I suspect from his description the difference is negligible for his case. However, there are ways around that anyway.
Assuming a fixed data set (as opposed to something like a streaming example, where there is no last element), one can take 3 passes to: 1. get the last element of each partition 2. take elements from each partition that fall before the last element of the previous partition, separate them from the rest of their partition 3. and add them to the previous (whichever previous is appropriate, in really degenerate cases, which it sounds like he doesn't have) in the right location On Fri, Oct 25, 2013 at 10:17 AM, Sebastian Schelter <[email protected]> wrote: > Using a local sort per partition only gives a correct result if the data > is already range partitioned. > > On 25.10.2013 16:11, Nathan Kronenfeld wrote: > > Since no one else has answered... > > I assume: > > > > data.mapPartitions(_.toList.sortBy(...).toIterator) > > > > would work, but I also suspect there's a better way. > > > > > > On Fri, Oct 25, 2013 at 5:01 AM, Arun Kumar <[email protected]> > wrote: > > > >> Hi, > >> > >> I am trying to process some logs and the data is sorted(*almost*) by > >> timestamp. > >> If I do a full sort it takes a lot of time. Is there some way to sort > more > >> efficiently (like restricting sort to per partition). > >> > >> Thanks in advance > >> > > > > > > > > -- Nathan Kronenfeld Senior Visualization Developer Oculus Info Inc 2 Berkeley Street, Suite 600, Toronto, Ontario M5A 4J5 Phone: +1-416-203-3003 x 238 Email: [email protected]
