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From Nicholas Chammas <nicholas.cham...@gmail.com>
Subject Re: PySpark RDD.partitionBy() requires an RDD of tuples
Date Wed, 02 Apr 2014 22:22:51 GMT
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 <mark@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.chammas@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.chammas@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 <ilikerps@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.chammas@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.
>>>>>
>>>>
>>>>
>>>
>>
>

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