Linearly Non-Separable Binary Classification Problem
First of all, this program isn' t working correctly for RBF ( gaussianKernel() ) and I want to fix it.
It is a non-linear SVM Demo to illustrate classifying 2 class with hard margin application.
Problem is about 2 dimensional radial random distrubuted data.
I used Quadratic Programming Solver to compute Lagrange multipliers (alphas)
xn = input .* (output*[1 1]); % xiyi
phi = gaussianKernel(xn, sigma2); % Radial Basis Function
k = phi * phi'; % Symmetric Kernel Matrix For QP Solver
gamma = 1; % Adjusting the upper bound of alphas
f = -ones(2 * len, 1); % Coefficient of sum of alphas
Aeq = output'; % yi
beq = 0; % Sum(ai*yi) = 0
A = zeros(1, 2* len); % A * alpha <= b; There isn't like this term
b = 0; % There isn't like this term
lb = zeros(2 * len, 1); % Lower bound of alphas
ub = gamma * ones(2 * len, 1); % Upper bound of alphas
alphas = quadprog(k, f, A, b, Aeq, beq, lb, ub);
To solve this non linear classification problem, I wrote some kernel functions such as gaussian (RBF), homogenous and non-homogenous polynomial kernel functions.
For RBF, I implemented the function in the image below:
Using Tylor Series Expansion, it yields:
And, I seperated the Gaussian Kernel like this:
K(x, x') = phi(x)' * phi(x')
The implementation of this thought is:
function phi = gaussianKernel(x, Sigma2)
gamma = 1 / (2 * Sigma2);
featDim = 10; % Length of Tylor Series; Gaussian Kernel Converge 0 so It doesn't have to Be Inf Dimension
phi = []; % Kernel Output, The Dimension will be (#Sample) x (featDim*2)
for k = 0 : (featDim - 1)
% Gaussian Kernel Trick Using Tylor Series Expansion
phi = [phi, exp( -gamma .* (x(:, 1)).^2) * sqrt(gamma^2 * 2^k / factorial(k)) .* x(:, 1).^k, ...
exp( -gamma .* (x(:, 2)).^2) * sqrt(gamma^2 * 2^k / factorial(k)) .* x(:, 2).^k];
end
end
*** I think my RBF implementation is wrong, but I don' t know how to fix it. Please help me here.
Here is what I got as output:
where,
1) The first image : Samples of Classes
2) The second image : Marking The Support Vectors of Classes
3) The third image : Adding Random Test Data
4) The fourth image : Classification
Also, I implemented Homogenous Polinomial Kernel " K(x, x') = ( )^2 ", code is:
function phi = quadraticKernel(x)
% 2-Order Homogenous Polynomial Kernel
phi = [x(:, 1).^2, sqrt(2).*(x(:, 1).*x(:, 2)), x(:, 2).^2];
end
And I got surprisingly nice output:
To sum up, the program is working correctly with using homogenous polynomial kernel but when I use RBF, it isn' t working correctly, there is something wrong with RBF implementation.
If you know about RBF (Gaussian Kernel) please let me know how I can make it right..
Edit: If you have same issue, use RBF directly that defined above and dont separe it by phi.
Why do you want to compute phi for Gaussian Kernel? Phi will be infinite dimensional vector and you are bounding the terms in your taylor series to 10 when we don't even know whether 10 is enough to approximate the kernel values or not! Usually, the kernel is computed directly instead of getting phi (and the computing k). For example [1].
Does this mean we should never compute phi for Gaussian? Not really, no, but we have to be slightly smarter about it. There have been recent works [2,3] which show how to compute phi for Gaussian so that you can compute approximate kernel matrices while having just finite dimensional phi's. Here [4] I give the very simple code to generate the approximate kernel using the trick from the paper. However, in my experiments I needed to generate anywhere from 100 to 10000 dimensional phi's to be able to get a good approximation of the kernel (depending upon on the number of features the original input had as well as the rate at which the eigenvalues of the original matrix tapers off).
For the moment, just use code similar to [1] to generate the Gaussian kernel and then observe the result of SVM. Also, play around with the gamma parameter, a bad gamma parameter can result in really bad classification.
