[Insight-users] Distinguishing distinct thresholded features

[Gaetan Lehmann]

On Wed, 21 Dec 2005 03:54:18 +0100, Zachary Pincus <zpincus at stanford.edu>  
wrote:

> There's a full list of all the classes available in ITK available here:
> http://www.itk.org/Doxygen/html/classes.html
>
> You may be interested in the ConnectedComponentImageFilter:
> http://www.itk.org/Doxygen/html/d3/d6a/ 
> classitk_1_1ConnectedComponentImageFilter.html
>
> I tend to modify this filter to achieve whatever results I want (e.g.  
> extracting each component into a separate image or whatever), but I  
> think the same results could be achieved (albeit less efficiently, but  
> these are fast operations we're talking about) with some combination of  
> ConnectedComponentImageFilter, RelabelComponentImageFilter,  
> ThresholdImageFilter, ExtractImageFilter, and some FloodFill iterator.
>
> I am actually surprised that there's no filter that takes a labeled  
> image (e.g. the result of ConnectedComponentImageFilter) and reports the  
> regions that contain each labeled region. Perhaps there is one but I  
> don't know of it...

LabelStatisticsImageFilter can give you the bounding box of each label.  
That's what you want ?
http://www.itk.org/Doxygen/html/d6/d8b/classitk_1_1LabelStatisticsImageFilter.html

>
> Zach
>
>
>
> On Dec 20, 2005, at 3:32 PM, Chris Farmer wrote:
>
>>
>>
>> Hi.  I’ve got a thresholded binary image which represents a couple  
>> disconnected features of interest.  I’d like to extract each of the  
>> distinct, disconnected features in my image and represent it on its own  
>> (as either a smaller binary image or some sort of mesh or pointset…  
>> doesn’t matter to me yet).  I’ve thought about just trudging through  
>> the main image with a neighborhood iterator and keeping track of  
>> “objects” I’ve seen, but I was hoping someone here could suggest  
>> whether there are established ways of accomplishing this.  I can’t seem  
>> to find any search terms that lead me in the right direction in the  
>> archives.
>>
>>
>>
>> Thanks,
>>
>> Chris
>>
>>
>>
>>
>>
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>
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