I'm using the FastIca toolbox (https://research.ics.aalto.fi/ica/fastica/) but am confused about the orientation of the resulting W (separating/unmixing) matrix.
Let X be a n x B matrix where n is the number of signals in a data set and B is the number of time points sampled at.
I've been calculating the W matrix using:
[A,W] = fastica(X)
However, because W is a n x n matrix I can't tell how it is oriented and whether to use W or its transpose (W.') in subsequent calculations, and can't seem to find a clear answer in the documentation.
To help me get my bearings, is fastica maximising the independence in the rows of Y where Y = W.'X or Y = WX?
It should be Y=W*X. To be sure, you can reduce the number of component to estimate and then W should no longer be square:
[A,W] = fastica(X,'numOfIC',2)
If W is of size 2xn then Y=W*X. Else W has a size nx2 and Y=W'*X
Related
The function to perform an N-dimensional convolution of arrays A and B in matlab is shown below:
C = convn(A,B) % returns the N-dimensional convolution of arrays A and B.
I am interested in a 3-D convolution with a Gaussian filter.
If A is a 3 x 5 x 6 matrix, what do the dimensions of B have to be?
The dimensions of B can be anything you want. There is no set restriction in terms of size. For the Gaussian filter, it can be 1D, 2D or 3D. In 1D, what will happen is that each row gets filtered independently. In 2D, what will happen is that each slice gets filtered independently. Finally, in 3D you will be doing what is expected in 3D convolution. I am assuming you would like a full 3D convolution, not just 1D or 2D.
You may be interested in the output size of convn. If you refer to the documentation, given the two N dimensional matrices, for each dimension k of the output and if nak is the size of dimension k for the matrix A and nbk is the size of dimension k for matrix B, the size of dimension of the output matrix C or nck is such that:
nck = max([nak + nbk - 1, nak, nbk])
nak + nbk - 1 is straight from convolution theory. The final output size of a dimension is simply the sum of the two sizes in dimension k subtracted by 1. However should this value be smaller than either of nak or nbk, we need to make sure that the output size is compatible so that any of the input matrices can fit in the final output. This is why you have the final output size and bounded by both A and B.
To make this easier, you can set the size of the filter guided by the standard deviation of the distribution. I would like to refer you to my previous Stack Overflow post: By which measures should I set the size of my Gaussian filter in MATLAB?
This determines what the output size of a Gaussian filter should be given a standard deviation.
In 2D, the dimensions of the filter are N x N, such that N = ceil(6*sigma + 1) with sigma being the desired standard deviation. Therefore, you would allocate a 3D matrix of size N x N x N with N = ceil(6*sigma + 1);.
Therefore, the code you would want to use to create a 3D Gaussian filter would be something like this:
% Example input
A = rand(3, 5, 6);
sigma = 0.5; % Example
% Find size of Gaussian filter
N = ceil(6*sigma + 1);
% Define grid of centered coordinates of size N x N x N
[X, Y, Z] = meshgrid(-N/2 : N/2);
% Compute Gaussian filter - note normalization step
B = exp(-(X.^2 + Y.^2 + Z.^2) / (2.0*sigma^2));
B = B / sum(B(:));
% Convolve
C = convn(A, B);
One final note is that if the filter you provide has any of its dimensions that are beyond the size of the input matrix A, you will get a matrix using the constraints of each nck value, but then the border elements will be zeroed due to zero-padding.
Lets say I have a samples matrix samples (n_samples x n1) and a labels vector labels (n_samples x 1), where the labels are in the range [1:n2]
I am looking for an efficient way to create an empirical joint probability matrix P in the size n2 x n1.
Where for every sample i, we add its row samples(i, :) to P in the location indicated by labels(i).
I.e. (pseudo code)
for i = 1:n_samples
P(l(i), :) += M(i, :)
Is there a killer matlab command for doing that? Rather than a for loop or arrayfun?
Following #BillBokeey comment: Here is the solution
[xx, yy] = ndgrid(labels,1:size(samples,2));
P = accumarray([xx(:) yy(:)],samples(:));
I am just wondering, how would I go about fitting a line to histogram, using the z-counts as weights? An example of this is shown below (although this post just discusses overlaying multiple plots), taken from Scatter plot with density in Matlab).
My initial thought is to make an array consisting of each pixel from the density plot, repeated n times to make a scatter plot (n == the number of counts), then do a linear polyfit. This seems awfully redundant though.
The other approach is to do a weighted least squares solution. You need the (x,y) location of each pixel and the number of counts n within each pixel. Then, I think that you'd do the weighted least-squares this way:
%gather your known data...have x,y, and n all in the same order as each other
A = [x(:) ones(length(x),1)]; %here are the x values from your histogram
b = y(:); %here are the y-values from your histogram
C = diag(n(:)); %counts from each pixel in your 2D histogram
%Define polynomial coefficients as p = [slope; y_offset];
%usual least-squares solution...written here for reference
% b = A*p; %remember, p = [slope; y_offset];
% p = inv(A'*A)*(A'*b); %remember, p = [slope; y_offset];
%We want to apply a weighting matrix, so incorporate the weighting matrix
% A' * b = A' * C * A * p;
p = inv(A' * C * A)*(A' * b); %remember, p = [slope; y_offset];
The biggest uncertainty for me with this solution is whether the C matrix should be made up of n or n.^2, I can never remember. Hopefully, someone can correct me in the comments, if needed.
If you have the original data, which is a collection of (x,y) points, you simply do a polyfit on the original data:
p = polyfit(x(:),y(:),1); %linear fit
That will give you a best fit (in the least-squares sense) to the original data, which is what you want.
