Apache spark 为什么在reduceByKey之后所有数据都会在一个分区中结束？

我有一个简单的spark程序。我想知道为什么所有数据最终都在一个分区中

val l = List((30002,30000), (50006,50000), (80006,80000), 
             (4,0), (60012,60000), (70006,70000), 
             (40006,40000), (30012,30000), (30000,30000),
             (60018,60000), (30020,30000), (20010,20000), 
             (20014,20000), (90008,90000), (14,0), (90012,90000),
             (50010,50000), (100008,100000), (80012,80000),
             (20000,20000), (30010,30000), (20012,20000), 
             (90016,90000), (18,0), (12,0), (70016,70000), 
             (20,0), (80020,80000), (100016,100000), (70014,70000),
             (60002,60000), (40000,40000), (60006,60000), 
             (80000,80000), (50008,50000), (60008,60000), 
             (10002,10000), (30014,30000), (70002,70000),
             (40010,40000), (100010,100000), (40002,40000),
             (20004,20000), 
             (10018,10000), (50018,50000), (70004,70000),
             (90004,90000), (100004,100000), (20016,20000))

val l_rdd = sc.parallelize(l, 2)

// print each item and index of the partition it belongs to
l_rdd.mapPartitionsWithIndex((index, iter) => {
   iter.toList.map(x => (index, x)).iterator
}).collect.foreach(println)

// reduce on the second element of the list.
// alternatively you can use aggregateByKey  
val l_reduced = l_rdd.map(x => {
                    (x._2, List(x._1))
                  }).reduceByKey((a, b) => {b ::: a})

// print the reduced results along with its partition index
l_reduced.mapPartitionsWithIndex((index, iter) => {
      iter.toList.map(x => (index, x._1, x._2.size)).iterator
}).collect.foreach(println)

---

当您运行此程序时，您将看到数据（

l_rdd

）分布到两个分区中。一旦我缩减，结果RDD（

l_reduced

）也有两个分区，但所有数据都在一个分区（索引0）中，而另一个分区为空。即使数据量很大（几GB），也会发生这种情况。

l_reduced

不也应该分配到两个分区中吗

除非您另外指定，否则分区将基于相关键的hashcode完成，并假设hashcode将导致相对均匀的分布。在本例中，您的哈希代码都是偶数，因此都将进入分区0

如果这确实代表了您的数据集，那么

reduceByKey

会有一个重载，它会使用分区器和reduce函数。我建议为这样的数据集提供一种替代的分区算法

val l_reduced = l_rdd.map(x => {
                    (x._2, List(x._1))
                  }).reduceByKey((a, b) => {b ::: a})

参考上面的代码片段，您正在按照RDD的第二个字段进行分区。第二个字段中的所有数字都以0结尾

调用HashPartitioner时，记录的分区号由以下内容决定：

而Utils.nonNegativeMod的定义如下：

因此，您的所有记录最终都位于分区0中

您可以通过重新分区来解决此问题：

val l_reduced = l_rdd.map(x => {
                    (x._2, List(x._1))
                  }).reduceByKey((a, b) => {b ::: a}).repartition(2)

其中：

(0,100000,4)
(0,10000,2)
(0,0,5)
(0,20000,6)
(0,60000,5)
(0,80000,4)
(1,50000,4)
(1,30000,6)
(1,90000,4)
(1,70000,5)
(1,40000,4)

或者，您可以创建一个

scala> l.map(_._2.hashCode % 2) // numPartitions = 2
res10: List[Int] = List(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)

val l_reduced = l_rdd.map(x => {
                    (x._2, List(x._1))
                  }).reduceByKey((a, b) => {b ::: a}).repartition(2)

(0,100000,4)
(0,10000,2)
(0,0,5)
(0,20000,6)
(0,60000,5)
(0,80000,4)
(1,50000,4)
(1,30000,6)
(1,90000,4)
(1,70000,5)
(1,40000,4)