Hi Andy, 

Thanks for the response. Just to further complicate things, let's say there is 
more than one feature and so my threshold results in multiple "islands" of 
data. Is there a filter or a way that I can loop over the discrete lumps of 
data to run the integration filter over each one? 

Alternatively, from the entire scalar field, I might also be interested in 
finding all of the local maxima, which may be an easier problem than 
thresholding and finding the center of mass of the independent volumes. 

I can already foresee issues trying to locate a center point/line when features 
merge/collide but I'm not going to worry about that one quite yet. 

I'm starting to get the feeling this might turn into a pretty big undertaking 
-- but maybe if I can figure something out I can send a filter upstream for 
others to use. 

Tim 

----- Original Message -----

From: "Andy Bauer" <andy.ba...@kitware.com> 
To: "tim gallagher" <tim.gallag...@gatech.edu> 
Cc: "ParaView list" <paraview@paraview.org> 
Sent: Tuesday, October 6, 2015 10:19:01 AM 
Subject: Re: [Paraview] Finding center points/lines of a scalar field 







Hi Tim, 

For 2D where you're just looking for a point, after you've done your threshold 
you can use the integrate variables to get the center point of the extracted 
domain. It should be the returned point location. This will also work in 3D for 
a center point. At this point in your Catalyst script you can just get the 
results from the integration and write your own csv file with the values (just 
have process 0 do this since in parallel I believe that only process 0 will 
have any data here). 

For 3D and looking for an arbitrarily shaped center line that's not known a 
priori, that's a tough one. The best I can think of here is after the threshold 
to compute some distance function over the thresholded domain, then maybe take 
the gradient of that, find the maximum point(s) and then follow the maximum 
gradient vector from those points to find your lines. As for doing this in 
ParaView, I can't think of any existing filter to compute the distance function 
so this would probably have to be a custom filter. For the lines from 
gradients, you can probably use the Python calculator or Python programmable 
filter to compute the maximum gradient vector at each point and then maybe use 
streamlines with custom inputs for the points to compute the lines you want. 

This sounds like a really interesting problem though so if you figure out how 
to do this, I'd be very interested in seeing your results. 

Cheers, 
Andy 







On Sat, Oct 3, 2015 at 8:25 PM, Tim Gallagher < tim.gallag...@gatech.edu > 
wrote: 


Hi everybody, 

I'm working on some flow feature identification and I want to track the (x,y,t) 
or (x,y,z,t) centers of the features I want to identify. I can come up with a 
scalar field, \phi, and a criterion for my feature, say \phi >= 1. I would like 
to find the center point of the feature if in 2D or the center-line of the 
feature if in 3D. 

This can be done for a variety of different features using different scalar 
fields, but for sake of argument, let's say I am looking at a vortex. In 2D, 
the vortex core is a point. In 3D, it may for a vortex sheet, or vortex tube, 
it may be arbitrary in shape (horseshoe, hairpin, etc.) and so there is no 
center point but really a center line for the core. I suppose this line may be 
composed of the center point of planes spanned by two of the principle 
components and the center line is oriented along the third, but I am not 
positive. 

At any rate, given a 2D or 3D scalar field and a threshold, what are the 
filters I should have in my pipeline to extract this information? I plan on 
doing this online with Catalyst. I would imagine that it would look like: 

MyData -> Threshold -> <Some filter I don't know to find the center 
points/lines> -> <Output each center point/line in a spreadsheet form> 

I'm not adverse to programming custom filters as needed, but I feel like this 
is something that a filter or set of filters may already exist to make this 
work. I don't want to reinvent the wheel because anything I come up with won't 
be as efficient as what is already there! 

Thanks, 

Tim 
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