I've been trying to implement this equation into matlab:
http://www.mathworks.com/matlabcentral/answers/uploaded_files/46747/Capture.PNG
Omega is some known region of the image u1.
and p1,i is:
http://www.mathworks.com/matlabcentral/answers/uploaded_files/46748/Capture1.PNG
Pay attention: u1 is an image (column ordered), an p1,i is supposed to be calculated in each pixel of this image.
Now, I've calculated p1,i in this way:
[f1,xi1] = ksdensity(u1(:),1:255);
p1_u1 = reshape(f1(floor(u1(:))+1),M,N);
Now my problem is to calculate the former equation.
I've tried with for loop but its taking forever..
Any other suggestions? Maybe there's a way using ksdensity and change the value inside the integral?
Thanks!
Related
I have a small project on Moving object detection in moving camera in which i have to use negative optical flow vector to minimize ego motion compensation. I have a video and some particular consecutive frames in which average of negative optical flow vector has to be computed. I have already calculated Optical flow between say, (k-1)th and kth frame. Also, I have calculated average of optical flow vector V=[u,v], where v is the average of horizontal optical flow and u is the average of vertical flow. Now, I have to apply inverse of optical flow vector i.e., -V to the (k-1)th frame. I'm new to matlab and i don't know much about it. Please help
I have tried this code segment to do so but the results aren't as expected
function I1=reverseOF(I,V)
R=I(:,:,1);
G=I(:,:,2);
B=I(:,:,3);
[m,n]=size(rgb2gray(I));
for i=1:m
for j=1:n
v1=[j i];
v2=-V;
v3=v1.*v2;
R(floor(1+abs(v3(1,2))),floor(1+abs(v3(1,2))))=R(i,j);
G(floor(1+abs(v3(1,2))),floor(1+abs(v3(1,2))))=G(i,j);
B(floor(1+abs(v3(1,2))),floor(1+abs(v3(1,2))))=B(i,j);
I1(floor(1+abs(v3(1,2))),floor(1+abs(v3(1,2))))=I(i,j);
end
end
I1=cat(3,R,G,B);
enter code here
I have used abs() function because otherwise some error was occuring like "attempted to access negative location; index must be a positive or logical".
Image A and Image B are the images that i have used to estimate the optical flow.enter image description here
This is the result that i am obtaining after applying the above function.
enter image description here
Unfortunately, you cant do this easily. This is a quite advanced research problem, because obtaining the inverse of a vector field on a mesh grid is not an easy problem, actually its quite hard.
Notice that your vector field (optical flow) start in a mesh grid, but it doesnt end in a mesh grid, it ends in random subpixel positions. If you just invert this field, doing -V is not enough! The result wont be the inverse!
This is a open research problem, look for example at this 2010 paper that addresses exactly this issue, and proposes a method to create "pseudoinverses".
Suppose you have that inverse, because you computed it somehow. Your code is quite bad for it, and the solutions (abs!) are showing (no offense) that you are not really understanding what you are doing. For a known vector field {Vx,Vy}, size equals to the image size (if its not, you can figure out easily how to interpolate it unsig interp2 ) the code would look something like:
newimg=zeros(size(I));
[ix,iy]=meshgrid(1:size(I,1),1:size(I,2));
newimg(:,:,1)=interp2(I(:,:,1),ix+Vx,iy+Vy); % this is your whole loop.
newimg(:,:,2)=interp2(I(:,:,3),ix+Vx,iy+Vy); % this is your whole loop.
newimg(:,:,3)=interp2(I(:,:,2),ix+Vx,iy+Vy); % this is your whole loop.
I have an RGB image and I am trying to calculate its Gaussian derivative.
Image is a greyscale image and the Gaussian window is 5x5,st is the standard deviation
This is the code i am using in order to find a 2D Gaussian derivative,in Matlab:
N=2
[X,Y]=meshgrid(-N:N,-N:N)
G=exp(-(x.^2+y.^2)/(2*st^2))/(2*pi*st)
G_x = -x.*G/(st^2);
G_x_s = G_x/sum(G_x(:));
G_y = -y.*G/(st^2);
G_y_s = G_y/sum(G_y(:));
where st is the standard deviation i am using. Before I proceed to the convolution of the Image using G_x_s and G_y_s, i have the following problem. When I use a standard deviation that is an even number(2,4,6,8) the program works and gives results as expected. But when i use an odd number for standard deviation (3 or 5) then the G_y_s value becomes Inf because sum(G_y(:))=0. I do not understand that behavior and I was wondering if there is some problem with the code or if in the above formula the standard deviation can only be an even number. Any help will be greatly appreciated.
Thank you.
Your program doesn't work at all. The results you find when using an even number is just because of some numerical errors.
