在Scala中迭代RDD迭代

Question

case class ID: Int,  Var1: Int, Var2: Int extends Serializable

  def foo(rdds: RDD[(ID, Iterable[(Var1, Var2)])]): RDD[(Var1, AvgVar2)] = {

    def average(as: Array[Var2]): AvgVar2 = {
       var sum = 0.0
       var i = 0.0
       while (i < as.length) {
           sum += Var2.val
           i += 1
      }
      sum/i
    }

    //My attempt at Scala
    rdds.map(x=> ((x._1),x._2)).groupByKey().map(x=>average(x._1)).collect()
}

null

//RDD[(ID,Iterable[(Var1,Var2)...])]
RDD[(1,[(1,3),(1,12),(1,6)])],
RDD[(2,[(2,5),(2,7)])]

一些示例输出数据：

//RDD[(Var1, AvgVar2)]
RDD[(1,7),(2,6)]

*编辑：工作的scala代码行：

rdd.map(x => (x._2.map(it => it._1).asInstanceOf[Var1], average(x._2.map(it => it._2).toArray)))

Answer 1

考虑到id=var1，简单的.map()将解决这个问题：

def foo(rdds: RDD[(Int, Iterable[(Int, Int)])]): RDD[(Int, Double)] = {

  def average(as: Iterable[(Int, Int)]): Double = {
    as.map(_._2).reduce(_+_)/as.size.toDouble
  }

  rdds.map(x => (x._1, average(x._2)))
}

产出：

val input = sc.parallelize(List((1,Iterable((1,3),(1,12),(1,6))), (2, Iterable((2,5),(2,7)))))

scala> foo(input).collect
res0: Array[(Int, Double)] = Array((1,7.0), (2,6.0))

编辑:(averary()，签名相同）：

def foo(rdds: RDD[(Int, Iterable[(Int, Int)])]): RDD[(Int, Double)] = {

  def average(as: Array[Int]): Double = {
    as.reduce(_+_)/as.size.toDouble
  }

  rdds.map(x => (x._1, average(x._2.map(tuple => tuple._2).toArray)))
}