[1] https://github.com/ssamot/causality/blob/master/matlab-code/Code/mfunc/indep/HSIC/rbf_dot.m
[2] http://www.eecs.berkeley.edu/~brecht/papers/07.rah.rec.nips.pdf
[3] http://www.eecs.berkeley.edu/~brecht/papers/08.rah.rec.nips.pdf
[4] https://github.com/aruniyer/misc/blob/master/rks.m
Since Gaussian kernel is often referred as mapping to infinity dimensions, I always have faith in its capacity. The problem here maybe due to a bad parameter while keeping in mind grid search is always needed for SVM training. Thus I propose you could take a look at here where you could find some tricks for parameter tuning. Exponentially increasing sequence is usually used as candidates.
Related
I have signal with peaks that has some structure on top of some background, and I'm trying to find a robust way to locate their positions and amplitudes.
For example, assume the peak has this form:
t=linspace(0,10,1e3);
w=0.25;
rw=#(t0) 2/(sqrt(3*w)*pi^0.25)*(1-((t-t0)/w).^2).*exp(-(t-t0).^2/(2*w.^2));
and I have several peaks a bit too close so their structure starts to interfere with the others, and one separated, so you can see it's structure:
pos=[ 2 2.5 3 8]; % positions
total_signal=0.*t;
for i=1:length(pos)
total_signal=total_signal+rw(pos(i));
end
plot(t,total_signal);
So the goal is to find all peaks found in total_signal and check that their positions agree with the original positions that were used to generate them in pos.
Your signal can be thought of as the convolution of a pulse train with the peak shape. Deconvolution can be used to retrieve the pulse train, whose peaks are the locations you are looking for. This assumes that the peak shape is known and constant.
I will start by rewriting your code as follows:
t = linspace(0,10,1e3);
w = 0.25;
rw = #(t) 2/(sqrt(3*w)*pi^0.25)*(1-((t)/w).^2).*exp(-(t).^2/(2*w.^2));
pos = [2, 2.5, 3, 8];
total_signal = zeros(size(t));
for i=1:length(pos)
total_signal = total_signal + rw(t-pos(i));
end
total_signal = total_signal + randn(size(t))*1e-2;
plot(t,total_signal);
I've changed rw to take t-t0, rather than just t0. This will allow me later to create a clean signal with just one peak in the middle. I have also added noise to the signal, for a more realistic problem. Without the noise, the problem is a lot easier to solve.
Wiener deconvolution is the simplest approach to solve this problem. In short, we assume that
total_signal = conv(pulse_train, shape)
In the frequency domain, this is written as
G = F .* H
(with .* the element-wise multiplication, using MATLAB syntax here). The Wiener deconvolution is:
F = (conj(H) .* G) ./ (abs(H).^2 + k)
with k some constant that we can tune to regularize the solution.
We implement this as follows:
shape = rw(linspace(-5,5,1e3));
G = fft(total_signal);
H = fft(ifftshift(shape)); % ifftshift moves the origin to sample #0, as expected by FFT.
k = 1;
F = (conj(H) .* G) ./ (abs(H).^2 + k);
pulse_train = ifft(F);
Now, findpeaks (requires the Signal Processing Toolbox) can be used to find the prominent peaks:
findpeaks(pulse_train, 'MinPeakProminence', 0.02)
Note how the four peaks are approximately the same height. It's not exact, because we're regularizing to deal with the noise. An exact solution is only possible in the noise-free case. Without noise, k=0, and the expression simplifies to F = G ./ H.
Also, the x-axis is off in the plot produced by findpeaks, but this shouldn't affect the results. The locations returned by it are indices into the array, those same indices can be used to index into t and find the actual locations of the peaks.
I have a set of frequency data with peaks to which I need to fit a Gaussian curve and then get the full width half maximum from. The FWHM part I can do, I already have a code for that but I'm having trouble writing code to fit the Gaussian.
Does anyone know of any functions that'll do this for me or would be able to point me in the right direction? (I can do least squares fitting for lines and polynomials but I can't get it to work for gaussians)
Also it would be helpful if it was compatible with both Octave and Matlab as I have Octave at the moment but don't get access to Matlab until next week.
Any help would be greatly appreciated!
