[Insight-users] Obtain overlapping pixel volume

[Luis Ibanez]

Hi Jonathan,

ITK has a full set of filters for comparing segmentations.


You may want to use one or several of the following:

KappaStatistics Metric:
http://www.itk.org/Insight/Doxygen/html/classitk_1_1KappaStatisticImageToImageMetric.html

Match Cardinality:
http://www.itk.org/Insight/Doxygen/html/classitk_1_1MatchCardinalityImageToImageMetric.html

HausdorffDistanceImageFilter:
http://www.itk.org/Insight/Doxygen/html/classitk_1_1HausdorffDistanceImageFilter.html

ContourMeanDistanceImageFilter
http://www.itk.org/Insight/Doxygen/html/classitk_1_1ContourMeanDistanceImageFilter.html

ContourDirectedMeanDistanceImageFilter
http://www.itk.org/Insight/Doxygen/html/classitk_1_1ContourDirectedMeanDistanceImageFilter.html


You may also want to try the program "Valmet" that was developed
at the University of North Carolina at Chapel Hill:

     http://www.ia.unc.edu/dev/download/valmet/index.htm

Unfortunately Valmet is not Open Source, but you can download
free binaries from the link above.


     Regards,


        Luis


=======================
Jonathan Wong wrote:
> Hi. The 3d images I have already are segmented and are therefore binary. 
> If i wanted to compare the the overlapping number of voxels how would i 
> do this? I've been looking around on the ITK mailing list and someone 
> has tried to use the SpatialObjecttoImageFilter, adding two images 
> together and then counting the number of pixels with intensity 2, and 
> used the LabelImageStatistics filter.
> 
> The LabelImageStatistics filter does not work for me (it compiles but 
> the LabelImageStatisticsFilter Test fails), and I'm running it off of 
> CVS. Also the SpatialObjectImageFilter and iterating through the volume 
> give me different results.
> 
> I don't really know what the spatialObjectImageFilter does and would 
> like to know how it works.
> 
> Also what modifications would i have to make, if i were comparing images 
> of the same size, but different resolutions also pre-segmented to be 
> binary?
> 
> Thanks
> -Jonathan Wong
> 
> 
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