Thanks everyone. As Nathan suggested, I ended up collecting the distinct
keys first and then assigning Ids to each key explicitly.
Regards
Sumit Chawla
On Fri, Jun 22, 2018 at 7:29 AM, Nathan Kronenfeld <
nkronenfeld@uncharted.software> wrote:
> On Thu, Jun 21, 2018 at 4:51 PM, Chawla,Sumit <sumitkcha...@gmail.com>
>>>> wrote:
>>>>
>>>>> Hi
>>>>>
>>>>> I have been trying to this simple operation. I want to land all
>>>>> values with one key in same partition, and not have any different key in
>>>>> the same partition. Is this possible? I am getting b and c always
>>>>> getting mixed up in the same partition.
>>>>>
>>>>>
>>>>>
> I think you could do something approsimately like:
>
> val keys = rdd.map(_.getKey).distinct.zipWithIndex
> val numKey = keys.map(_._2).count
> rdd.map(r => (r.getKey, r)).join(keys).partitionBy(new Partitioner()
> {def numPartitions=numKeys;def getPartition(key: Any) =
> key.asInstanceOf[Long].toInt})
>
> i.e., key by a unique number, count that, and repartition by key to the
> exact count. This presumes, of course, that the number of keys is <MAXINT.
>
> Also, I haven't tested this code, so don't take it as anything more than
> an approximate idea, please :-)
>
> -Nathan Kronenfeld
>
