I have a 1000x1000x3 3d matrix. I want to calculate the cosine of the angle between each 3d vector of a certain set and each 3d vector I can vertically extract from the 3d matrix. Then I should be able to create a 1000x1000 matrix with the index of the vector that has the maximum cosine similarity (i.e. minimum angle) with the original data.
How can I vectorize this calculation or at least some parts of it? Currently I do it with nested for loops (huge amount of time and overhead).
I couldn't find a function that takes the norm over the third dimension, but this works as well I think.
a = rand(1000,1000,3)-.5; %dataset
na = sqrt(a(:,:,1).^2+a(:,:,2).^2+a(:,:,3).^2); %the norm of each vector
b = [1.2,1,3]; %vector to compare angle against
nb = norm(b); %the norm of the compare vector
b = repmat(b,[1000,1000]);
b = reshape(b,[1000,1000,3]); %1000x1000 copies of b
pl = a.*nb + na.*b;
mn = a.*nb - na.*b;
npl = sqrt(pl(:,:,1).^2+pl(:,:,2).^2+pl(:,:,3).^2);
nmn = sqrt(mn(:,:,1).^2+mn(:,:,2).^2+mn(:,:,3).^2);
theta = 2 * atan(nmn./npl); %angle between [0 and pi] as expected
The math is an adaptation of this formula:
theta = 2 * atan(norm(x*norm(y) - norm(x)*y) / norm(x * norm(y) + norm(x) * y))
Related
What I am currently doing is computing the euclidean distance between all elements in a vector (the elements are pixel locations in a 2D image) to see if the elements are close to each other. I create a reference vector that takes on the value of each index within the vector incrementally. The euclidean distance between the reference vector and all the elements in the pixel location vector is computed using the MATLAB function "pdist2" and the result is applied to some conditions; however, upon running the code, this function seems to be taking the longest to compute (i.e. for one run, the function was called upon 27,245 times and contributed to about 54% of the overall program's run time). Is there a more efficient method to do this and speed up the program?
[~, n] = size(xArray); %xArray and yArray are same size
%Pair the x and y coordinates of the interest pixels
pairLocations = [xArray; yArray].';
%Preallocate cells with the max amount (# of interest pixels)
p = cell(1,n);
for i = 1:n
ref = [xArray(i), yArray(i)];
d = pdist2(ref,pairLocations,'euclidean');
d = d < dTh;
d = find(d==1);
[~,k] = size(d);
if (k >= num)
p{1,i} = d;
end
end
For squared Euclidean distance, there is a trick using matrix dot product:
||a-b||² = <a-b, a-b> = ||a||² - 2<a,b> + ||b||²
Let C = [xArray; yArray]; a 2×n matrix of all locations, then
n2 = sum(C.^2); % sq norm of coordinates
D = bsxfun(#plus, n2, n2.') - 2 * C.' * C;
Now D(ii,jj) holds the square distance between point ii and point jj.
Should run quite quickly.
I have two images which one of them is the Original image and the second one is Transformed image.
I have to find out how many degrees Transformed image was rotated using 3x3 transformation matrix. Plus, I need to find how far translated from origin.
Both images are grayscaled and held in matrix variables. Their sizes are same [350 500].
I have found a few lecture notes like this.
Lecture notes say that I should use the following matrix formula for rotation:
For translation matrix the formula is given:
Everything is good. But there are two problems:
I could not imagine how to implement the formulas using MATLAB.
The formulas are shaped to find x',y' values but I already have got x,x',y,y' values. I need to find rotation angle (theta) and tx and ty.
I want to know the equivailence of x, x', y, y' in the the matrix.
I have got the following code:
rotationMatrix = [ cos(theta) sin(theta) 0 ; ...
-sin(theta) cos(theta) 0 ; ...
0 0 1];
translationMatrix = [ 1 0 tx; ...
0 1 ty; ...
0 0 1];
But as you can see, tx, ty, theta variables are not defined before used. How can I calculate theta, tx and ty?
PS: It is forbidden to use Image Processing Toolbox functions.
This is essentially a homography recovery problem. What you are doing is given co-ordinates in one image and the corresponding co-ordinates in the other image, you are trying to recover the combined translation and rotation matrix that was used to warp the points from the one image to the other.
You can essentially combine the rotation and translation into a single matrix by multiplying the two matrices together. Multiplying is simply compositing the two operations together. You would this get:
H = [cos(theta) -sin(theta) tx]
[sin(theta) cos(theta) ty]
[ 0 0 1]
The idea behind this is to find the parameters by minimizing the error through least squares between each pair of points.