If you do not have the original data, and you only have the 2D histogram, the approach that you defined (which basically recreates a facsimile of the original data) will give a similar answer as if you did the polyfit on the original data.
I'm trying to reconstruct a 3d image from two calibrated cameras. One of the steps involved is to calculate the 3x3 essential matrix E, from two sets of corresponding (homogeneous) points (more than the 8 required) P_a_orig and P_b_orig and the two camera's 3x3 internal calibration matrices K_a and K_b.
We start off by normalizing our points with
P_a = inv(K_a) * p_a_orig
and
P_b = inv(K_b) * p_b_orig
We also know the constraint
P_b' * E * P_a = 0
I'm following it this far, but how do you actually solve that last problem, e.g. finding the nine values of the E matrix? I've read several different lecture notes on this subject, but they all leave out that crucial last step. Likely because it is supposedly trivial math, but I can't remember when I last did this and I haven't been able to find a solution yet.
This equation is actually pretty common in geometry algorithms, essentially, you are trying to calculate the matrix X from the equation AXB=0. To solve this, you vectorise the equation, which means,
vec() means vectorised form of a matrix, i.e., simply stack the coloumns of the matrix one over the another to produce a single coloumn vector. If you don't know the meaning of the scary looking symbol, its called Kronecker product and you can read it from here, its easy, trust me :-)
Now, say I call the matrix obtained by Kronecker product of B^T and A as C.
Then, vec(X) is the null vector of the matrix C and the way to obtain that is by doing the SVD decomposition of C^TC (C transpose multiplied by C) and take the the last coloumn of the matrix V. This last coloumn is nothing but your vec(X). Reshape X to 3 by 3 matrix. This is you Essential matrix.
In case you find this maths too daunting to code, simply use the following code by Y.Ma et.al:
% p are homogenius coordinates of the first image of size 3 by n
% q are homogenius coordinates of the second image of size 3 by n
function [E] = essentialDiscrete(p,q)
n = size(p);
NPOINTS = n(2);
% set up matrix A such that A*[v1,v2,v3,s1,s2,s3,s4,s5,s6]' = 0
A = zeros(NPOINTS, 9);
if NPOINTS < 9
error('Too few mesurements')
return;
end
for i = 1:NPOINTS
A(i,:) = kron(p(:,i),q(:,i))';
end
r = rank(A);
if r < 8
warning('Measurement matrix rank defficient')
T0 = 0; R = [];
end;
[U,S,V] = svd(A);
% pick the eigenvector corresponding to the smallest eigenvalue
e = V(:,9);
e = (round(1.0e+10*e))*(1.0e-10);
% essential matrix
E = reshape(e, 3, 3);
You can do several things:
The Essential matrix can be estimated using the 8-point algorithm, which you can implement yourself.
You can use the estimateFundamentalMatrix function from the Computer Vision System Toolbox, and then get the Essential matrix from the Fundamental matrix.
Alternatively, you can calibrate your stereo camera system using the estimateCameraParameters function in the Computer Vision System Toolbox, which will compute the Essential matrix for you.
I have a high dimensional Gaussian with mean M and covariance matrix V. I would like to calculate the distance from point p to M, taking V into consideration (I guess it's the distance in standard deviations of p from M?).
Phrased differentially, I take an ellipse one sigma away from M, and would like to check whether p is inside that ellipse.
If V is a valid covariance matrix of a gaussian, it then is symmetric positive definite and therefore defines a valid scalar product. By the way inv(V) also does.
Therefore, assuming that M and p are column vectors, you could define distances as:
d1 = sqrt((M-p)'*V*(M-p));
d2 = sqrt((M-p)'*inv(V)*(M-p));
the Matlab way one would rewrite d2as (probably some unnecessary parentheses):
d2 = sqrt((M-p)'*(V\(M-p)));
The nice thing is that when V is the unit matrix, then d1==d2and it correspond to the classical euclidian distance. To find wether you have to use d1 or d2is left as an exercise (sorry, part of my job is teaching). Write the multi-dimensional gaussian formula and compare it to the 1D case, since the multidimensional case is only a particular case of the 1D (or perform some numerical experiment).
NB: in very high dimensional spaces or for very many points to test, you might find a clever / faster way from the eigenvectors and eigenvalues of V (i.e. the principal axes of the ellipsoid and their corresponding variance).
Hope this helps.
A.
Consider computing the probability of the point given the normal distribution:
M = [1 -1]; %# mean vector
V = [.9 .4; .4 .3]; %# covariance matrix
p = [0.5 -1.5]; %# 2d-point
prob = mvnpdf(p,M,V); %# probability P(p|mu,cov)
The function MVNPDF is provided by the Statistics Toolbox
Maybe I'm totally off, but isn't this the same as just asking for each dimension: Am I inside the sigma?
PSEUDOCODE:
foreach(dimension d)
(M(d) - sigma(d) < p(d) < M(d) + sigma(d)) ?
Because you want to know if p is inside every dimension of your gaussian. So actually, this is just a space problem and your Gaussian hasn't have to do anything with it (except for M and sigma which are just distances).
In MATLAB you could try something like:
all(M - sigma < p < M + sigma)
A distance to that place could be, where I don't know the function for the Euclidean distance. Maybe dist works:
dist(M, p)
Because M is just a point in space and p as well. Just 2 vectors.
And now the final one. You want to know the distance in a form of sigma's:
% create a distance vector and divide it by sigma
M - p ./ sigma
I think that will do the trick.