Thanks for the suggestion! That does look really helpful, I see what you
mean about it being more general than fold. I think I will replace my fold
with aggregate - it should give me more control over the process.
I think the problem will still exist though - which is that I can't get the
correct partitioning I need. When I change my key to user_id, I lose the
timestamp partitioning. My problem is that I'm trying to retain a parent
RDD's partitioning in an RDD that no longer has the same keys as its
parent.
On Thu, Oct 16, 2014 at 11:46 AM, Cheng Lian <lian.cs....@gmail.com> wrote:
> Hi Michael,
>
> I'm not sure I fully understood your question, but I think RDD.aggregate
> can be helpful in your case. You can see it as a more general version of
> fold.
>
> Cheng
>
>
>
> On 10/16/14 11:15 PM, Michael Misiewicz wrote:
>
> Hi,
>
> I'm working on a problem where I'd like to sum items in an RDD *in order
> (*approximately*)*. I am currently trying to implement this using a fold,
> but I'm having some issues because the sorting key of my data is not the
> same as the folding key for my data. I have data that looks like this:
>
> user_id, transaction_timestamp, transaction_amount
>
> And I'm interested in doing a foldByKey on user_id to sum transaction
> amounts - taking care to note approximately when a user surpasses a total
> transaction threshold. I'm using RangePartitioner to make sure that data
> is ordered sequentially between partitions, and I'd also make sure that
> data is sorted within partitions, though I'm not sure how to do this
> exactly (I was going to look at the code for sortByKey to figure this out
> - I believe sorting in place in a mapPartitions should work). What do you
> think about the approach? Here's some sample code that demonstrates what
> I'm thinking:
>
> def myFold(V1:Float, V2:Float) : Float = {
> val partialSum = V1 + V2
> if (partialSum >= 500) {
> // make a note of it, do things
> }
> return partialSum
> }
>
> val rawData = sc.textFile("hdfs://path/to/data").map{ x => // load data
> l = x.split()
> (l(0).toLong, l(1).toLong, l(2).toFloat) // user_id:long,
> transaction_timestamp:long, transaction_amount:float
> }
> val keyByTimestamp = rawData.map(x=> (x._2, (x._1, x._3))) // rearrange
> to make timestamp the key (for sorting), convert to PairRDD
> val sortedByTimestamp = keyByTimestamp.sortByKey()
> val partitionedByTimestamp = sortedByTimestamp.partitionBy(
> new org.apache.spark.RangePartitioner(partitions=500,
> rdd=sortedByTimestamp)).persist()
> // By this point, the RDD should be sorted and partitioned according to
> the timestamp. However, I need to now make user_id the key,
> // because the output must be per user. At this point, since I change the
> keys of the PairRDD, I understand that I lose the partitioning
> // the consequence of this is that I can no longer be sure in my fold
> function that the ordering is retained.
>
> val keyByUser = partitionedByTimestamp.map(x => (x._2._1, x._2._2))
> val finalResult = keyByUser.foldByKey(zeroValue=0)(myFold)
> finalResult.saveAsTextFile("hdfs://...")
>
> The problem as you'd expect takes place in the folding function, after
> I've re-arranged my RDD to no longer be keyed by timestamp (when I produce
> keyByUser, I lose the correct partitioning). As I've read in the
> documentation, partitioning is not preserved when keys are changed (which
> makes sense).
>
> Reading this thread:
> https://groups.google.com/forum/#!topic/spark-users/Fx7DNtWiSx4 it
> appears that one possible solution might be to subclass RDD (à la
> MappedValuesRDD) to define my own RDD that retains the partitions of its
> parent. This seems simple enough, but I've never done anything like that
> before, but I'm not sure where to start. I'm also willing to write my own
> custom partitioner class, but it appears that the getPartition method
> only accepts a "key" argument - and since the value I need to partition on
> in the final step (the timestamp) would be in the Value, my
> partitioner class doesn't have the data it needs to make the right
> decision. I cannot have timestamp in my key.
>
> Alternatively, has anyone else encountered a problem like this (i.e. an
> approximately ordered sum) and did they find a good solution? Does my
> approach of subclassing RDD make sense? Would there be some way to
> finagle a custom partitioner into making this work? Perhaps this might be a
> job for some other tool, like spark streaming?
>
> Thanks,
> Michael
>
>
>
