> > 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