[Insight-users] Documentation for itkHistogramMatchingImageFilter

Miller, James V (Research) millerjv at crd.ge.com
Wed Jun 22 09:38:33 EDT 2005


The histogram matching image filter basically transforms the histogram of one image into the histogram of the other
image.  This is done by mapping intensities through the CDF of one image and inverse CDF of the other.

If you take intensity x from image I1, and evaluate the CDF of I1 at x, you get a number y between 0 and 1.

	y = CDF_{I1}(x)

If you then take this value y and run it through the inverse CDF of image I2, you determine corresponding intensity
value from image 2.

	x' = CDF_{I2}^{-1}(y) = CDF_{I2}^{-1}( CDF_{I1}(x) )

If you transform all the intensities of Image 1 through this mapping, you will get an image that has approximately
the same histogram as Image 2.  I say "approximately" because we are dealing with sampled random variables and 
histograms as opposed to dealing with real density functions. 

The HistogramMatchingImageFilter performs this intensity transformation.  To perform the inverse CDF mapping, it
samples the range of the CDF at a user specified number of quantiles, then uses linear interpolation within each 
of these quantiles to map back to the domain.

I am not sure of a reference for this.  For a keyword search I would look at histogram equalization, and distribution whitening.  They are all related.

Jim


-----Original Message-----
From: insight-users-bounces+millerjv=crd.ge.com at itk.org
[mailto:insight-users-bounces+millerjv=crd.ge.com at itk.org]On Behalf Of
Jan Schreiber
Sent: Wednesday, June 22, 2005 6:09 AM
To: insight-users at itk.org
Subject: [Insight-users] Documentation for
itkHistogramMatchingImageFilter


Hi, 

does anyone have or know some detailed description on what happens in the itkHistogramMatchingImageFilter?
I want to use it in a project for my master's thesis and need some more information on what is going on, than I could find in the SoftwareGuide or in the Doxygen documentation. 
Is there a related publication?

Thanks, 
Jan


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