[jira] [Commented] (SPARK-21436) Take advantage of known partioner for distinct on RDDs
[ https://issues.apache.org/jira/browse/SPARK-21436?page=com.atlassian.jira.plugin.system.issuetabpanels:comment-tabpanel=16570573#comment-16570573 ] Apache Spark commented on SPARK-21436: -- User 'holdenk' has created a pull request for this issue: https://github.com/apache/spark/pull/22010 > Take advantage of known partioner for distinct on RDDs > -- > > Key: SPARK-21436 > URL: https://issues.apache.org/jira/browse/SPARK-21436 > Project: Spark > Issue Type: Improvement > Components: Spark Core >Affects Versions: 2.3.0 >Reporter: holdenk >Priority: Minor > > If we have a known partitioner we should be able to avoid the shuffle. -- This message was sent by Atlassian JIRA (v7.6.3#76005) - To unsubscribe, e-mail: issues-unsubscr...@spark.apache.org For additional commands, e-mail: issues-h...@spark.apache.org
[jira] [Commented] (SPARK-21436) Take advantage of known partioner for distinct on RDDs
[
https://issues.apache.org/jira/browse/SPARK-21436?page=com.atlassian.jira.plugin.system.issuetabpanels:comment-tabpanel=16570517#comment-16570517
]
holdenk commented on SPARK-21436:
-
@[~podongfeng]
So distinct triggers a `map` first (e.g. it is `map(x => (x,
null)).reduceByKey((x, y) => x, numPartitions).map(_._1)`) and the `map` will
throw away our partition information.
I did a quick test in the shell and you can see it causes a second shuffle if
you run the same commands:
{code:java}
scala> sc.parallelize(1.to(100))
res0: org.apache.spark.rdd.RDD[Int] = ParallelCollectionRDD[0] at parallelize
at :25
scala> sc.parallelize(1.to(100)).count()
res1: Long = 100
scala> sc.parallelize(1.to(100))
res2: org.apache.spark.rdd.RDD[Int] = ParallelCollectionRDD[2] at parallelize
at :25
scala> res2.sort()
:26: error: value sort is not a member of org.apache.spark.rdd.RDD[Int]
res2.sort()
^
scala> res2.map(x => (x, x))
res4: org.apache.spark.rdd.RDD[(Int, Int)] = MapPartitionsRDD[3] at map at
:26
scala> res4.sortByKey()
res5: org.apache.spark.rdd.RDD[(Int, Int)] = ShuffledRDD[4] at sortByKey at
:26
scala> res5.cache()
res6: res5.type = ShuffledRDD[4] at sortByKey at :26
scala> res6.count()
res7: Long = 100
scala> res6.distinct()
res8: org.apache.spark.rdd.RDD[(Int, Int)] = MapPartitionsRDD[7] at distinct at
:26
scala> res8.count()
res9: Long = 100
{code}
> Take advantage of known partioner for distinct on RDDs
> --
>
> Key: SPARK-21436
> URL: https://issues.apache.org/jira/browse/SPARK-21436
> Project: Spark
> Issue Type: Improvement
> Components: Spark Core
>Affects Versions: 2.3.0
>Reporter: holdenk
>Priority: Minor
>
> If we have a known partitioner we should be able to avoid the shuffle.
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[jira] [Commented] (SPARK-21436) Take advantage of known partioner for distinct on RDDs
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https://issues.apache.org/jira/browse/SPARK-21436?page=com.atlassian.jira.plugin.system.issuetabpanels:comment-tabpanel=16558013#comment-16558013
]
zhengruifeng commented on SPARK-21436:
--
[~holdenk] It looks like that \{distinct} already utilized the known
partitioner. \{distinct} calls
{{color:#ffc66d}{color:#33}combineByKeyWithClassTag} internally, and will
avoid the shuffle if rdd's partitioner is equal to a hash partitioner with same
#partitions.
{color}{color}
{color:#ffc66d}{color:#33}Or do you mean that we need to expose some API
like {distinct(partitioner: {color}{color}Partitioner)} for other kinds of
partitioners like \{RangePartitioner}?
> Take advantage of known partioner for distinct on RDDs
> --
>
> Key: SPARK-21436
> URL: https://issues.apache.org/jira/browse/SPARK-21436
> Project: Spark
> Issue Type: Improvement
> Components: Spark Core
>Affects Versions: 2.3.0
>Reporter: holdenk
>Priority: Minor
>
> If we have a known partitioner we should be able to avoid the shuffle.
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