unifrnd, matrix

Hi friends, can you help me out with this

I have a matrix(or a frame(image) in my case) called 'z', say its a 3x3 frame.

x = [1.2, 0.45, 0.2, 0.25, 20]';

n=3;

m=3;

blur = 0.7;

a = 3.*randn(n,m);

for i=1:n

 for j=1:m

if i-x(1) < 3 && j-x(3) <3

      h(i,j) = x(5)/(2 * pi * (blur^2)) * exp(-((i-x(1))^2 + (j-x(3))^2)/ (2* (blur^2)))

else

      h(i,j) = 0
  end

 end

end

z = a + h;

So, 'z' will be a 3x3 frame having certain values for the 9 pixels.

'x' is a state vector having are [position along x-axis, velocity along x-axis, position along y-axis, velocity along y-axis, Intensity].

So the pixels in frame 'z' that fall in the vicinity of the position given by 'x' in x-y direction ie x(1) & x(3) will have higher values(Intensity than others)

now, what I want is, the position of the particles should be initialized from the distribution q(x,y | z) is a uniform density over those regions in 'z' for which z(i,j) > T

T = threshold = 2

I need to draw values from that distribution.

Can you help me out with this?

[Image Analyst]

Probably could, if I could figure out what you mean. I fail to see why z would be higher at locations where x and y are listed in your 'x' state vector. Why would that be? What guarantees or forces that? Also, shouldn't this really have been continued on your original thread http://www.mathworks.com/matlabcentral/answers/18395-uniform-density-matrix

[PChoppala]

Well, if you run could the code I put in, you can see that 'h' will be a matrix that has more intensity at locations given by (x(1),x(3)). It can be obtained from the Gaussian expression where we plug in x(1), x(3) and x(5).

a is just noise

So, z = a+h(noise + signal)

Now I want the vicinity of 'z' where it crosses the threshold, and then initialize a new position.

this new position will be drawn from a uniform density over those regions, where the pixel values(intensities) are greater than T=2

I posted it again, because I could not get a reply for a long time from any of the Matlab friends, and I assume, with this I should be able to run my program correctly, for which I am due tomorrow.

[PChoppala]

z =

0.8097 -0.2496 0.0621

5.8031 -2.2792 -0.1705

3.5348 11.1277 -3.6788

I have a better way of asking, my friend!

Consider the above matrix.

I want to find the indices of the rows in which atleast one element is greater than 2

Similarly, I want to find the indices of the columns in which atleast one element is greater than 2

Please help

[Image Analyst]

Well I didn't see it because x(1) = 1.2 and x(3) = 0.2 and you can't have a z value at the (1.2th, 0.2th) location, unless you make another array that is interpolated to give values there. But whatever...see my answer below, which answers your last question and thankfully doesn't depend on understanding the first part.

Answer 1

[Image Analyst]

Just do

[rows columns] = find(z>2);

The find() function is usually one of the first ones people learn (along with min, max, mean, and sum). You'll also need sum if you want to find which rows or columns have more than 2 elements above the threshold. It's a really really easy one-liner. Write back if you can't figure it out.

[PChoppala]

I used

Arr=[];

for n1=1:n

p = find(h(n1,:)>2);

if isempty(p) == 0

Arr = [Arr n1];

end

end

Brr=[];

for m1=1:m

q = find(h(:,m1)>2);

if isempty(q) == 0

Brr = [Brr m1];

end

end

Any better solution?

[PChoppala]

n=rows=3

m=columns=3

[Image Analyst]

Not sure what all that was. This is what I was thinking:

columnsWithMoreThan2 = sum(z>2, 1);

rowsWithMoreThan2 = sum(z>2, 2);