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From Christopher Nguyen <...@adatao.com>
Subject Re: almost sorted data
Date Mon, 28 Oct 2013 17:26:15 GMT
Nathan: why iterator semantics could be more appropriate than a
materialized list: the partition data could be sitting on disk, which could
be streamed into RAM upon access, and which could be left untouched if the
algo decided by some condidtional logic that it didn't need it. And you
could always get to a list from an iterator.

Best,
--
Christopher T. Nguyen
Co-founder & CEO, Adatao <http://adatao.com>
linkedin.com/in/ctnguyen



On Mon, Oct 28, 2013 at 7:44 AM, Nathan Kronenfeld <
nkronenfeld@oculusinfo.com> wrote:

> I'm not sure what you're asking.
>
> At some level, all RDDs only do partition-wise operations - they all only
> operate on one partition at a time.
>
> I suspect what you're looking for is something where you could just write:
>
> data.mapPartitions(_.sortBy(...))
>
> If that's what you want, then no - but only because Iterator has no sortBy
> method.  I'm not sure why mapPartitions hands one an iterator rather than a
> list.  Presumably so one can avoid having to have the whole partition in
> memory at once - but equally presumably, one already has the whole
> partition in memory at once, so that seems odd to me.  Anyone know why?
> Perhaps to allow for worst-case scenarios?
>
>              -Nathan
>
>
>
> On Mon, Oct 28, 2013 at 4:54 AM, Arun Kumar <arunpatala@gmail.com> wrote:
>
>> I will try using per partition sorted data. Can I also use groupBy and
>> join per partition? Basically I want to restrict the computation per
>> partition like using this data.mapPartitions(_.toList.sortBy(...).toIterator).
>> Is there a more direct way to create a RDD that does partition wise
>> operations?
>>
>>
>> On Sat, Oct 26, 2013 at 3:50 AM, Aaron Davidson <ilikerps@gmail.com>wrote:
>>
>>> Currently, our sortByKey should be using Java's native Timsort
>>> implementation, which is an adaptive sort. That should also mean sorting is
>>> very fast for almost-sorted data. The overhead you're seeing might be
>>> caused by reshuffling everything during the range partitioning step *before
>>> *the sort, which has to serialize all your data.
>>>
>>> Nathan's solution might then work out nicely for you, as it will avoid
>>> shuffling the data.
>>>
>>>
>>> On Fri, Oct 25, 2013 at 9:18 AM, Josh Rosen <rosenville@gmail.com>wrote:
>>>
>>>> Adaptive sorting algorithms (
>>>> https://en.wikipedia.org/wiki/Adaptive_sort) can benefit from
>>>> presortedness in their inputs, so that might be a helpful search term
>>>> for researching this problem.
>>>>
>>>>
>>>> On Fri, Oct 25, 2013 at 7:23 AM, Nathan Kronenfeld <
>>>> nkronenfeld@oculusinfo.com> wrote:
>>>>
>>>>> I suspect from his description the difference is negligible for his
>>>>> case.  However, there are ways around that anyway.
>>>>>
>>>>> Assuming a fixed data set (as opposed to something like a streaming
>>>>> example, where there is no last element), one can take 3 passes to:
>>>>>
>>>>>    1. get the last element of each partition
>>>>>    2. take elements from each partition that fall before the last
>>>>>    element of the previous partition, separate them from the rest of
their
>>>>>    partition
>>>>>    3. and add them to the previous (whichever previous is
>>>>>    appropriate, in really degenerate cases, which it sounds like he doesn't
>>>>>    have) in the right location
>>>>>
>>>>>
>>>>>
>>>>>
>>>>> On Fri, Oct 25, 2013 at 10:17 AM, Sebastian Schelter <ssc@apache.org>wrote:
>>>>>
>>>>>> Using a local sort per partition only gives a correct result if the
>>>>>> data
>>>>>> is already range partitioned.
>>>>>>
>>>>>> On 25.10.2013 16:11, Nathan Kronenfeld wrote:
>>>>>> > Since no one else has answered...
>>>>>> > I assume:
>>>>>> >
>>>>>> >     data.mapPartitions(_.toList.sortBy(...).toIterator)
>>>>>> >
>>>>>> > would work, but I also suspect there's a better way.
>>>>>> >
>>>>>> >
>>>>>> > On Fri, Oct 25, 2013 at 5:01 AM, Arun Kumar <arunpatala@gmail.com>
>>>>>> wrote:
>>>>>> >
>>>>>> >> Hi,
>>>>>> >>
>>>>>> >> I am trying to process some logs and the data is sorted(*almost*)
>>>>>> by
>>>>>> >> timestamp.
>>>>>> >> If I do a full sort it takes a lot of time. Is there some
way to
>>>>>> sort more
>>>>>> >> efficiently (like restricting sort to per partition).
>>>>>> >>
>>>>>> >> Thanks in advance
>>>>>> >>
>>>>>> >
>>>>>> >
>>>>>> >
>>>>>>
>>>>>>
>>>>>
>>>>>
>>>>> --
>>>>> Nathan Kronenfeld
>>>>> Senior Visualization Developer
>>>>> Oculus Info Inc
>>>>> 2 Berkeley Street, Suite 600,
>>>>> Toronto, Ontario M5A 4J5
>>>>> Phone:  +1-416-203-3003 x 238
>>>>> Email:  nkronenfeld@oculusinfo.com
>>>>>
>>>>
>>>>
>>>
>>
>
>
> --
> Nathan Kronenfeld
> Senior Visualization Developer
> Oculus Info Inc
> 2 Berkeley Street, Suite 600,
> Toronto, Ontario M5A 4J5
> Phone:  +1-416-203-3003 x 238
> Email:  nkronenfeld@oculusinfo.com
>

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