akpatnam25 commented on a change in pull request #33644:
URL: https://github.com/apache/spark/pull/33644#discussion_r683031421
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File path: core/src/main/scala/org/apache/spark/rdd/RDD.scala
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@@ -1233,6 +1233,21 @@ abstract class RDD[T: ClassTag](
(i, iter) => iter.map((i % curNumPartitions, _))
}.foldByKey(zeroValue, new
HashPartitioner(curNumPartitions))(cleanCombOp).values
}
+ if (conf.get(ENABLE_EXECUTOR_TREE_AGGREGATE) &&
partiallyAggregated.partitions.length > 1) {
+ // define a new partitioner that results in only 1 partition
+ val constantPartitioner = new Partitioner {
+ override def numPartitions: Int = 1
+
+ override def getPartition(key: Any): Int = 0
+ }
+ // map the partially aggregated rdd into a key-value rdd
+ // do the computation in the single executor with one partition
+ // get the new RDD[U]
+ partiallyAggregated = partiallyAggregated
+ .map(v => (0.toByte, v))
+ .foldByKey(zeroValue, constantPartitioner)(cleanCombOp)
+ .values
+ }
val copiedZeroValue = Utils.clone(zeroValue,
sc.env.closureSerializer.newInstance())
Review comment:
no it would not submit another job, because there is only one partition.
spark would do the fold internal the partition (we only have 1 in this case),
and return.
As described on Apache Spark's website the fold definition states:
```
"Aggregate the elements of each partition, and then the results for all the
partitions, using a given associative function and a neutral "zero value"."
```
Thus fold operates by folding first each element of partition and then the
result of partition
it is partition dependent
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