Basically, what you want to find is the following relationship:
xi_after = H*xi_before
H is the combined rotation and translation matrix required to map the co-ordinates from the one image to the other. H is also a 3 x 3 matrix, and knowing that the lower right entry (row 3, column 3) is 1, it makes things easier. Also, assuming that your points are in the augmented co-ordinate system, we essentially want to find this relationship for each pair of co-ordinates from the first image (x_i, y_i) to the other (x_i', y_i'):
[p_i*x_i'] [h11 h12 h13] [x_i]
[p_i*y_i'] = [h21 h22 h23] * [y_i]
[ p_i ] [h31 h32 1 ] [ 1 ]
The scale of p_i is to account for homography scaling and vanishing points. Let's perform a matrix-vector multiplication of this equation. We can ignore the 3rd element as it isn't useful to us (for now):
p_i*x_i' = h11*x_i + h12*y_i + h13
p_i*y_i' = h21*x_i + h22*y_i + h23
Now let's take a look at the 3rd element. We know that p_i = h31*x_i + h32*y_i + 1. As such, substituting p_i into each of the equations, and rearranging to solve for x_i' and y_i', we thus get:
x_i' = h11*x_i + h12*y_i + h13 - h31*x_i*x_i' - h32*y_i*x_i'
y_i' = h21*x_i + h22*y_i + h23 - h31*x_i*y_i' - h32*y_i*y_i'
What you have here now are two equations for each unique pair of points. What we can do now is build an over-determined system of equations. Take each pair and build two equations out of them. You will then put it into matrix form, i.e.:
Ah = b
A would be a matrix of coefficients that were built from each set of equations using the co-ordinates from the first image, b would be each pair of points for the second image and h would be the parameters you are solving for. Ultimately, you are finally solving this linear system of equations reformulated in matrix form:
You would solve for the vector h which can be performed through least squares. In MATLAB, you can do this via:
h = A \ b;
A sidenote for you: If the movement between images is truly just a rotation and translation, then h31 and h32 will both be zero after we solve for the parameters. However, I always like to be thorough and so I will solve for h31 and h32 anyway.
NB: This method will only work if you have at least 4 unique pairs of points. Because there are 8 parameters to solve for, and there are 2 equations per point, A must have at least a rank of 8 in order for the system to be consistent (if you want to throw in some linear algebra terminology in the loop). You will not be able to solve this problem if you have less than 4 points.
If you want some MATLAB code, let's assume that your points are stored in sourcePoints and targetPoints. sourcePoints are from the first image and targetPoints are for the second image. Obviously, there should be the same number of points between both images. It is assumed that both sourcePoints and targetPoints are stored as M x 2 matrices. The first columns contain your x co-ordinates while the second columns contain your y co-ordinates.
numPoints = size(sourcePoints, 1);
%// Cast data to double to be sure
sourcePoints = double(sourcePoints);
targetPoints = double(targetPoints);
%//Extract relevant data
xSource = sourcePoints(:,1);
ySource = sourcePoints(:,2);
xTarget = targetPoints(:,1);
yTarget = targetPoints(:,2);
%//Create helper vectors
vec0 = zeros(numPoints, 1);
vec1 = ones(numPoints, 1);
xSourcexTarget = -xSource.*xTarget;
ySourcexTarget = -ySource.*xTarget;
xSourceyTarget = -xSource.*yTarget;
ySourceyTarget = -ySource.*yTarget;
%//Build matrix
A = [xSource ySource vec1 vec0 vec0 vec0 xSourcexTarget ySourcexTarget; ...
vec0 vec0 vec0 xSource ySource vec1 xSourceyTarget ySourceyTarget];
%//Build RHS vector
b = [xTarget; yTarget];
%//Solve homography by least squares
h = A \ b;
%// Reshape to a 3 x 3 matrix (optional)
%// Must transpose as reshape is performed
%// in column major format
h(9) = 1; %// Add in that h33 is 1 before we reshape
hmatrix = reshape(h, 3, 3)';
Once you are finished, you have a combined rotation and translation matrix. If you want the x and y translations, simply pick off column 3, rows 1 and 2 in hmatrix. However, we can also work with the vector of h itself, and so h13 would be element 3, and h23 would be element number 6. If you want the angle of rotation, simply take the appropriate inverse trigonometric function to rows 1, 2 and columns 1, 2. For the h vector, this would be elements 1, 2, 4 and 5. There will be a bit of inconsistency depending on which elements you choose as this was solved by least squares. One way to get a good overall angle would perhaps be to find the angles of all 4 elements then do some sort of average. Either way, this is a good starting point.
References
I learned about homography a while ago through Leow Wee Kheng's Computer Vision course. What I have told you is based on his slides: http://www.comp.nus.edu.sg/~cs4243/lecture/camera.pdf. Take a look at slides 30-32 if you want to know where I pulled this material from. However, the MATLAB code I wrote myself :)
I have two images which one of them is the Original image and the second one is Transformed image.
I have to find out how many degrees Transformed image was rotated using 3x3 transformation matrix. Plus, I need to find how far translated from origin.
