I am trying to implement Hough transform algorithm. Algorithm works, but it's slow.
Currently i calculate rho, by this equation in two for loops:
for i = 1 : length(x)
j=1;
for theta = -pi/2:nBinsTheta:pi/2-nBinsTheta
ro =round(x(i).*cos(theta) + y(i).*sin(theta));
....
j = j + 1;
end
end
How can i simplify this, to work without for loops?
I need to calculate ro without loops, but how can i do this, to cover all possible theta's?
EDIT: Now i need to know how to add 1, to designated cell's in accumulator matrix given x and y coordinate vector. For example let's say that i have vectors like:
x: [1 2 1 3]
y: [1 3 1 4]
I'd like to solve this problem without loops. I know that i need to convert to linear indices using sub2ind, but the problem is that there'll be a lot of same linear indices for example that i gave, there will be 2x1 (1,1 coordinate is repeated twice). If you try to add 1 like so:
A([1 1]) = A([1 1]) + 1;
it'll add 1 only once, that's my problem.
Assuming x and y to be row vectors, you can use the following code to pre-calculate all ro values in a 2D matrix which hopefully should speed things up for you inside the nested loops for the rest of the work you might be doing involving the ro values -
theta_vec = [-pi/2:nBinsTheta:pi/2-nBinsTheta].'; %//'
ro_vals = round( cos(theta_vec)*x + sin(theta_vec)*y );
assert(all(size(x) == size(y)), 'dimension mismatch: x, y')
theta = (-pi/2:nBinsTheta:pi/2-nBinsTheta)';
assert(all(size(theta) == size(y)), 'dimension mismatch: theta, y')
rho = x.*cos(theta) + y.*sin(theta);
rho_rounded = round(rho);
do you really need j?
PS: the previous answer might not work because of matrix multiplication operator * instead of elementwise .*
Related
I have a 3D matrix:
A = [5 7 8; 0 1 9; 4 3 6];
A(:,:,2) = [1 0 4; 3 5 6; 9 8 7]
I want to apply a 3D FFT in this matrix using decomposition of 1D FFT. I read that it I should apply 1D FFT in each dimension.
How can I do this?
For x and y, I do this:
for k=0:2
y1 = A(:,k+1,:);
A(:,k+1,:) = fft(y1);
end
for k=0:2
y2 = A(k+1,:,:);
A(k+1,:,:) = fft(y2);
end
For the dimension z, I don't know how to do this.
The fft function accepts a third input specifiying dimension, and is vectorized with respect to the other dimensions. So you can simply use:
result = fft(fft(fft(A, [], 1), [], 2), [], 3);
First, your loops should look like this:
for k=1:size(A,2)
y = A(:,k,:);
A(:,k,:) = fft(y);
end
Second, the loop above is identical to (as #Luis Mendo said in his answer):
A = fft(A,[],2);
There is no need to write a loop at all.
Third, to compute the 1D FFT along the 3rd dimension, you use:
fft(A,[],3);
You could write this as a loop (just to answer your explicit question, I don't recommend you do this):
for k=1:size(A,3)
y = A(:,:,k);
A(:,:,k) = fft(y);
end
If, for some reason, that doesn't work in your version of MATLAB because of the shape of y, you can reshape y to be a column vector:
... fft(y(:));
Finally, to compute the 3D FFT using 1D decompositions, you can simply write
A = fftn(A);
This follows the exact same process you are trying to implement, except it does it much faster.
I have a matrix a and I want to calculate the distance from one point to all other points. So really the outcome matrix should have a zero (at the point I have chosen) and should appear as some sort of circle of numbers around that specific point.
This is what I have already but I cant seem to get the correct outcome.
a = [1 2 3 4 5 6 7 8 9 10]
for i = 2:20
a(i,:) = a(i-1,:) + 1;
end
N = 10
for I = 1:N
for J = 1:N
dx = a(I,1)-a(J,1);
dy = a(I,2)-a(J,2);
distance(I,J) = sqrt(dx^2 + dy^2)
end
end
Your a matrix is a 1D vector and is incompatible with the nested loop, which computes distance in 2D space from each point to each other point. So the following answer applies to the problem of finding all pairwise distances in a N-by-D matrix, as your loop does for the case of D=2.
