Questions about matlab median filter commands - matlab

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 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

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

Related

Matlab Dimensions Swapped in Meshgrid

In something which made me spent several hours, I have found an inconsistency in how Matlab deals with dimensions. I somebody can explain to me OR indicate me how to report this to Matlab, please enlight me.
For size, ones,zeros,mean, std, and most every other old and common commands existing inside Matlab, the dimension arrangement is like the classical one and like the intended standard (as per the size of every dimension), the first dimension is along the column vector, the second dimension is along the row vector, and the following are the non graphical following indexes.
>x(:,:,1)=[1 2 3 4;5 6 7 8];
>x(:,:,2)=[9 10 11 12;13 14 15 16];
>m=mean(x,1)
m(:,:,1) = 3 4 5 6
m(:,:,2) = 11 12 13 14
m=mean(x,2)
m(:,:,1) =
2.5000
6.5000
m(:,:,2) =
10.5000
14.5000
m=mean(x,3)
m = 5 6 7 8
9 10 11 12
d=size(m)
d = 2 4 2
However, for graphical commands like stream3,streamline and others relying on the meshgrid output format, the dimensions 1 and 2 are swapped!: the first dimension is the row vector and the second dimension is the column vector, and the following (third) is the non graphical index.
> [x,y]=meshgrid(1:2,1:3)
x = 1 2
1 2
1 2
y = 1 1
2 2
3 3
Then, for stream3 to operate with classically arranged matrices, we should use permute(XXX,[2 1 3]) at every 3D argument:
xyz=stream3(permute(x,[2 1 3]),permute(y,[2 1 3]),permute(z,[2 1 3])...
,permute(u,[2 1 3]),permute(v,[2 1 3]),permute(w,[2 1 3])...
,xs,ys,zs);
If anybody can explain why this happens, and can indicate to me why this is not a bug, welcome.
This behavior is not a bug because it is clearly documented as the intended behavior: https://www.mathworks.com/help/matlab/ref/meshgrid.html. Specifically:
[X,Y,Z]= meshgrid(x,y,z) returns 3-D grid coordinates defined by the vectors x, y, and z. The grid represented by X, Y, and Z has size length(y)-by-length(x)-by-length(z).
Without speaking to the original authors, the exact motivation may be a bit obscure, but I suspect it has to do with the fact that the y-axis is generally associated with the rows of an image, while x is the columns.
Columns are either "j" or "x" in the documentation, rows are either "i" or "y".
Some functions deal with spatial coordinates. The documentation will refer to "x, y, z". These functions will thus take column values before row values as input arguments.
Some functions deal with array indices. The documentation will refer to "i, j" (or sometimes "i1, i2, i3, ..., in", or using specific names instead of "i" before the dimension number). These functions will thus take row values before column values as input arguments.
Yes, this can be confusing. But if you pay attention to the names of the variables in the documentation, you will quickly figure out the right order.
With meshgrid in particular, if the "x, y, ..." order of arguments is confusing, use ndgrid instead, which takes arguments in array indexing order.

Vector with specific number of equally spaced values

I am familiar with the command:
(-5:0.1:5)
which creates a vector of values, equally spaced by 0.1, from -5 to 5.
However, is there a way to produce a vector of values, equally spaced from -5 to 5 such that there are, say, 100 values in the vector.
(-5:0.1:5) gives a vector with 101 values, however, is there a way to get a vector of 100 values without manually calculating the step size?
Yes there is. Use function linspace. See documentation here
linspace(1,10,10)
ans =
1 2 3 4 5 6 7 8 9 10
also the question is a duplicate of this question

Average on contiguos segments of a vector

I'm sure this is a trivial question for a signals person. I need to find the function in Matlab that outputs averaging of contiguous segments of windowsize= l of a vector, e.g.
origSignal: [1 2 3 4 5 6 7 8 9];
windowSize = 3;
output = [2 5 8]; % i.e. [(1+2+3)/3 (4+5+6)/3 (7+8+9)/3]
EDIT: Neither one of the options presented in How can I (efficiently) compute a moving average of a vector? seems to work because I need that the window of size 3 slides, and doesnt include any of the previous elements... Maybe I'm missing it. Take a look at my example...
Thanks!
If the size of the original data is always a multiple of widowsize:
mean(reshape(origSignal,windowSize,[]));
Else, in one line:
mean(reshape(origSignal(1:end-mod(length(origSignal),windowSize)),windowSize,[]))
This is the same as before, but the signal is only taken to the end minus the extra values less than windowsize.

Matlab Co-occurrence Matrix

I'm sure this is a very simple mistake by me somewhere! But when I use Matlab's graycomatrix function, I don't get the expected result. Instead of a matrix output I expect, I always get an 8 x 8 (nearly) zero matrix with one entry in the bottom right - usually equal to 16. I haven't changed the default settings or used 'offset', so I'm not too sure what the problem is.
That's because your image is not normalized!
Your image should be range 0-1, so:
I = [1 1 2; 2 2 3; 1 2 5]; %or any other I
glcm = graycomatrix(I/max(I(:))); % or I/255 , but it would not work for this example
should do the job.
In your case, Matlab interprets that everything avobe 1 is 1, therefore the co-occurrence matrix gives you a unique value in the max position.

Calculating the Local Ternary Pattern of an depth image

I found the detail and implementation of Local Ternary Pattern (LTP) on Calculating the Local Ternary Pattern of an image?. I want to ask more details that what the best way to choose the threshold t and also I have confusion in understand the role of reorder_vector = [8 7 4 1 2 3 6 9];
Unfortunately there isn't a good way to figure out what the threshold is using LTPs. It's mostly trial and error or by experimentation. However, I could suggest to make the threshold adaptive. You can use Otsu's algorithm to dynamically determine the best threshold of your image. This is assuming that the distribution of your intensities in the image is bimodal. In other words, there is a clear separation between objects and background. MATLAB has an implementation of this by the graythresh function. However, this generates a threshold between 0 and 1, so you will need to multiply the result by 255, assuming that the type of your image is uint8.
Therefore, do:
t = 255*graythresh(im);
im is the image that you desire to compute the LTPs. Now, I can certainly provide insight on what the reorder_vector is doing. Look at the following figure on how to calculate LTPs:
(source: hindawi.com)
When we generate the ternary code matrix (matrix in the middle), we need to generate an 8 element sequence that doesn't include the middle of the neighbourhood. We start from the east most element (row 2, column 3), then traverse the elements in counter-clockwise order. The reorder_vector variable allows you to select those specific elements that respect that order. If you recall, MATLAB can access matrices using column-major linear indices. Specifically, given a 3 x 3 matrix, we can access an element using a number from 1 to 9 and the memory is laid out like so:
1 4 7
2 5 8
3 6 9
Therefore, the first element of reorder_vector is index 8, which is the east most element. Next is index 7, which is the top right element, then index 4 which is the north facing element, then 1, 2, 3, 6 and finally 9.
If you follow these numbers, you will determine how I got the reorder_vector:
reorder_vector = [8 7 4 1 2 3 6 9];
By using this variable for accessing each 3 x 3 local neighbourhood, we would thus generate the correct 8 element sequence that respects the ordering of the ternary code so that we can proceed with the next stage of the algorithm.