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https://issues.apache.org/jira/browse/SPARK-7181?page=com.atlassian.jira.plugin.system.issuetabpanels:all-tabpanel
]
Qiping Li updated SPARK-7181:
-----------------------------
Description:
In the function {{mergeWithAggregation}} of {{ExternalSorter.scala}}, when
there is a total ordering for keys K, values of the same key in the sorted
iterator should be combined. Currently this is done by this:
{code}
val elem = sorted.next()
val k = elem._1
var c = elem._2
while (sorted.hasNext && sorted.head._1 == k) {
c = mergeCombiners(c, sorted.head._2)
}
{code}
This will go to an infinity loop when there are more than 1 values with the
same key. `sorted.next()` should be called to fix this.
was:
In the function {{mergeWithAggregation}} of {{ExternalSorter.scala}}, when
there is a total ordering for keys K, values of the same key in the sorted
iterator should be combined. Currently this is done by this:
{code}
val elem = sorted.next()
val k = elem._1
var c = elem._2
while (sorted.hasNext && sorted.head._1 == k) {
c = mergeCombiners(c, sorted.head._2)
}
{code}
This will only combine the first two values with the same key, which is not
what expected.So in the while loop, we should call {{sorted.next()}} to combine
other values too.
> External Sorter merge with aggregation doesn't aggregate all the values for
> the same key when we have a total ordering
> ----------------------------------------------------------------------------------------------------------------------
>
> Key: SPARK-7181
> URL: https://issues.apache.org/jira/browse/SPARK-7181
> Project: Spark
> Issue Type: Bug
> Components: Spark Core
> Affects Versions: 1.3.1
> Reporter: Qiping Li
> Priority: Critical
> Fix For: 1.3.2, 1.4.0
>
>
> In the function {{mergeWithAggregation}} of {{ExternalSorter.scala}}, when
> there is a total ordering for keys K, values of the same key in the sorted
> iterator should be combined. Currently this is done by this:
> {code}
> val elem = sorted.next()
> val k = elem._1
> var c = elem._2
> while (sorted.hasNext && sorted.head._1 == k) {
> c = mergeCombiners(c, sorted.head._2)
> }
> {code}
> This will go to an infinity loop when there are more than 1 values with the
> same key. `sorted.next()` should be called to fix this.
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