Both images are grayscaled and held in matrix variables. Their sizes are same [350 500].
I have found a few lecture notes like this.
Lecture notes say that I should use the following matrix formula for rotation:
For translation matrix the formula is given:
Everything is good. But there are two problems:
I could not imagine how to implement the formulas using MATLAB.
The formulas are shaped to find x',y' values but I already have got x,x',y,y' values. I need to find rotation angle (theta) and tx and ty.
I want to know the equivailence of x, x', y, y' in the the matrix.
I have got the following code:
rotationMatrix = [ cos(theta) sin(theta) 0 ; ...
-sin(theta) cos(theta) 0 ; ...
0 0 1];
translationMatrix = [ 1 0 tx; ...
0 1 ty; ...
0 0 1];
But as you can see, tx, ty, theta variables are not defined before used. How can I calculate theta, tx and ty?
PS: It is forbidden to use Image Processing Toolbox functions.
This is essentially a homography recovery problem. What you are doing is given co-ordinates in one image and the corresponding co-ordinates in the other image, you are trying to recover the combined translation and rotation matrix that was used to warp the points from the one image to the other.
You can essentially combine the rotation and translation into a single matrix by multiplying the two matrices together. Multiplying is simply compositing the two operations together. You would this get:
H = [cos(theta) -sin(theta) tx]
[sin(theta) cos(theta) ty]
[ 0 0 1]
The idea behind this is to find the parameters by minimizing the error through least squares between each pair of points.
Basically, what you want to find is the following relationship:
xi_after = H*xi_before
H is the combined rotation and translation matrix required to map the co-ordinates from the one image to the other. H is also a 3 x 3 matrix, and knowing that the lower right entry (row 3, column 3) is 1, it makes things easier. Also, assuming that your points are in the augmented co-ordinate system, we essentially want to find this relationship for each pair of co-ordinates from the first image (x_i, y_i) to the other (x_i', y_i'):
[p_i*x_i'] [h11 h12 h13] [x_i]
[p_i*y_i'] = [h21 h22 h23] * [y_i]
[ p_i ] [h31 h32 1 ] [ 1 ]
The scale of p_i is to account for homography scaling and vanishing points. Let's perform a matrix-vector multiplication of this equation. We can ignore the 3rd element as it isn't useful to us (for now):
p_i*x_i' = h11*x_i + h12*y_i + h13
p_i*y_i' = h21*x_i + h22*y_i + h23
Now let's take a look at the 3rd element. We know that p_i = h31*x_i + h32*y_i + 1. As such, substituting p_i into each of the equations, and rearranging to solve for x_i' and y_i', we thus get:
x_i' = h11*x_i + h12*y_i + h13 - h31*x_i*x_i' - h32*y_i*x_i'
y_i' = h21*x_i + h22*y_i + h23 - h31*x_i*y_i' - h32*y_i*y_i'
What you have here now are two equations for each unique pair of points. What we can do now is build an over-determined system of equations. Take each pair and build two equations out of them. You will then put it into matrix form, i.e.:
Ah = b
A would be a matrix of coefficients that were built from each set of equations using the co-ordinates from the first image, b would be each pair of points for the second image and h would be the parameters you are solving for. Ultimately, you are finally solving this linear system of equations reformulated in matrix form:
You would solve for the vector h which can be performed through least squares. In MATLAB, you can do this via:
h = A \ b;
A sidenote for you: If the movement between images is truly just a rotation and translation, then h31 and h32 will both be zero after we solve for the parameters. However, I always like to be thorough and so I will solve for h31 and h32 anyway.
NB: This method will only work if you have at least 4 unique pairs of points. Because there are 8 parameters to solve for, and there are 2 equations per point, A must have at least a rank of 8 in order for the system to be consistent (if you want to throw in some linear algebra terminology in the loop). You will not be able to solve this problem if you have less than 4 points.
If you want some MATLAB code, let's assume that your points are stored in sourcePoints and targetPoints. sourcePoints are from the first image and targetPoints are for the second image. Obviously, there should be the same number of points between both images. It is assumed that both sourcePoints and targetPoints are stored as M x 2 matrices. The first columns contain your x co-ordinates while the second columns contain your y co-ordinates.
numPoints = size(sourcePoints, 1);
%// Cast data to double to be sure
sourcePoints = double(sourcePoints);
targetPoints = double(targetPoints);
%//Extract relevant data
xSource = sourcePoints(:,1);
ySource = sourcePoints(:,2);
xTarget = targetPoints(:,1);
yTarget = targetPoints(:,2);
%//Create helper vectors
vec0 = zeros(numPoints, 1);
vec1 = ones(numPoints, 1);
xSourcexTarget = -xSource.*xTarget;
ySourcexTarget = -ySource.*xTarget;
xSourceyTarget = -xSource.*yTarget;
ySourceyTarget = -ySource.*yTarget;
%//Build matrix
A = [xSource ySource vec1 vec0 vec0 vec0 xSourcexTarget ySourcexTarget; ...
