Spark处理数据排序问题如何避免OOM

 val cidAndSidCountGrouped: RDD[(Long, Iterable[(String, Int)])] = cidAndSidCount.groupByKey()
    // 4. 排序, 取top10
    val result: RDD[(Long, List[(String, Int)])] = cidAndSidCountGrouped.map {
      case (cid, sidCountIt) = 
        // sidCountIt 排序, 取前10
        // Iterable转成容器式集合的时候, 如果数据量过大, 极有可能导致oom
        (cid, sidCountIt.toList.sortBy(-_._2).take(5))
    }

 //把long（即key值）提取出来
    val cids: List[Long] = categoryCountList.map(_.cid.toLong)
    val buffer: ListBuffer[(Long, List[(String, Int)])] = ListBuffer[(Long, List[(String, Int)])]()
    //根据每个key来过滤RDD
    for (cid <- cids) {
      /*
      List((15,(632972a4-f811-4000-b920-dc12ea803a41,10)), (15,(f34878b8-1784-4d81-a4d1-0c93ce53e942,8)), (15,(5e3545a0-1521-4ad6-91fe-e792c20c46da,8)), (15,(66a421b0-839d-49ae-a386-5fa3ed75226f,8)), (15,(9fa653ec-5a22-4938-83c5-21521d083cd0,8)))
      目标:
      (9,List((199f8e1d-db1a-4174-b0c2-ef095aaef3ee,9), (329b966c-d61b-46ad-949a-7e37142d384a,8), (5e3545a0-1521-4ad6-91fe-e792c20c46da,8), (e306c00b-a6c5-44c2-9c77-15e919340324,7), (bed60a57-3f81-4616-9e8b-067445695a77,7)))
       */
      val arr: Array[(String, Int)] = cidAndSidCount.filter(cid == _._1)
        .sortBy(-_._2._2)
        .take(5)
        .map(_._2)
      buffer += ((cid, arr.toList))
    }
    buffer.foreach(println)

 def statCategoryTop10Session_3(sc: SparkContext,
                  categoryCountList: List[CategroyCount],
                  userVisitActionRDD: RDD[UserVisitAction]) = {
    // 1. 过滤出来 top10品类的所有点击记录
    // 1.1 先map出来top10的品类id
    val cids = categoryCountList.map(_.cid.toLong)
    val topCategoryActionRDD: RDD[UserVisitAction] = userVisitActionRDD.filter(action =  cids.contains(action.click_category_id))


    // 2. 计算每个品类 下的每个session 的点击量 rdd ((cid, sid) ,1)
    val cidAndSidCount: RDD[(Long, (String, Int))] = topCategoryActionRDD
      .map(action =  ((action.click_category_id, action.session_id), 1))
      // 使用自定义分区器 重点理解分区器的原理
      .reduceByKey(new CategoryPartitioner(cids), _ + _)
      .map {
        case ((cid, sid), count) =  (cid, (sid, count))
      }
    
    // 3. 排序取top10
//因为已经按key分好了区，所以用Mappartitions ，在每个分区中新建一个TreeSet即可
    val result: RDD[(Long, List[SessionInfo])] = cidAndSidCount.mapPartitions((it: Iterator[(Long, (String, Int))]) =  {
//new 一个TreeSet，并同时指定排序规则
   var treeSet: mutable.TreeSet[CategorySession] = new mutable.TreeSet[CategorySession]()(new Ordering[CategorySession] {
          override def compare(x: CategorySession, y: CategorySession): Int = {
            if (x.clickCount  = y.clickCount) -1 else 1
          }
        })
   var id = 0l
  iter.foreach({
    case (l, session) =  {
      id = l
      treeSet.add(session)
    if (treeSet.size   10) treeSet = treeSet.take(10)
          }
        })
        Iterator(id, treeSet)
      })
  
    result.collect.foreach(println)
    
    Thread.sleep(1000000)
  }
}

/*
根据传入的key值来决定分区号，让相同key进入相同的分区，能够避免多次shuffle
 */
class CategoryPartitioner(cids: List[Long]) extends Partitioner {
  // 用cid索引, 作为将来他的分区索引.
  private val cidWithIndex: Map[Long, Int] = cids.zipWithIndex.toMap
  
  // 返回集合的长度
  override def numPartitions: Int = cids.length
  
  // 根据key返回分区的索引
  override def getPartition(key: Any): Int = {
    key match {
      // 根据品类id返回分区的索引!  0-9
      case (cid: Long, _) = 
        cidWithIndex(cid)
    }
  }
}