Your G will be a matrix symmetrical to the center. x and y are both point symmetrical to the center. So the multiplication (G times x or y) will result in a matrix with a sum of zero. So a division by that sum is a division by zero. Everything else you observe is because of some roundoff errors. Here, I see a sum og G_xof about 1.3e-17.
I think your error is in the multiplication x.*G and y.*G. I can not figure out why you would do that.
I assume you want to do edge detection rigth? You can use fspecialto create several edge filters. Laplace of gaussian for instance. You could also create two gaussian filters with different standard deviations and subtract them from another to get an edge filter.
I have some data which I wish to model in order to be able to get relatively accurate values in the same range as the data.
To do this I used polyfit to fit a 6th order polynomial and due to my x-axis values it suggested I centred and scaled it to get a more accurate fit which I did.
However, now I want to find the derivative of this function in order to model the velocity of my model.
But I am not sure how the polyder function interacts with the scaled and fitted polyfit which I have produced. (I don't want to use the unscaled model as this is not very accurate).
Here is some code which reproduces my problem. I attempted to rescale the x values before putting them into the fit for the derivative but this still did no fix the problem.
x = 0:100;
y = 2*x.^2 + x + 1;
Fit = polyfit(x,y,2);
[ScaledFit,s,mu] = polyfit(x,y,2);
Deriv = polyder(Fit);
ScaledDeriv = polyder(ScaledFit);
plot(x,polyval(Deriv,x),'b.');
hold on
plot(x,polyval(ScaledDeriv,(x-mu(1))/mu(2)),'r.');
Here I have chosen a simple polynomial so that I could fit it accurate and produce the actual derivative.
Any help would be greatly appreciated thanks.
I am using Matlab R2014a BTW.
Edit.
Just been playing about with it and by dividing the resulting points for the differential by the standard deviation mu(2) it gave a very close result within the range -3e-13 to about 5e-13.
polyval(ScaledDeriv,(x-mu(1))/mu(2))/mu(2);
Not sure quite why this is the case, is there another more elegant way to solve this?
Edit2. Sorry for another edit but again was mucking around and found that for a large sample x = 1:1000; the deviation became much bigger up to 10. I am not sure if this is due to a bad polyfit even though it is centred and scaled or due to the funny way the derivative is plotted.
Thanks for your time
A simple application of the chain rule gives
Since by definition
it follows that
Which is exactly what you have verified numerically.
The lack of accuracy for large samples is due to the global, rather then local, Lagrange polynomial interpolation which you have done. I would suggest that you try to fit your data with splines, and obtain the derivative with fnder(). Another option is to apply the polyfit() function locally, i.e. to a moving small set of points, and then apply polyder() to all the fitted polynomials.
I implemented the Basic Lucas Kanade Optical Flow algorithm in Matlab.
I used the algorithm from Wikipedia.
Since I want to improve this Basic optical flow algorithm, I tried adding a weightening function which makes certain Pixels in the beighbourhood more important or less important (see also Wikipedia).
I basically calculated the following for every Pixel in the beighbourhood and the Center Pixel itself.
for: Center Pixel and every neighbourhood-pixel
sigma = 10;
weight(s) = (1/(2*pi*sigma^2)) * exp(-((first-x)^2+(second-y)^2)/(2*sigma^2))
x,y ist the Center Point pixel, it always stays the same.
first,second is the current neighbourhood-pixel
Since I am using a 5x5 neighbourhood, (first-x) or (second-y) will always be one of these: "0,1,-1,2,-2"
I then apply the weight-values in each part of the sum.
Problem: With Sigma = 10 I don't get a better result for the optical flow than without the weightening function.
With smaller Sigmas it's not better. Afterall there is no difference between the Output vectors with or without the gaussian function
Is there a way to improve this Gaussian function to actually make the vectors more acurate than without weightening?
Thank you sooo much.
I'm not sure how you apply the values, but it usually should make a little difference.
For a better optical flow you could :
presmooth the images with a gaussian
use a spatiotemporal Lucas-Kanade method
or use a more advanced algorithm
If I estimate the entropy of a vector of standard normal random variables using the Matlab entropy() function, I get an answer somewhere in the region of 4, whereas the actual entropy should be 0.5 * log(2*pi*e*sigma^2) which is approximately equal to 1.4.
Does anyone know where the discrepancy is coming from?
Note: To save time here is the Matlab code
for i = 1:1000
X(i) = randn();
end
'The entropy of X is'
entropy(X)
Please read the help (help entropy) or documentation for entropy. You'll see that it's designed for images and uses a histogram technique rather than calculating the it analytically. You'll need to create your own function if you want the formula from Wikipedia, but as the formula is so simple, that should be no problem.
I believe that the reason that you're getting such divergent answers is that entropy scales the bins of the histogram by the number of elements. If you want to uses such an estimation technique you'll want to use hist and scale the bins by area. See this StackOverflow question.