When you call rdd.take() or rdd.first(), it may[1] executor the job
locally (in driver),
otherwise, all the jobs are executed in cluster.

There is config called `spark.localExecution.enabled` (since 1.1+) to
change this,
it's not enabled by default, so all the functions will be executed in cluster.
If you change set this to `true`, then you get the same behavior as 1.0.

[1] If it did not get enough items from the first partitions, it will
try multiple partitions
in a time, so they will be executed in cluster.

On Thu, Oct 9, 2014 at 12:14 PM, esamanas <evan.sama...@gmail.com> wrote:
> Hi,
>
> I am using pyspark and I'm trying to support both Spark 1.0.2 and 1.1.0 with
> my app, which will run in yarn-client mode.  However, it appears when I use
> 'map' to run a python lambda function over an RDD, they appear to be run on
> different machines, and this is causing problems.
>
> In both cases, I am using a Hadoop cluster that runs linux on all of its
> nodes.  I am submitting my jobs with a machine running Mac OS X 10.9.  As a
> reproducer, here is my script:
>
> import platform
> print sc.parallelize([1]).map(lambda x: platform.system()).take(1)[0]
>
> The answer in Spark 1.1.0:
> 'Linux'
>
> The answer in Spark 1.0.2:
> 'Darwin'
>
> In other experiments I changed the size of the list that gets parallelized,
> thinking maybe 1.0.2 just runs jobs on the driver node if they're small
> enough.  I got the same answer (with only 1 million numbers).
>
> This is a troubling difference.  I would expect all functions run on an RDD
> to be executed on my worker nodes in the Hadoop cluster, but this is clearly
> not the case for 1.0.2.  Why does this difference exist?  How can I
> accurately detect which jobs will run where?
>
> Thank you,
>
> Evan
>
>
>
>
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