Ah, now I see what Aaron was referring to. So I'm guessing we will get this in the next release or two. Thank you.
On Wed, Apr 2, 2014 at 6:09 PM, Mark Hamstra <m...@clearstorydata.com>wrote: > There is a repartition method in pyspark master: > https://github.com/apache/spark/blob/master/python/pyspark/rdd.py#L1128 > > > On Wed, Apr 2, 2014 at 2:44 PM, Nicholas Chammas < > nicholas.cham...@gmail.com> wrote: > >> Update: I'm now using this ghetto function to partition the RDD I get >> back when I call textFile() on a gzipped file: >> >> # Python 2.6 >> def partitionRDD(rdd, numPartitions): >> counter = {'a': 0} >> def count_up(x): >> counter['a'] += 1 >> return counter['a'] >> return (rdd.keyBy(count_up) >> .partitionBy(numPartitions) >> .map(lambda (counter, data): data)) >> >> If there's supposed to be a built-in Spark method to do this, I'd love to >> learn more about it. >> >> Nick >> >> >> On Tue, Apr 1, 2014 at 7:59 PM, Nicholas Chammas < >> nicholas.cham...@gmail.com> wrote: >> >>> Hmm, doing help(rdd) in PySpark doesn't show a method called >>> repartition(). Trying rdd.repartition() or rdd.repartition(10) also >>> fail. I'm on 0.9.0. >>> >>> The approach I'm going with to partition my MappedRDD is to key it by a >>> random int, and then partition it. >>> >>> So something like: >>> >>> rdd = sc.textFile('s3n://gzipped_file_brah.gz') # rdd has 1 partition; >>> minSplits is not actionable due to gzip >>> keyed_rdd = rdd.keyBy(lambda x: randint(1,100)) # we key the RDD so we >>> can partition it >>> partitioned_rdd = keyed_rdd.partitionBy(10) # rdd has 10 partitions >>> >>> Are you saying I don't have to do this? >>> >>> Nick >>> >>> >>> >>> On Tue, Apr 1, 2014 at 7:38 PM, Aaron Davidson <ilike...@gmail.com>wrote: >>> >>>> Hm, yeah, the docs are not clear on this one. The function you're >>>> looking for to change the number of partitions on any ol' RDD is >>>> "repartition()", which is available in master but for some reason doesn't >>>> seem to show up in the latest docs. Sorry about that, I also didn't realize >>>> partitionBy() had this behavior from reading the Python docs (though it is >>>> consistent with the Scala API, just more type-safe there). >>>> >>>> >>>> On Tue, Apr 1, 2014 at 3:01 PM, Nicholas Chammas < >>>> nicholas.cham...@gmail.com> wrote: >>>> >>>>> Just an FYI, it's not obvious from the >>>>> docs<http://spark.incubator.apache.org/docs/latest/api/pyspark/pyspark.rdd.RDD-class.html#partitionBy>that >>>>> the following code should fail: >>>>> >>>>> a = sc.parallelize([1,2,3,4,5,6,7,8,9,10], 2) >>>>> a._jrdd.splits().size() >>>>> a.count() >>>>> b = a.partitionBy(5) >>>>> b._jrdd.splits().size() >>>>> b.count() >>>>> >>>>> I figured out from the example that if I generated a key by doing this >>>>> >>>>> b = a.map(lambda x: (x, x)).partitionBy(5) >>>>> >>>>> then all would be well. >>>>> >>>>> In other words, partitionBy() only works on RDDs of tuples. Is that >>>>> correct? >>>>> >>>>> Nick >>>>> >>>>> >>>>> ------------------------------ >>>>> View this message in context: PySpark RDD.partitionBy() requires an >>>>> RDD of >>>>> tuples<http://apache-spark-user-list.1001560.n3.nabble.com/PySpark-RDD-partitionBy-requires-an-RDD-of-tuples-tp3598.html> >>>>> Sent from the Apache Spark User List mailing list >>>>> archive<http://apache-spark-user-list.1001560.n3.nabble.com/>at >>>>> Nabble.com. >>>>> >>>> >>>> >>> >> >
