can you help me, I have 480(rows)*256(columns) which extracted by LBP operator.so i need to get the similarity matrices to apply the verification scenario.
e.g vector one with itself will give zero and vector one with vector two will give score and so on
why I am doing this, is because I need to calculate false accept rate and false reject rate
(FAR,FRR) by threshold.
thanks in advance
Use the pdist function. Note that it considers rows as instances (so you might want to transpose the matrix if you want to apply it to column vectors).
Related
I have a non-negative function f defined on a unit square S = [0,1] x [0,1] such that
My question is, how can I use MATLAB to generate a 2D random vector from S according to the probability density function f?
Rejection Sampling
The suggestion Luis Mendo made is very good because it applies to nearly all distribution functions. Based on this answer I wrote code for m.
An important point when using rejection sampling this way is that you must know the maximum of your pdf within the range. If you over-estimate the maximum your code will only run slower. If you under-estimate it it will create wrong numbers!
The idea is that you sample many uniform distributed points and accept depending on the probability density for the points.
pdf=#(x).5.*x(:,1)+3./2.*x(:,2);
maximum=2; %Right maximum for THIS EXAMPLE.
%If you are unable to determine the maximum of your
%function within the [0,1]x[0,1] range, please give an example.
result=[];
n=10;
while (size(result,1)<n)
%1. sample random point:
val=rand(1,2);
%2. Accept with probability pdf(val)/maximum
if rand<pdf(val)/maximum
%append to solution
result(end+1,:)=val;
end
end
I know that this solution is not a fast implementation, but I wanted to start with an implementation as simple as possible to make sure that the concept of rejection sampling becomes clear.
ICDF
Besides rejection sampling there is a different approach to address this issue on a more mathematical level, but you need to sit down and do some math first to end up with a better solution. For 1 dimensional distributions you typically sample using the ICDF (inverted cumulative density function) function simply using ICDF(rand(n,1)) to get random samples.
If you manage to do the math, you could instead for your PDF function define two functions ICDF1 (ICDF for the first dimension) and ICDF2 (ICDF for the second dimension) in matlab.
The first ICDF1 would map unifrom random distributed samples to sample values for the first dimension of your random distribution.
The second ICDF2 would map the output if ICDF1 and uniform distributed samples to your intended solution.
Here is some matlab code assuming you already defined ICDF1 and ICDF2
samples=ICDF1(rand(n,1));
samples(:,2)=ICDF2(samples,rand(n,1));
The great advantage of this solution is, that it does not reject any samples, being potentially much faster.
I have a matrix X, the size of which is 100*2000 double. I want to know which kind of scaling technique is applied to matrix X in the following command, and why it does not use z-score to do scaling?
X = X./repmat(sqrt(sum(X.^2)),size(X,1),1);
That scaling comes from linear algebra. That's what we call normalizing by producing a unit vector. Assuming that each row is an observation and each column is a feature, what's happening here is that we are going through every observation that you collected and normalizing each feature value over all observations such that the overall length / magnitude of a particular feature for all observations is set to 1.
The bottom division takes a look at each feature and determines the norm or magnitude of the feature over all observations. Once you find these magnitudes, you then take each feature for each observation and divide by their respective magnitudes.
The reason why unit vectors are often employed is to describe a point in feature space with respect to a set of basis vectors. Normalizing by producing unit vectors gives you the smallest possible way to represent one component in feature space and so what's probably happening here is that the observations are now being transformed such that each component / feature is being represented in terms of a set of basis vectors. Each basis vector is one feature in the data.
Check out the Wikipedia article on Unit Vectors for more details: http://en.wikipedia.org/wiki/Unit_vector
I performed PCA on a 63*2308 matrix and obtained a score and a co-efficient matrix. The score matrix is 63*2308 and the co-efficient matrix is 2308*2308 in dimensions.
How do i extract the column names for the top 100 features which are most important so that i can perform regression on them?
PCA should give you both a set of eigenvectors (your co-efficient matrix) and a vector of eigenvalues (1*2308) often referred to as lambda). You might been to use a different PCA function in matlab to get them.
The eigenvalues indicate how much of your data each eigenvector explains. A simple method for selecting features would be to select the 100 features with the highest eigen values. This gives you a set of feature which explain most of the variance in the data.
If you need to justify your approach for a write up you can actually calculate the amount of variance explained per eigenvector and cut of at, for example, 95% variance explained.
Bear in mind that selecting based solely on eigenvalue, might not correspond to the set of features most important to your regression, so if you don't get the performance you expect you might want to try a different feature selection method such as recursive feature selection. I would suggest using google scholar to find a couple of papers doing something similar and see what methods they use.
A quick matlab example of taking the top 100 principle components using PCA.
[eigenvectors, projected_data, eigenvalues] = princomp(X);
[foo, feature_idx] = sort(eigenvalues, 'descend');
selected_projected_data = projected(:, feature_idx(1:100));
Have you tried with
B = sort(your_matrix,2,'descend');
C = B(:,1:100);
Be careful!
With just 63 observations and 2308 variables, your PCA result will be meaningless because the data is underspecified. You should have at least (rule of thumb) dimensions*3 observations.
With 63 observations, you can at most define a 62 dimensional hyperspace!
How can i randomly pick a number from the given following matrix below?
A=[0.06 0.47 0.47]
I just want to randomly pick a number from the matrix above. I am doing this in matlab enviornment. please help.
Also, Is it possible assume a variable in matlab that tends to zero, like we do in limits?
If your matrix is M then to pick a random element with uniform probability you can use randi:
M(randi(numel(M)))
Yes, using randi:
A(randi(numel(A)))
Say, I have a cube of dimensions 1x1x1 spanning between coordinates (0,0,0) and (1,1,1). I want to generate a random set of points (assume 10 points) within this cube which are somewhat uniformly distributed (i.e. within certain minimum and maximum distance from each other and also not too close to the boundaries). How do I go about this without using loops? If this is not possible using vector/matrix operations then the solution with loops will also do.
Let me provide some more background details about my problem (This will help in terms of what I exactly need and why). I want to integrate a function, F(x,y,z), inside a polyhedron. I want to do it numerically as follows:
$F(x,y,z) = \sum_{i} F(x_i,y_i,z_i) \times V_i(x_i,y_i,z_i)$
Here, $F(x_i,y_i,z_i)$ is the value of function at point $(x_i,y_i,z_i)$ and $V_i$ is the weight. So to calculate the integral accurately, I need to identify set of random points which are not too close to each other or not too far from each other (Sorry but I myself don't know what this range is. I will be able to figure this out using parametric study only after I have a working code). Also, I need to do this for a 3D mesh which has multiple polyhedrons, hence I want to avoid loops to speed things out.
Check out this nice random vectors generator with fixed sum FEX file.
The code "generates m random n-element column vectors of values, [x1;x2;...;xn], each with a fixed sum, s, and subject to a restriction a<=xi<=b. The vectors are randomly and uniformly distributed in the n-1 dimensional space of solutions. This is accomplished by decomposing that space into a number of different types of simplexes (the many-dimensional generalizations of line segments, triangles, and tetrahedra.) The 'rand' function is used to distribute vectors within each simplex uniformly, and further calls on 'rand' serve to select different types of simplexes with probabilities proportional to their respective n-1 dimensional volumes. This algorithm does not perform any rejection of solutions - all are generated so as to already fit within the prescribed hypercube."
Use i=rand(3,10) where each column corresponds to one point, and each row corresponds to the coordinate in one axis (x,y,z)