20060410-Quick tip: Determining uniqueness of local maximum

原文：http://blogs.mathworks.com/steve/2006/04/10/quick-tip-determining-uniqueness-of-neighborhood-maximum/

Let's review first the concept of a local maximum filter. This image operator replaces every pixel value with the maximum pixel value found in a neighborhood surrounding the pixel. For example:

a = magic(5)

What should the output be for a local maximum filter using 3-by-3 neighborhoods?

neighborhood = a(2:4, 1:3)

Local maximum filtering is the same as morphological dilation, so I usually use imdilate to compute the result for the entire image:

b = imdilate(a, ones(3,3))

I suggest using ordfilt2 twice. This function is for ''two-dimensional order-statistic filtering.'' This operation replaces every pixel value with the n-th sorted pixel value in its neighborhood. It uses an ascending sort order, so if n is the number of pixels in the neighborhood, the operation is exactly equivalent to the local maximum filter.

The specific procedure is:

1. Call ordfilt2 to get the highest pixel values in each neighborhood.
2. Call ordfilt2 again to get the second-highest pixel values in each neighborhood.
3. Determine where the outputs of steps 1 and 2 are equal. Wherever they are equal, the local maximum is not unique.

Let's try an example. First, modify our original matrix a bit:

ap = a;
ap(3,4) = 22

Get the highest pixel values in each 3-by-3 neighborhood:

highest = ordfilt2(ap, 9, ones(3,3))

Get the second-highest pixel values:

second_highest = ordfilt2(ap, 8, ones(3,3))

Determine where the highest and second highest pixel values are equal:

non_unique_maxima_mask = (highest == second_highest)