[IPython-User] ipython parallel task depencencies

Peter Prettenhofer peter.prettenhofer@gmail....
Wed Sep 25 14:20:08 CDT 2013


Hi Min,

2013/9/25 MinRK <benjaminrk@gmail.com>

> [..]
>
>>
> I think these probably can be expressed with the dependencies, but I need
> more information about the relationship between tasks. Specifically, what
> information needs to be conveyed from one task to the other, and what
> restriction is applied to the dependent task - must it run in the same
> location, or should it run anywhere, as long as it is after the earlier
> task completes?
>
> 1. what information do extract-transform tasks need from the fetch result?
>

The fetch task writes a file to a (distributed) file system;
extract-transform gets as input the file name (netcdf file) and an
identifier of the chunk within the file to process (there are k chunks per
file)

2. how to the extract-transform tasks relate to their parent fetch tasks?
> Do they want to run on the same engine, or can they run anywhere, as long
> as fetch happened first?
>

Each extract-transform task relates to a single parent fetch task. It can
run anywhere as long as fetch happens first if I use a distributed file
system but given that locality is supported this would be my preferred
solution.


> 3. Barrier is easy - asyncresult.wait and/or client.wait([list, of,
> asyncresults])
>

ok


> 4. what is m? Is it n * k (one load task per extract task)? Do the load
> tasks want to happen in the same place as the extract, or can they run
> anywhere, again as long as it is later? What information does load need
> from extract-transform?
>

No, m is not n * k but rather the number of aggregates I want to compute;
each load task takes as input (n / m) * k intermediate results (generated
by extract-transform) and aggregates them.
The inputs are three dimensional numpy arrays (roughly 500mb in size).

I've read the task dependence section in the user guide more carefully and
it seems I can express this computation easily in a DAG; even without the
barrier.

Thanks for the help and the great project!

best,
 Peter


>
>
>>
>> thanks,
>>  Peter
>>
>> --
>> Peter Prettenhofer
>>
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-- 
Peter Prettenhofer
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