spark技术分享Case Study: Number of Partitions for groupBy Aggregation
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Important
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As it fairly often happens in my life, right after I had described the discovery I found out I was wrong and the “Aha moment” was gone. Until I thought about the issue again and took the shortest path possible. See Case 4 for the definitive solution. I’m leaving the page with no changes in-between so you can read it and learn from my mistakes. |
The goal of the case study is to fine tune the number of partitions used for groupBy aggregation.
Given the following 2-partition dataset the task is to write a structured query so there are no empty partitions (or as little as possible).
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// 2-partition dataset val ids = spark.range(start = 0, end = 4, step = 1, numPartitions = 2) scala> ids.show +---+ | id| +---+ | 0| | 1| | 2| | 3| +---+ scala> ids.rdd.toDebugString res1: String = (2) MapPartitionsRDD[8] at rdd at <console>:26 [] | MapPartitionsRDD[7] at rdd at <console>:26 [] | MapPartitionsRDD[6] at rdd at <console>:26 [] | MapPartitionsRDD[5] at rdd at <console>:26 [] | ParallelCollectionRDD[4] at rdd at <console>:26 [] |
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Note
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By default Spark SQL uses spark.sql.shuffle.partitions number of partitions for aggregations and joins, i.e. That often leads to explosion of partitions for nothing that does impact the performance of a query since these 200 tasks (per partition) have all to start and finish before you get the result. Less is more remember? |
Case 1: Default Number of Partitions — spark.sql.shuffle.partitions Property
This is the moment when you learn that sometimes relying on defaults may lead to poor performance.
Think how many partitions the following query really requires?
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val groupingExpr = 'id % 2 as "group" val q = ids. groupBy(groupingExpr). agg(count($"id") as "count") |
You may have expected to have at most 2 partitions given the number of groups.
Wrong!
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scala> q.explain == Physical Plan == *HashAggregate(keys=[(id#0L % 2)#17L], functions=[count(1)]) +- Exchange hashpartitioning((id#0L % 2)#17L, 200) +- *HashAggregate(keys=[(id#0L % 2) AS (id#0L % 2)#17L], functions=[partial_count(1)]) +- *Range (0, 4, step=1, splits=2) scala> q.rdd.toDebugString res5: String = (200) MapPartitionsRDD[16] at rdd at <console>:30 [] | MapPartitionsRDD[15] at rdd at <console>:30 [] | MapPartitionsRDD[14] at rdd at <console>:30 [] | ShuffledRowRDD[13] at rdd at <console>:30 [] +-(2) MapPartitionsRDD[12] at rdd at <console>:30 [] | MapPartitionsRDD[11] at rdd at <console>:30 [] | MapPartitionsRDD[10] at rdd at <console>:30 [] | ParallelCollectionRDD[9] at rdd at <console>:30 [] |
When you execute the query you should see 200 or so partitions in use in web UI.
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scala> q.show +-----+-----+ |group|count| +-----+-----+ | 0| 2| | 1| 2| +-----+-----+ |
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Note
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The number of Succeeded Jobs is 5. |
Case 2: Using repartition Operator
Let’s rewrite the query to use repartition operator.
repartition operator is indeed a step in a right direction when used with caution as it may lead to an unnecessary shuffle (aka exchange in Spark SQL’s parlance).
Think how many partitions the following query really requires?
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val groupingExpr = 'id % 2 as "group" val q = ids. repartition(groupingExpr). // <-- repartition per groupBy expression groupBy(groupingExpr). agg(count($"id") as "count") |
You may have expected 2 partitions again?!
Wrong!
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scala> q.explain == Physical Plan == *HashAggregate(keys=[(id#6L % 2)#105L], functions=[count(1)]) +- Exchange hashpartitioning((id#6L % 2)#105L, 200) +- *HashAggregate(keys=[(id#6L % 2) AS (id#6L % 2)#105L], functions=[partial_count(1)]) +- Exchange hashpartitioning((id#6L % 2), 200) +- *Range (0, 4, step=1, splits=2) scala> q.rdd.toDebugString res1: String = (200) MapPartitionsRDD[57] at rdd at <console>:30 [] | MapPartitionsRDD[56] at rdd at <console>:30 [] | MapPartitionsRDD[55] at rdd at <console>:30 [] | ShuffledRowRDD[54] at rdd at <console>:30 [] +-(200) MapPartitionsRDD[53] at rdd at <console>:30 [] | MapPartitionsRDD[52] at rdd at <console>:30 [] | ShuffledRowRDD[51] at rdd at <console>:30 [] +-(2) MapPartitionsRDD[50] at rdd at <console>:30 [] | MapPartitionsRDD[49] at rdd at <console>:30 [] | MapPartitionsRDD[48] at rdd at <console>:30 [] | ParallelCollectionRDD[47] at rdd at <console>:30 [] |
Compare the physical plans of the two queries and you will surely regret using repartition operator in the latter as you did cause an extra shuffle stage (!)
