[Paraview] Numpy masking (via programm filter) not quite working in parallel

George Zagaris george.zagaris at kitware.com
Mon Aug 20 21:36:20 EDT 2012


Hello,

I have done some work in VTK towards that. Currently, this works only
with structured data.

If your data is structured, you can use vtkPStructuredGridConnectivity
to generate a bitmask (an unsigned char associated with each mesh
node). The bitmask can then be queried using the vtkGhostArray (See
http://www.vtk.org/doc/nightly/html/classvtkGhostArray.html) to see
which properties are "turned on" for the given mesh node. If the
IGNORE property is ON, that means that the node is a duplicate with
respect to the given process and should not be taken into account. The
implementation assigns ownership to the lowest rank of a mesh node --
on all other processes, the IGNORE property is set to ON.

TestPStructuredGridConnectivity illustrates how this can be used. The
test ensures that:
(1) Ensuring that the total number of nodes computed is invariant to
the number of processes/partitions
(2) Ensuring that computing the average for a given field is also
invariant to the numer of processes/partitions

If this could be useful in your work, I can elaborate more if you have
any questions on how to use it.

Best,
George


On Fri, Aug 17, 2012 at 6:18 PM, Sohail Shafii <sohailshafii at yahoo.com> wrote:
> Very well. I think that should be fine for now.
>
> Sohail
>
> ________________________________
> From: David Thompson <david.thompson at kitware.com>
> To: Sohail Shafii <sohailshafii at yahoo.com>
> Cc: David Thompson <david.thompson at kitware.com>; "paraview at paraview.org"
> <paraview at paraview.org>
> Sent: Friday, August 17, 2012 4:15 PM
>
> Subject: Re: [Paraview] Numpy masking (via programm filter) not quite
> working in parallel
>
> Hi Sohail,
>
>> ... Will this point sharing information become available in later versions
>> of ParaView? i.e. would it ever be easy to identify and count duplicates?
>
> I don't know of any plans to implement this (but that doesn't mean there
> aren't).
>
>     David
>
>
>
>
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