[Insight-users] Segmentation based on discrete points
Xuan Zhao
xuanaux at gmail.com
Sun Nov 15 13:13:35 EST 2009
Hi, Luis
Thank you very much for your comments. I only have dense points. I will try
the connected components filter you mentioned. What I did is simplifying the
problem from 3D to 2D by taking every points on each slice and building a
polygon. Then I called function isInside(point) to determine whether a point
is inside or outside of the polygon boundary determined by the dense points.
It works although inefficient.
Luis Ibanez wrote:
>
> Hi Xuan,
>
> Do you have only the dense cloud or points ?
>
> or
> do you have also a topological surface that connects these points ?
> (e.g. polygonal faces defining the surface)
>
> ---
>
> If you only have the cloud of points, then, this is
> an incompletely defined problem, and the best
> you can hope to get is a "reasonable" partition
> of space, but it will be impossible to guarantee
> that a correct segmentation can be constructed.
>
> ---
>
> Along the lines of "imperfect" but plausible methods:
>
> One option that you may want to explore is to use
> to dense set of points to turn on pixels in an underlying
> image. If the cloud is dense enough, you may get lucky
> and be able to create a closed surface of pixels.
>
> If you get there, then partitioning the space will be
> as simple as using the connected components filter
> in ITK, or if you don't mind providing a seed point,
> you could use the region growing filters.
>
>
> Regards,
>
>
> Luis
>
>
> -------------------------------------------------------------------------
> On Fri, Nov 13, 2009 at 12:27 PM, Xuan Zhao <xuanaux at gmail.com> wrote:
>>
>> Dear all,
>>
>> Is there a filter in ITK to do segmentation based on discrete points that
>> specify the surface of a volume of interest(VOI)? Or in other word, is
>> there
>> a function in ITK that can discriminate whether a point is inside or
>> outside
>> the surface determined by densely distributed discrete points?
>>
>> Thank you in advance,
>>
>> Xuan
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>>
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