Fitting a single 1D Gaussian directly is a non-linear fitting problem. You'll find ready-made implementations here, or here, or here for 2D, or here (if you have the statistics toolbox) (have you heard of Google? :)
Anyway, there might be a simpler solution. If you know for sure your data y will be well-described by a Gaussian, and is reasonably well-distributed over your entire x-range, you can linearize the problem (these are equations, not statements):
y = 1/(σ·√(2π)) · exp( -½ ( (x-μ)/σ )² )
ln y = ln( 1/(σ·√(2π)) ) - ½ ( (x-μ)/σ )²
= Px² + Qx + R
where the substitutions
P = -1/(2σ²)
Q = +2μ/(2σ²)
R = ln( 1/(σ·√(2π)) ) - ½(μ/σ)²
have been made. Now, solve for the linear system Ax=b with (these are Matlab statements):
% design matrix for least squares fit
xdata = xdata(:);
A = [xdata.^2, xdata, ones(size(xdata))];
% log of your data
b = log(y(:));
% least-squares solution for x
x = A\b;
The vector x you found this way will equal
x == [P Q R]
which you then have to reverse-engineer to find the mean μ and the standard-deviation σ:
mu = -x(2)/x(1)/2;
sigma = sqrt( -1/2/x(1) );
Which you can cross-check with x(3) == R (there should only be small differences).
Perhaps this has the thing you are looking for? Not sure about compatability:
http://www.mathworks.com/matlabcentral/fileexchange/11733-gaussian-curve-fit
From its documentation:
[sigma,mu,A]=mygaussfit(x,y)
[sigma,mu,A]=mygaussfit(x,y,h)
this function is doing fit to the function
y=A * exp( -(x-mu)^2 / (2*sigma^2) )
the fitting is been done by a polyfit
the lan of the data.
h is the threshold which is the fraction
from the maximum y height that the data
is been taken from.
h should be a number between 0-1.
if h have not been taken it is set to be 0.2
as default.
i had similar problem.
this was the first result on google, and some of the scripts linked here made my matlab crash.
finally i found here that matlab has built in fit function, that can fit Gaussians too.
it look like that:
>> v=-30:30;
>> fit(v', exp(-v.^2)', 'gauss1')
ans =
General model Gauss1:
ans(x) = a1*exp(-((x-b1)/c1)^2)
Coefficients (with 95% confidence bounds):
a1 = 1 (1, 1)
b1 = -8.489e-17 (-3.638e-12, 3.638e-12)
c1 = 1 (1, 1)
I found that the MATLAB "fit" function was slow, and used "lsqcurvefit" with an inline Gaussian function. This is for fitting a Gaussian FUNCTION, if you just want to fit data to a Normal distribution, use "normfit."
Check it
% % Generate synthetic data (for example) % % %
nPoints = 200; binSize = 1/nPoints ;
fauxMean = 47 ;fauxStd = 8;
faux = fauxStd.*randn(1,nPoints) + fauxMean; % REPLACE WITH YOUR ACTUAL DATA
xaxis = 1:length(faux) ;fauxData = histc(faux,xaxis);
yourData = fauxData; % replace with your actual distribution
xAxis = 1:length(yourData) ;
gausFun = #(hms,x) hms(1) .* exp (-(x-hms(2)).^2 ./ (2*hms(3)^2)) ; % Gaussian FUNCTION
% % Provide estimates for initial conditions (for lsqcurvefit) % %
height_est = max(fauxData)*rand ; mean_est = fauxMean*rand; std_est=fauxStd*rand;
x0 = [height_est;mean_est; std_est]; % parameters need to be in a single variable
options=optimset('Display','off'); % avoid pesky messages from lsqcurvefit (optional)
[params]=lsqcurvefit(gausFun,x0,xAxis,yourData,[],[],options); % meat and potatoes
lsq_mean = params(2); lsq_std = params(3) ; % what you want
% % % Plot data with fit % % %
myFit = gausFun(params,xAxis);
figure;hold on;plot(xAxis,yourData./sum(yourData),'k');
plot(xAxis,myFit./sum(myFit),'r','linewidth',3) % normalization optional
xlabel('Value');ylabel('Probability');legend('Data','Fit')
My approach
fun = #(y) (1/sqrt(pi))*exp(-(y-1).^2).*log(1 + exp(-4*y))
integral(fun,-Inf,Inf)
This gives NaN.
So I tried plotting it.
y= -10:0.1:10;
plot(y,exp(-(y-1).^2).*log(1 + exp(-4*y)))
Then understood that domain (siginificant part) is from -4 to +4.