Option 1 - pdist
I think you are looking for pdist with the 'euclidean' distance option.
a = randn(10, 2); %// 2D, 10 samples
D = pdist(a,'euclidean'); %// euclidean distance
Follow that by squareform to get the square matrix with zero on the diagonal as you want it:
distances = squareform(D);
Option 2 - bsxfun
If you don't have pdist, which is in the Statistics Toolbox, you can do this easily with bsxfun:
da = bsxfun(#minus,a,permute(a,[3 2 1]));
distances = squeeze(sqrt(sum(da.^2,2)));
Option 3 - reformulated equation
You can also use an alternate form of Euclidean (2-norm) distance,
||A-B|| = sqrt ( ||A||^2 + ||B||^2 - 2*A.B )
Writing this in MATLAB for two data arrays u and v of size NxD,
dot(u-v,u-v,2) == dot(u,u,2) + dot(v,v,2) - 2*dot(u,v,2) % useful identity
%// there are actually small differences from floating point precision, but...
abs(dot(u-v,u-v,2) - (dot(u,u,2) + dot(v,v,2) - 2*dot(u,v,2))) < 1e-15
With the reformulated equation, the solution becomes:
aa = a*a';
a2 = sum(a.*a,2); % diag(aa)
a2 = bsxfun(#plus,a2,a2');
distances = sqrt(a2 - 2*aa);
You might use this method if Option 2 eats up too much memory.
Timings
For a random data matrix of size 1e3-by-3 (N-by-D), here are timings for 100 runs (Core 2 Quad, 4GB DDR2, R2013a).
Option 1 (pdist): 1.561150 sec (0.560947 sec in pdist)
Option 2 (bsxfun): 2.695059 sec
Option 3 (bsxfun alt): 1.334880 sec
Findings: (i) Do computations with bsxfun, use the alternate formula. (ii) the pdist+squareform option has comparable performance. (iii) The reason why squareform takes twice as much time as pdist is probably because pdist only computes the triangular matrix since the distance matrix is symmetric. If you can do without the square matrix, then you can avoid squareform and do your computations in about 40% of the time required to do it manually with bsxfun (0.5609/1.3348).
This is what i was looking for, but thanks for all the suggestions.
A = rand(5, 5);
select_cell = [3 3];
distance = zeros(size(A, 1), size(A, 2));
for i = 1:size(A, 1)
for j = 1:size(A, 2)
distance(i, j) = sqrt((i - select_cell(1))^2 + (j - select_cell(2))^2);
end
end
disp(distance)
Also you can improve it by using vectorisation:
distances = sqrt((x-xCenter).^2+(y-yCenter).^2
IMPORTANT: data_matrix is D X N, where D is number of dimensions and N is number of data points!
final_dist_pairs=data_matrix'*data_matrix;
norms = diag(final_dist_pairs);
final_dist_pairs = bsxfun(#plus, norms, norms') - 2 * final_dist_pairs;
Hope it helps!
% Another important thing,
Never use pdist function of MATLAB. It is a sequential evaluation, that is something like for loops and takes a lot of time, maybe in O(N^2)
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 trying to get an vector optimized version of the linear rectifier. i.e. y = max(0,x). So what it should compute its element wise max of zero and x_i. I obviously implemented:
function [ y ] = rectSig( x )
%rectSig computes vector-wise rectified linear function
% computes y = [..., max(0,x_i), ...]
n=length(x);
y = zeros(1,n);
for i=1:1:length(x);
y(i) = max(0,x(i));
end
end
however, I know that looping like this in MATLAB is ill advised. So I was wondering if there was a better way to do this or if obviously matlab had its own implementation of a vectorized version of such a function? I always try to avoid loops if I can in matlab if there is a way to vectorize my code. It usually tends to speed things up.
Btw, I obviously tried googling it but didn't really get the result I expected...
The solution is as simple as
y = max(x,0);
This works for x being a column, row vector, matrix, higher dimensional matrix, etc. On the other hand
y = max(zeros(1,length(x)),x);
only works for x being a row vector. It fails when x is a column vector or matrix.
max accepts matrix inputs:
x = -5:5;
comparisonvector = zeros(size(x));
y = max(comparisonvector, x);
Returns:
y =
0 0 0 0 0 0 1 2 3 4 5
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 :)