on the PMC list, as
> > well as call for an official vote on it on a public list. Basically, as
> the
> > Spark project scales up, we need to define a model to make sure there is
> > still great oversight of key components (in particular internal
> > architecture and public APIs), and to this end I've proposed
> implementing a
> > maintainer model for some of these components, similar to other large
> > projects.
> >
> >
> > -
> > To unsubscribe, e-mail: dev-unsubscr...@spark.apache.org
> > For additional commands, e-mail: dev-h...@spark.apache.org
> >
> >
>
--
Liquan Pei
Department of Physics
University of Massachusetts Amherst
as to be positive definite...
>
> I think the tests are not running any 0.0 regularization tests otherwise we
> should have caught it as well...
>
> For the sparse coding NMF variant that I am running, I have to turn off L2
> regularization when I run a L1 on products to extract sparse topics...
>
> Thanks.
>
> Deb
>
--
Liquan Pei
Department of Physics
University of Massachusetts Amherst
u Wang wrote:
>
> Liquan, yes, for full outer join, one hash table on both sides is more
> efficient.
>
> For the left/right outer join, it looks like one hash table should be
> enought.
>
> --
> *From:* Liquan Pei [mailto:liquan...@gm
uot;
> side, so Spark can iterate through the left side and find matches in the
> right side from the hash table efficiently. Please comment and suggest,
> thanks again!
>
>
> ------
>
> *From:* Liquan Pei [mailto:liquan...@gmail.com]
> *Se
n the partition is big. And it
> doesn't reduce the iteration on streamed relation, right?
>
> Thanks!
>
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Hi
What is the current status of Sparrow integration with Spark? I would like
to integrate Sparrow with Spark 1.0 on a 100 node cluster. Any suggestions?
Thanks a lot for your help!
Liquan
Hi
I am currently implementing an algorithm involving matrix multiplication.
Basically, I have matrices represented as RDD[Array[Double]]. For example,
If I have A:RDD[Array[Double]] and B:RDD[Array[Double]] and what would be
the most efficient way to get C = A * B
Both A and B are large, so it w
