repartition() puts all values with the same key in one partition, but,
multiple other keys can be in the same partition. It sounds like you want
groupBy, not repartition, if you want to handle these separately.
On Mon, Jun 20, 2022 at 8:26 AM DESCOTTE Loic - externe
<loic-externe.desco...@edf.fr.invalid> wrote:
> Hi,
>
>
>
> I have a data type like this :
>
> case class Data(col: String, ...)
>
>
>
> and a Dataset[Data] ds. Some rows have columns filled with value 'a', and
> other with value 'b', etc.
>
>
>
> I want to process separately all data with a 'a', and all data with a 'b'.
> But I also need to have all the 'a' in the same partition.
>
>
>
> If I do : ds.repartition(col("col")).mapPartition(data => ???)
>
>
>
> Is it guaranteed by default that I will have all the 'a' in a single
> partition, and no 'b' mixed with it in this partition?
>
>
>
> My understanding is that all the 'a' will be in the same partition, but
> 'a' and 'b' may be mixed. So I should do :
>
>
>
> val nbDistinct = ds.select("col").distinct.count
>
> ds.repartition(col("col")).mapPartition{ data =>
>
> // split mixed values in a single partition with group by :
>
> data.groupBy(_.col).flatMap { case (col, rows) => ??? }
>
> }
>
>
>
> Is that correct?
>
>
>
>
>
> I can also do this to force the number of partitions with the number of
> distinct values :
>
> val nbDistinct = ds.select("col").distinct.count
>
> ds.repartition(nbDistinct , col("col")).mapPartition(data => ???)
>
>
>
> But is it useful?
>
> But it adds an action that may be expensive in some cases, and sometimes
> it seems to use less partitions than it should.
>
>
>
> Example (spark shell) :
>
> scala> (Range(0,10).map(_ => Data("a")) union Range(0,10).map(_ =>
> Data("b")) union Range(0,10).map(_ =>
> Data("c"))).toList.toDS.repartition(col("col")).rdd.foreachPartition(part
> => println(part.mkString))
>
>
>
> output :
>
> Data(b)Data(b)Data(b)Data(b)Data(b)Data(b)Data(b)Data(b)Data(b)Data(b)
>
> Data(a)Data(a)Data(a)Data(a)Data(a)Data(a)Data(a)Data(a)Data(a)Data(a)
>
> Data(c)Data(c)Data(c)Data(c)Data(c)Data(c)Data(c)Data(c)Data(c)Data(c)
>
>
>
> I have 3 partitions + a lot of empty partitions (blank lines)
>
> If I do :
>
>
>
> scala> (Range(0,10).map(_ => Data("a")) union Range(0,10).map(_ =>
> Data("b")) union Range(0,10).map(_ =>
> Data("c"))).toList.toDS.repartition(3,col("col")).rdd.foreachPartition(part
> => println(part.mkString))
>
> then output is :
>
>
> Data(a)Data(a)Data(a)Data(a)Data(a)Data(a)Data(a)Data(a)Data(a)Data(a)Data(b)Data(b)Data(b)Data(b)Data(b)Data(b)Data(b)Data(b)Data(b)Data(b)Data(c)Data(c)Data(c)Data(c)Data(c)Data(c)Data(c)Data(c)Data(c)Data(c)
>
>
>
> I would like to have 3 partitions with all ‘a’ (and only the ‘a’), then
> all ‘b’ etc. but I have 2 empty partitions and a partition with all the
> ‘a’, ‘b’ and ‘c’ of my dataset.
>
>
>
> So, is there a good way to guarantee a dataset is split over unique
> partitions for a column value?
>
>
>
> Thanks,
>
> Loïc
>
>
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