vec0 vec0 vec0 xSource ySource vec1 xSourceyTarget ySourceyTarget];
%//Build RHS vector
b = [xTarget; yTarget];
%//Solve homography by least squares
h = A \ b;
%// Reshape to a 3 x 3 matrix (optional)
%// Must transpose as reshape is performed
%// in column major format
h(9) = 1; %// Add in that h33 is 1 before we reshape
hmatrix = reshape(h, 3, 3)';
Once you are finished, you have a combined rotation and translation matrix. If you want the x and y translations, simply pick off column 3, rows 1 and 2 in hmatrix. However, we can also work with the vector of h itself, and so h13 would be element 3, and h23 would be element number 6. If you want the angle of rotation, simply take the appropriate inverse trigonometric function to rows 1, 2 and columns 1, 2. For the h vector, this would be elements 1, 2, 4 and 5. There will be a bit of inconsistency depending on which elements you choose as this was solved by least squares. One way to get a good overall angle would perhaps be to find the angles of all 4 elements then do some sort of average. Either way, this is a good starting point.
References
I learned about homography a while ago through Leow Wee Kheng's Computer Vision course. What I have told you is based on his slides: http://www.comp.nus.edu.sg/~cs4243/lecture/camera.pdf. Take a look at slides 30-32 if you want to know where I pulled this material from. However, the MATLAB code I wrote myself :)
I was aquainted in using the fft in matlab with the code
fft(signal,[],n)
where n tells the dimension on which to apply the fft as from Matlab documentation:
http://www.mathworks.it/it/help/matlab/ref/fft.html
I would like to do the same with dct.
Is this possible? I could not find any useful information around.
Thanks for the help.
Luigi
dct does not have the option to pick a dimension like fft. You will have to either transpose your input to operate on rows or pick one vector from your signal and operate on that.
yep!
try it:
matrix = dctmtx(n);
signal_dct = matrix * signal;
Edit
Discrete cosine transform.
Y = dct(X) returns the discrete cosine transform of X.
The vector Y is the same size as X and contains the discrete cosine transform coefficients.
Y = dct(X,N) pads or truncates the vector X to length N before transforming.
If X is a matrix, the dct operation is applied to each column. This transform can be inverted using IDCT.
% Example:
% Find how many dct coefficients represent 99% of the energy
% in a sequence.
x = (1:100) + 50*cos((1:100)*2*pi/40); % Input Signal
X = dct(x); % Discrete cosine transform
[XX,ind] = sort(abs(X)); ind = fliplr(ind);
num_coeff = 1;
while (norm([X(ind(1:num_coeff)) zeros(1,100-num_coeff)])/norm(X)<.99)
num_coeff = num_coeff + 1;
end;
num_coeff
I have the following code for calculating the result of a linear combination of Gaussian functions. What I'd really like to do is to vectorize this somehow so that it's far more performant in Matlab.
Note that y is a column vector (output), x is a matrix where each column corresponds to a data point and each row corresponds to a dimension (i.e. 2 rows = 2D), variance is a double, gaussians is a matrix where each column is a vector corresponding to the mean point of the gaussian and weights is a row vector of the weights in front of each gaussian. Note that the length of weights is 1 bigger than gaussians as weights(1) is the 0th order weight.
function [ y ] = CalcPrediction( gaussians, variance, weights, x )
basisFunctions = size(gaussians, 2);
xvalues = size(x, 2);
if length(weights) ~= basisFunctions + 1
ME = MException('TRAIN:CALC', 'The number of weights should be equal to the number of basis functions plus one');
throw(ME);
end
y = weights(1) * ones(xvalues, 1);
for xIdx = 1:xvalues
for i = 1:basisFunctions
diff = x(:, xIdx) - gaussians(:, i);
y(xIdx) = y(xIdx) + weights(i+1) * exp(-(diff')*diff/(2*variance));
end
end
end
You can see that at the moment I simply iterate over the x vectors and then the gaussians inside 2 for loops. I'm hoping that this can be improved - I've looked at meshgrid but that seems to only apply to vectors (and I have matrices)
Thanks.
Try this
diffx = bsxfun(#minus,x,permute(gaussians,[1,3,2])); % binary operation with singleton expansion
diffx2 = squeeze(sum(diffx.^2,1)); % dot product, shape is now [XVALUES,BASISFUNCTIONS]
weight_col = weights(:); % make sure weights is a column vector
y = exp(-diffx2/2/variance)*weight_col(2:end); % a column vector of length XVALUES
Note, I changed diff to diffx since diff is a builtin. I'm not sure this will improve performance as allocating arrays will offset increase by vectorization.