Comments on: Nonlinear operations using imfilter

By: Steve

Steve — Fri, 03 Dec 2010 15:32:32 +0000

Sean—A lot of the mathematical details are omitted in this post. See the places, for example, where I include a note that says “scale factors omitted.” There is a significant difference between stdfilt and the code in this post because, well, they don’t compute the same thing.

By: Sean de Wolski

Sean de Wolski — Thu, 02 Dec 2010 22:20:56 +0000

Okay, I guess by opening stdfilt you can see the difference between the two. So why the discrepancy?

By: Sean de Wolski

Sean de Wolski — Thu, 02 Dec 2010 22:01:36 +0000

Hi Steve,
I realize this is old but I have a question about it. Why is there such a significant difference between your imfilter method and stdfilt with numbers in the range 0-255. Is it just numerical round off? The thing that irks me is that the greatest difference between the two is in the middle where there is a significant standard deviation and that the standard deviation from your imfilter example doesn’t seem reasonable.

Thanks!
-Sean

%%%
I = double(imread(‘cameraman.tif’));

h = ones(11,11);
Istd1 = sqrt(imfilter(I.^2, h, ‘symmetric’) – imfilter(I, h, ‘symmetric’).^2 / numel(h));

Istd2 = stdfilt(I,h);

imshow(Istd1-Istd2,[])

norm(Istd1(:)-Istd2(:))
%{
ans = 75950
%}

max([Istd1(:), Istd2(:)])
%{
ans = 1092.4 99.719
%}

By: Steve

Steve — Tue, 20 Jul 2010 11:51:13 +0000

Diego—It doesn't matter; the procedure still gives the correct answer. Note that the geometric mean of a set of values containing a 0 is 0. Try this:

exp(-inf)

By: Diego

Diego — Tue, 20 Jul 2010 02:58:24 +0000

I am aware this post is old, but just to note:

At this line:
geo_mean = imfilter(log(I), h, ‘replicate’);

You are calculating the log() of an array of doubles where 0 <= x <= 1.

That will produce an array of negative values, and possibly -Inf.

also replicating to the image as black squares.

By: Steve

Steve — Thu, 10 Jul 2008 16:58:19 +0000

Francisco—At last count, there were 13,273,474 journal and conference papers describing different methods of edge detection. :-)

Think about regions of differing texture. The local standard deviation operator would have a relatively higher response in regions of “busier” texture, and it would have a relatively lower response in “smoother” regions. This is different than what most people think an edge detector does.

By: francisco

francisco — Wed, 09 Jul 2008 16:56:05 +0000

Hi Steve,

You mention that the local standard deviation technique measures the ‘business’ of an image. How is this different, if at all, from edge detection? It looks like your first example does a really amazing job at finding edges, albeit they are rather thick.

thanks
-francisco

By: Steve

Steve — Mon, 07 Jul 2008 15:58:31 +0000

Luca—Thanks for submitting your code to the File Exchange!

By: Luca Balbi

Luca Balbi — Mon, 07 Jul 2008 15:21:18 +0000

This very same “sum of squares minus sqaure of sum” consideration drove me to re-implement the Kuwahara nonlinear filter in a faster way, though I did not use imfilter but just straightforward convolution, as I did not want the Image Processing Toolbox to be needed:

http://www.mathworks.com/matlabcentral/fileexchange/loadFile.do?objectId=15027&objectType=FILE

Nice example by the way Steve! And thanks for pointing to the stdfilt function.