You can use DStream.transform() to do any arbitrary RDD transformations on
the RDDs generated by a DStream.
val coalescedDStream = myDStream.transform { _.coalesce(...) }
On Tue, Mar 3, 2015 at 1:47 PM, Saiph Kappa <saiph.ka...@gmail.com> wrote:
> Sorry I made a mistake in my code. Please ignore my question number 2.
> Different numbers of partitions give *the same* results!
>
>
> On Tue, Mar 3, 2015 at 7:32 PM, Saiph Kappa <saiph.ka...@gmail.com> wrote:
>
>> Hi,
>>
>> I have a spark streaming application, running on a single node,
>> consisting mainly of map operations. I perform repartitioning to control
>> the number of CPU cores that I want to use. The code goes like this:
>>
>> val ssc = new StreamingContext(sparkConf, Seconds(5))
>>> val distFile = ssc.textFileStream("/home/myuser/spark-example/dump")
>>> val words = distFile.repartition(cores.toInt).flatMap(_.split(" "))
>>> .filter(_.length > 3)
>>>
>>> val wordCharValues = words.map(word => {
>>> var sum = 0
>>> word.toCharArray.foreach(c => {sum += c.toInt})
>>> sum.toDouble / word.length.toDouble
>>> }).foreachRDD(rdd => {
>>> println("MEAN: " + rdd.mean())
>>> })
>>>
>>
>> I have 2 questions:
>> 1) How can I use coalesce in this code instead of repartition?
>>
>> 2) Why, using the same dataset (which is a small file processed within a
>> single batch), the result that I obtain for the mean varies with the number
>> of partitions? If I don't call the repartition method, the result is always
>> the same for every execution, as it should be. But repartitioning for
>> instance in 2 partitions gives a different mean value than using 8
>> partitions. I really don't understand why given that my code is
>> deterministic. Can someone enlighten me on this?
>>
>> Thanks.
>>
>
>