So changed the limits to
integral(fun,-10,10)
However I do not want to always plot the graph and then know its limits. So is there any way to know the integral directly from -Inf to Inf.
Discussion
If your integrals are always of the form
I would use a high-order Gauss–Hermite quadrature rule.
It's similar to the Gauss-Legendre-Kronrod rule that forms the basis for quadgk but is specifically tailored for integrals over the real line with a standard Gaussian multiplier.
Rewriting your equation with the substitution x = y-1, we get
.
The integral can then be computed using the Gauss-Hermite rule of arbitrary order (within reason):
>> order = 10;
>> [nodes,weights] = GaussHermiteRule(order);
>> f = #(x) log(1 + exp(-4*(x+1)))/sqrt(pi);
>> sum(f(nodes).*weights)
ans =
0.1933
I'd note that the function below builds a full order x order matrix to compute nodes, so it shouldn't be made too large.
There is a way to avoid this by explicitly computing the weights, but I decided to be lazy.
Besides, event at order 100, the Gaussian multiplier is about 2E-98, so the integrand's contribution is extremely minimal.
And while this isn't inherently adaptive, a high-order rule should be sufficient in most cases ... I hope.
Code
function [nodes,weights] = GaussHermiteRule(n)
% ------------------------------------------------------------------------------
% Find the nodes and weights for a Gauss-Hermite Quadrature integration.
%
if (n < 1)
error('There is no Gauss-Hermite rule of order 0.');
elseif (n < 0) || (abs(n - round(n)) > eps())
error('Given order ''n'' must be a strictly positive integer.');
else
n = round(n);
end
% Get the nodes and weights from the Golub-Welsch function
n = (0:n)' ;
b = n*0 ;
a = b + 0.5 ;
c = n ;
[nodes,weights] = GolubWelsch(a,b,c,sqrt(pi));
end
function [xk,wk] = GolubWelsch(ak,bk,ck,mu0)
%GolubWelsch
% Calculate the approximate* nodes and weights (normalized to 1) of an orthogonal
% polynomial family defined by a three-term reccurence relation of the form
% x pk(x) = ak pkp1(x) + bk pk(x) + ck pkm1(x)
%
% The weight scale factor mu0 is the integral of the weight function over the
% orthogonal domain.
%
% Calculate the terms for the orthonormal version of the polynomials
alpha = sqrt(ak(1:end-1) .* ck(2:end));
% Build the symmetric tridiagonal matrix
T = full(spdiags([[alpha;0],bk,[0;alpha]],[-1,0,+1],length(alpha),length(alpha)));
% Calculate the eigenvectors and values of the matrix
[V,xk] = eig(T,'vector');
% Calculate the weights from the eigenvectors - technically, Golub-Welsch requires
% a normalization, but since MATLAB returns unit eigenvectors, it is omitted.
wk = mu0*(V(1,:).^2)';
end
I've had success with transforming such infinite-bounded integrals using a numerical variable transformation, as explained in Numerical Recipes 3e, section 4.5.3. Basically, you substitute in y=c*tan(t)+b and then numerically integrate over t in (-pi/2,pi/2), which sweeps y from -infinity to infinity. You can tune the values of c and b to optimize the process. This approach largely dodges the question of trying to determine cutoffs in the domain, but for this to work reliably using quadrature you have to know that the integrand does not have features far from y=b.
A quick and dirty solution would be to look for a position, where your function is sufficiently small enough and then taking it as limits. This assumes that for x>0 the function fun decreases montonically and fun(x) is roughly the same size as fun(-x) for all x.
%// A small number
epsilon = eps;
%// Stepsize for searching bound
stepTest = 1;
%// Starting position for searching bound
position = 0;
%// Not yet small enough
smallEnough = false;
%// Search bound
while ~smallEnough
smallEnough = (fun(position) < eps);
position = position + stepTest;
end
%// Calculate integral
integral(fun, -position, position)
If your were happy with plotting the function, deciding by eye where you can cut, then this code will suffice, I guess.
I'm writing a code to detect anomalys using Gaussian distribution.
This is the code i wrote to compute the probability density function:
function p = multivariateGaussian(X, mu, Sigma2)
%MULTIVARIATEGAUSSIAN Computes the probability density function of the
%multivariate gaussian distribution.