Case 3: Using repartition Operator With Explicit Number of Partitions
The discovery of the day is to notice that repartition operator accepts an additional parameter for…the number of partitions (!)
As a matter of fact, there are two variants of repartition operator with the number of partitions and the trick is to use the one with partition expressions (that will be used for grouping as well as…hash partitioning).
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repartition(numPartitions: Int, partitionExprs: Column*): Dataset[T] |
Can you think of the number of partitions the following query uses? I’m sure you have guessed correctly!
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val groupingExpr = 'id % 2 as "group" val q = ids. repartition(numPartitions = 2, groupingExpr). // <-- repartition per groupBy expression groupBy(groupingExpr). agg(count($"id") as "count") |
You may have expected 2 partitions again?!
Correct!
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scala> q.explain == Physical Plan == *HashAggregate(keys=[(id#6L % 2)#129L], functions=[count(1)]) +- Exchange hashpartitioning((id#6L % 2)#129L, 200) +- *HashAggregate(keys=[(id#6L % 2) AS (id#6L % 2)#129L], functions=[partial_count(1)]) +- Exchange hashpartitioning((id#6L % 2), 2) +- *Range (0, 4, step=1, splits=2) scala> q.rdd.toDebugString res14: String = (200) MapPartitionsRDD[78] at rdd at <console>:30 [] | MapPartitionsRDD[77] at rdd at <console>:30 [] | MapPartitionsRDD[76] at rdd at <console>:30 [] | ShuffledRowRDD[75] at rdd at <console>:30 [] +-(2) MapPartitionsRDD[74] at rdd at <console>:30 [] | MapPartitionsRDD[73] at rdd at <console>:30 [] | ShuffledRowRDD[72] at rdd at <console>:30 [] +-(2) MapPartitionsRDD[71] at rdd at <console>:30 [] | MapPartitionsRDD[70] at rdd at <console>:30 [] | MapPartitionsRDD[69] at rdd at <console>:30 [] | ParallelCollectionRDD[68] at rdd at <console>:30 [] |
Congratulations! You are done.
Not quite. Read along!
Case 4: Remember spark.sql.shuffle.partitions Property? Set It Up Properly
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import org.apache.spark.sql.internal.SQLConf.SHUFFLE_PARTITIONS spark.sessionState.conf.setConf(SHUFFLE_PARTITIONS, 2) // spark.conf.set(SHUFFLE_PARTITIONS.key, 2) scala> spark.sessionState.conf.numShufflePartitions res8: Int = 2 val q = ids. groupBy(groupingExpr). agg(count($"id") as "count") |
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scala> q.explain == Physical Plan == *HashAggregate(keys=[(id#0L % 2)#40L], functions=[count(1)]) +- Exchange hashpartitioning((id#0L % 2)#40L, 2) +- *HashAggregate(keys=[(id#0L % 2) AS (id#0L % 2)#40L], functions=[partial_count(1)]) +- *Range (0, 4, step=1, splits=2) scala> q.rdd.toDebugString res10: String = (2) MapPartitionsRDD[31] at rdd at <console>:31 [] | MapPartitionsRDD[30] at rdd at <console>:31 [] | MapPartitionsRDD[29] at rdd at <console>:31 [] | ShuffledRowRDD[28] at rdd at <console>:31 [] +-(2) MapPartitionsRDD[27] at rdd at <console>:31 [] | MapPartitionsRDD[26] at rdd at <console>:31 [] | MapPartitionsRDD[25] at rdd at <console>:31 [] | ParallelCollectionRDD[24] at rdd at <console>:31 [] |
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Note
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The number of Succeeded Jobs is 2. |
Congratulations! You are done now.
