：事实证明 usingaccumulableCollection 确实有价值：您不需要定义 anAccumulableParam和以下作品。我将保留这两种解决方案，以防它们对人们有用。

case class Employee(id:String, name:String, dept:String)

val conf = new SparkConf().setAppName("Employees") setMaster ("local[4]")
val sc = new SparkContext(conf)

val empAccu = sc.accumulableCollection(mutable.HashMap[String,Employee]())

val employees = List(
  Employee("10001", "Tom", "Eng"),
  Employee("10002", "Roger", "Sales"),
  Employee("10003", "Rafael", "Sales"),
  Employee("10004", "David", "Sales"),
  Employee("10005", "Moore", "Sales"),
  Employee("10006", "Dawn", "Sales"),
  Employee("10007", "Stud", "Marketing"),
  Employee("10008", "Brown", "QA")
)

System.out.println("employee count " + employees.size)

sc.parallelize(employees).foreach(e => {
  // notice this is different from the previous solution
  empAccu += e.id -> e
})

println("empAccumulator size " + empAccu.value.size)
empAccu.value.foreach(entry =>
  println("emp id = " + entry._1 + " name = " + entry._2.name))

两种解决方案均使用 Spark 1.0.2 进行了测试。

于 2014-09-19T04:44:18.973 回答