I need to generate N random coordinates for a 2D plane. The distance between any two points are given (number of distance is N(N - 1) / 2). For example, say I need to generate 3 points i.e. A, B, C. I have the distance between pair of them i.e. distAB, distAC and distBC.
Is there any built-in function in MATLAB that can do this? Basically, I'm looking for something that is the reverse of pdist() function.
My initial idea was to choose a point (say A is the origin). Then, I can randomly find B and C being on two different circles with radii distAB and distAC. But then the distance between B and C might not satisfy distBC and I'm not sure how to proceed if this happens. And I think this approach will get very complicated if N is a large number.
Elaborating on Ansaris answer I produced the following. It assumes a valid distance matrix provided, calculates positions in 2D based on cmdscale, does a random rotation (random translation could be added also), and visualizes the results:
%Distance matrix
D = [0 2 3; ...
2 0 4; ...
3 4 0];
%Generate point coordinates based on distance matrix
Y = cmdscale(D);
[nPoints dim] = size(Y);
%Add random rotation
randTheta = 2*pi*rand(1);
Rot = [cos(randTheta) -sin(randTheta); sin(randTheta) cos(randTheta) ];
Y = Y*Rot;
%Visualization
figure(1);clf;
plot(Y(:,1),Y(:,2),'.','markersize',20)
hold on;t=0:.01:2*pi;
for r = 1 : nPoints - 1
for c = r+1 : nPoints
plot(Y(r,1)+D(r,c)*sin(t),Y(r,2)+D(r,c)*cos(t));
plot(Y(c,1)+D(r,c)*sin(t),Y(c,2)+D(r,c)*cos(t));
end
end
You want to use a technique called classical multidimensional scaling. It will work fine and losslessly if the distances you have correspond to distances between valid points in 2-D. Luckily there is a function in MATLAB that does exactly this: cmdscale. Once you run this function on your distance matrix, you can treat the first two columns in the first output argument as the points you need.
Related
I would like to calculate the mahalanobis distance of input feature vector Y (1x14) to all feature vectors in matrix X (18x14). Each 6 vectors of X represent one class (So I have 3 classes). Then based on mahalanobis distances I will choose the vector that is the nearest to the input and classify it to one of the three classes as well.
My problem is when I use the following code I got only one value. How can I get mahalanobis distance between the input Y and every vector in X. So at the end I have 18 values and then I choose the smallest one. Any help will be appreciated. Thank you.
Note: I know that mahalanobis distance is a measure of the distance between a point P and a distribution D, but I don't how could this be applied in my situation.
Y = test1; % Y: 1x14 vector
S = cov(X); % X: 18x14 matrix
mu = mean(X,1);
d = ((Y-mu)/S)*(Y-mu)'
I also tried to separate the matrix X into 3; so each one represent the feature vectors of one class. This is the code, but it doesn't work properly and I got 3 distances and some have negative value!
Y = test1;
X1 = Action1;
S1 = cov(X1);
mu1 = mean(X1,1);
d1 = ((Y-mu1)/S1)*(Y-mu1)'
X2 = Action2;
S2 = cov(X2);
mu2 = mean(X2,1);
d2 = ((Y-mu2)/S2)*(Y-mu2)'
X3= Action3;
S3 = cov(X3);
mu3 = mean(X3,1);
d3 = ((Y-mu3)/S3)*(Y-mu3)'
d= [d1,d2,d3];
MahalanobisDist= min(d)
One last thing, when I used mahal function provided by Matlab I got this error:
Warning: Matrix is close to singular or badly scaled. Results may be inaccurate.
If you have to implement the distance yourself (school assignment for instance) this is of absolutely no use to you, but if you just need to calculate the distance as an intermediate step for other calculations I highly recommend d = Pdist2(a,b, distance_measure) the documentation is on matlabs site
It computes the pairwise distance between a vector (or even a matrix) b and all elements in a and stores them in vector d where the columns correspond to entries in b and the rows are entries from a. So d(i,j) is the distance between row j in b and row i in a (hope that made sense). If you want it could even parameters to find the k nearest neighbors, it's a great function.
in your case you would use the following code and you'd end up with the distance between elements, and the index as well
%number of neighbors
K = 1;
% X=18x14, Y=1x14, dist=18x1
[dist, iidx] = pdist2(X,Y,'mahalanobis','smallest',K);
%to find the class, you can do something like this
num_samples_per_class = 6;
matching_class = ceil(iidx/ num_samples_per_class);
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 am trying to calculate the distance between nearest neighbours within a nx2 matrix like the one shown below
point_coordinates =
11.4179 103.1400
16.7710 10.6691
16.6068 119.7024
25.1379 74.3382
30.3651 23.2635
31.7231 105.9109
31.8653 36.9388
%for loop going from the top of the vector column to the bottom
for counter = 1:size(point_coordinates,1)
%current point defined selected
current_point = point_coordinates(counter,:);
%math to calculate distance between the current point and all the points
distance_search= point_coordinates-repmat(current_point,[size(point_coordinates,1) 1]);
dist_from_current_point = sqrt(distance_search(:,1).^2+distance_search(:,2).^2);
%line to omit self subtraction that gives zero
dist_from_current_point (dist_from_current_point <= 0)=[];
%gives the shortest distance calculated for a certain vector and current_point
nearest_dist=min(dist_from_current_point);
end
%final line to plot the u,v vectors and the corresponding nearest neighbour
%distances
matnndist = [point_coordinates nearest_dist]
I am not sure how to structure the 'for' loop/nearest_neighbour line to be able to get the nearest neighbour distance for each u,v vector.
I would like to have, for example ;
for the first vector you could have the coordinates and the corresponding shortest distance, for the second vector another its shortest distance, and this goes on till n
Hope someone can help.