% p = MULTIVARIATEGAUSSIAN(X, mu, Sigma2) Computes the probability
% density function of the examples X under the multivariate gaussian
% distribution with parameters mu and Sigma2. If Sigma2 is a matrix, it is
% treated as the covariance matrix. If Sigma2 is a vector, it is treated
% as the \sigma^2 values of the variances in each dimension (a diagonal
% covariance matrix)
%
k = length(mu);
if (size(Sigma2, 2) == 1) || (size(Sigma2, 1) == 1)
Sigma2 = diag(Sigma2);
end
X = bsxfun(#minus, X, mu(:)');
p = (2 * pi) ^ (- k / 2) * det(Sigma2) ^ (-0.5) * ...
exp(-0.5 * sum(bsxfun(#times, X * pinv(Sigma2), X), 2));
end
My first question: are there a fastter and clever way to compute this? I have a little matlab cluster setuped here with 2 pcs, but in this case, i have no clue how to parallelize this.
My second question:In one of the matrices i using as trainning set is [42712X19700], even having 24 gb of ram, im getting out of memory error. Is it possible use a technique like random forest(slice the trainning set, and then combine de results?)? Or any other way to circumvent this problem?
I appreciate any help. Tks in advance!
Divide the data into small chunks and apply parfor for processing of each chunk. That is simply my choose for large scale processing. Or you might use process based parallelization and read one chunk as another process is computing on another chunk of data.
I try to write an algorithm which determine $\mu$, $\sigma$,$\pi$ for each class from a mixture multivariate normal distribution.
I finish with the algorithm partially, it works when I set the random guess values($\mu$, $\sigma$,$\pi$) near from the real value. But when I set the values far from the real one, the algorithm does not converge. The sigma goes to 0 $(2.30760684053766e-24 2.30760684053766e-24)$.
I think the problem is my covarience calculation, I am not sure that this is the right way. I found this on wikipedia.
I would be grateful if you could check my algorithm. Especially the covariance part.
Have a nice day,
Thanks,
2 mixture gauss
size x = [400, 2] (400 point 2 dimension gauss)
mu = 2 , 2 (1 row = first gauss mu, 2 row = second gauss mu)
for i = 1 : k
gaussEvaluation(i,:) = pInit(i) * mvnpdf(x,muInit(i,:), sigmaInit(i, :) * eye(d));
gaussEvaluationSum = sum(gaussEvaluation(i, :));
%mu calculation
for j = 1 : d
mu(i, j) = sum(gaussEvaluation(i, :) * x(:, j)) / gaussEvaluationSum;
end
%sigma calculation methode 1
%for j = 1 : n
% v = (x(j, :) - muNew(i, :));
% sigmaNew(i) = sigmaNew(i) + gaussEvaluation(i,j) * (v * v');
%end
%sigmaNew(i) = sigmaNew(i) / gaussEvaluationSum;
%sigma calculation methode 2
sub = bsxfun(#minus, x, mu(i,:));
sigma(i,:) = sum(gaussEvaluation(i,:) * (sub .* sub)) / gaussEvaluationSum;
%p calculation
p(i) = gaussEvaluationSum / n;
Two points: you can observe this even when you implement gaussian mixture EM correctly, but in your case, the code does seem to be incorrect.
First, this is just a problem that you have to deal with when fitting mixtures of gaussians. Sometimes one component of the mixture can collapse on to a single point, resulting in the mean of the component becoming that point and the variance becoming 0; this is known as a 'singularity'. Hence, the likelihood also goes to infinity.
Check out slide 42 of this deck: http://www.cs.ubbcluj.ro/~csatol/gep_tan/Bishop-CUED-2006.pdf
The likelihood function that you are evaluating is not log-concave, so the EM algorithm will not converge to the same parameters with different initial values. The link I gave above also gives some solutions to avoid this over-fitting problem, such as putting a prior or regularization term on your parameters. You can also consider running multiple times with different starting parameters and discarding any results with variance 0 components as having over-fitted, or just reduce the number of components you are using.
In your case, your equation is right; the covariance update calculation on Wikipedia is the same as the one on slide 45 of the above link. However, if you are in a 2d space, for each component the mean should be a length 2 vector and the covariance should be a 2x2 matrix. Hence your code (for two components) is wrong because you have a 2x2 matrix to store the means and a 2x2 matrix to store the covariances; it should be a 2x2x2 matrix.