Hi Ivan,
I think you can use MapPartition for that.
So basically:
dataset // assuming some partitioning that can be reused to avoid a shuffle
.sortPartition(1, Order.DESCENDING)
.mapPartition(new ReturnFirstTen())
.sortPartition(1, Order.DESCENDING).parallelism(1)
.mapPartition(new ReturnFirstTen())
Best, Fabian
2017-01-24 10:10 GMT+01:00 Ivan Mushketyk <ivan.mushke...@gmail.com>:
> Hi,
>
> I have a dataset of tuples with two fields ids and ratings and I need to
> find 10 elements with the highest rating in this dataset. I found a
> solution, but I think it's suboptimal and I think there should be a better
> way to do it.
>
> The best thing that I came up with is to partition dataset by rating, sort
> locally and write the partitioned dataset to disk:
>
> dataset
> .partitionCustom(new Partitioner<Double>() {
> @Override
> public int partition(Double key, int numPartitions) {
> return key.intValue() % numPartitions;
> }
> }, 1) . // partition by rating
> .setParallelism(5)
> .sortPartition(1, Order.DESCENDING) // locally sort by rating
> .writeAsText("..."); // write the partitioned dataset to disk
>
> This will store tuples in sorted files with names 5, 4, 3, ... that
> contain ratings in ranges (5, 4], (4, 3], and so on. Then I can read sorted
> data from disk and and N elements with the highest rating.
> Is there a way to do the same but without writing a partitioned dataset to
> a disk?
>
> I tried to use "first(10)" but it seems to give top 10 items from a random
> partition. Is there a way to get top N elements from every partition? Then
> I could locally sort top values from every partition and find top 10 global
> values.
>
> Best regards,
> Ivan.
>
>
>
