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From Antoine Wendlinger <awendlin...@mytraffic.fr>
Subject Re: Renaming a DataFrame column makes Spark lose partitioning information
Date Wed, 05 Aug 2020 07:51:27 GMT
Well that's great ! Thank you very much :)


Antoine

On Tue, Aug 4, 2020 at 11:22 PM Terry Kim <yuminkim@gmail.com> wrote:

> This is fixed in Spark 3.0 by https://github.com/apache/spark/pull/26943:
>
> scala> :paste
> // Entering paste mode (ctrl-D to finish)
>
> Seq((1, 2))
>       .toDF("a", "b")
>       .repartition($"b")
>       .withColumnRenamed("b", "c")
>       .repartition($"c")
>       .explain()
>
> // Exiting paste mode, now interpreting.
>
> == Physical Plan ==
> *(1) Project [a#7, b#8 AS c#11]
> +- Exchange hashpartitioning(b#8, 200), false, [id=#12]
>    +- LocalTableScan [a#7, b#8]
>
> Thanks,
> Terry
>
> On Tue, Aug 4, 2020 at 6:26 AM Antoine Wendlinger <
> awendlinger@mytraffic.fr> wrote:
>
>> Hi,
>>
>> When renaming a DataFrame column, it looks like Spark is forgetting the
>> partition information:
>>
>>     Seq((1, 2))
>>       .toDF("a", "b")
>>       .repartition($"b")
>>       .withColumnRenamed("b", "c")
>>       .repartition($"c")
>>       .explain()
>>
>> Gives the following plan:
>>
>>     == Physical Plan ==
>>     Exchange hashpartitioning(c#40, 10)
>>     +- *(1) Project [a#36, b#37 AS c#40]
>>        +- Exchange hashpartitioning(b#37, 10)
>>           +- LocalTableScan [a#36, b#37]
>>
>> As you can see, two shuffles are done, but the second one is unnecessary.
>> Is there a reason I don't know for this behavior ? Is there a way to work
>> around it (other than not renaming my columns) ?
>>
>> I'm using Spark 2.4.3.
>>
>>
>> Thanks for your help,
>>
>> Antoine
>>
>

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