Questions about matlab median filter commands

This is a question about Matlab/Octave. I am seeing some results of the medfilt1(1D Median filter command in matlab) computation by which I am confused.

EDIT:Sorry forgot to mention:I am开发者_高级运维 using Octave for Windows 3.2.4. This is where i see this behavior.

Please see the questions below, and point if I am missing something.

1] I have a 1D data array b=[ 3 5 -8 6 0];

out=medfilt1(b,3);

I expected the output to be [3 3 5 0 0] but it is showing the output as [4 3 5 0 3]

How come? What is wrong here? FYI-Help says it pads the data at boundaries by 0(zero).

2] How does medfilt2(2D median filter command in matlab) work.

Help says "Each output pixel contains the median value in the m-by-n neighborhood around the corresponding pixel in the input image".

For m=3,n=3, So does it calculate a 3x3 matrix MAT for each of input pixels placed at its center and do median(median(MAT)) to compute its median value in the m-by-n neighbourhood?

Any pointers will help.

thank you. -AD

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I was not able to replicate your error with Matlab 7.11.0, but from the information in your question it seems like your version of medfilt1 does not differentiate between an odd or even n.

When finding the median in a vector of even length, one usually take the mean of the two median values,

median([1 3 4 5]) = (3+4)/2 = 3.5

This seems to be what happens in your case. Instead of treating n as odd, and setting the value to be 3, n is treated as even and your first out value is calculated to be

median([0 3 5]) = (3+5)/2 = 4

and so on.. EDIT: This only seems to happen in the endpoints, which suggest that the padding with zeros is not properly working in your Octave code.

For your second question, you are almost right, it calculates a 3x3 matrix in each center, but it does not do median(median(MAT)), but median(MAT(:)). There is a difference!

A = [1     2     3
     14     5    33
     11     7    13];

median(median(A)) = 11
median(A(:)) = 7