Thanks
I understand you want to obtain the minimum distance between different points.
You can compute the distance for each pair of points with bsxfun; remove self-distances; minimize. It's more computationally efficient to work with squared distances, and take the square root only at the end.
n = size(point_coordinates,1);
dist = bsxfun(#minus, point_coordinates(:,1), point_coordinates(:,1).').^2 + ...
bsxfun(#minus, point_coordinates(:,2), point_coordinates(:,2).').^2;
dist(1:n+1:end) = inf; %// remove self-distances
min_dist = sqrt(min(dist(:)));
Alternatively, you could use pdist. This avoids computing each distance twice, and also avoids self-distances:
dist = pdist(point_coordinates);
min_dist = min(dist(:));
If I can suggest a built-in function, use knnsearch from the statistics toolbox. What you are essentially doing is a K-Nearest Neighbour (KNN) algorithm, but you are ignoring self-distances. The way you would call knnsearch is in the following way:
[idx,d] = knnsearch(X, Y, 'k', k);
In simple terms, the KNN algorithm returns the k closest points to your data set given a query point. Usually, the Euclidean distance is the distance metric that is used. For MATLAB's knnsearch, X is a 2D array that consists of your dataset where each row is an observation and each column is a variable. Y would be the query points. Y is also a 2D array where each row is a query point and you need to have the same number of columns as X. We would also specify the flag 'k' to denote how many closest points you want returned. By default, k = 1.
As such, idx would be a N x K matrix, where N is the total number of query points (number of rows of Y) and K would be those k closest points to the dataset for each query point we have. idx indicates the particular points in your dataset that were closest to each query. d is also a N x K matrix that returns the smallest distances for these corresponding closest points.
As such, what you want to do is find the closest point for your dataset to each of the other points, ignoring self-distances. Therefore, you would set both X and Y to be the same, and set k = 2, discarding the first column of both outputs to get the result you're looking for.
Therefore:
[idx,d] = knnsearch(point_coordinates, point_coordinates, 'k', 2)
idx = idx(:,2);
d = d(:,2);
We thus get for idx and d:
>> idx
idx =
3
5
1
1
7
3
5
>> d
d =
17.3562
18.5316
17.3562
31.9027
13.7573
20.4624
13.7573
As such, this tells us that for the first point in your data set, it matched with point #3 the best. This matched with the closest distance of 17.3562. For the second point in your data set, it matched with point #5 the best with the closest distance being 18.5316. You can continue on with the rest of the results in a similar pattern.
If you don't have access to the statistics toolbox, consider reading my StackOverflow post on how I compute KNN from first principles.
Finding K-nearest neighbors and its implementation
In fact, it is very similar to Luis Mendo's post to you earlier.
Good luck!
I have a vector of 3D points lets say A as shown below,
A=[
-0.240265581092000 0.0500598627544876 1.20715641293013
-0.344503191645519 0.390376667574812 1.15887540716612
-0.0931248606994074 0.267137193112796 1.24244644549763
-0.183530493218807 0.384249186312578 1.14512014134276
-0.0201358671977785 0.404732019283683 1.21816745283019
-0.242108038906952 0.229873488902244 1.24229940627651
-0.391349107031230 0.262170158259873 1.23856838565023
]
what I want to do is to connect 3D points with lines which only have distance less than a specific threshold T. I want to get a list of pairs of points needed to be connected. Such as,
[
( -0.240265581092000 0.0500598627544876 1.20715641293013), (-0.344503191645519 0.390376667574812 1.15887540716612);
(-0.0931248606994074 0.267137193112796 1.24244644549763),(-0.183530493218807 0.384249186312578 1.14512014134276),.....
]
So as shown, I'll have a vector of pairs of points needed to be connected. So if anyone could please advise how this can be done in Matlab.
The following example demonstrates how to accomplish this.
%# Build an example matrix
A = [1 2 3; 0 0 0; 3 1 3; 2 0 2; 0 1 0];
Threshold = 3;
%# Calculate distance between all points
D = pdist2(A, A);
%# Discard any points with distance greater than threshold
D(D > Threshold) = nan;
If you wish to extract an index of all observation pairs that are linked by a distance less than (or equal to) Threshold, as well as the corresponding distance (your question didn't specify what form you wanted the output to take, so I am essentially guessing here), then instead use the following:
%# Obtain a list of linear indices of observations less than or equal to TH
I1 = find(D <= Threshold);
%#Extract the actual distances, as well as the corresponding observation indices from A
[Obs1Index, Obs2Index] = ind2sub(size(D), I1);
DList = [Obs1Index, Obs2Index, D(I1)];
Note, pdist2 uses Euclidean distance by default, but there are other options - see the documentation here.
UPDATE: Based on the OP's comments, the following code will express the output as a K*6 matrix, where K is the number of distance measures less than the threshold value, and the first three columns of each row is the first data point (3 dimensions) and the second three columns of each row is the connected data point.
DList2 = [A(Obs1Index, :), A(Obs2Index, :)];
SECOND UPDATE: I have not made any assumptions on the distance measure in this answer. That is, I'm deliberately using pdist2 in case your distance measure is not symmetric. However, if you are using a symmetric distance measure, then you could probably speed up the run-time by using pdist instead, although my indexing code would need to be adjusted accordingly.
Plot3 and pdist2 can be used to achieve what you want.
D=pdist2(A,A);
T=0.2;
for i=1:7
for j=i+1:7
if D(i,j)<T & D(i,j)~=0
i
j
plot3(A([i j],1),A([i j],2),A([i j],3));
hold on;
fprintf('line is plotted\n');
pause;
end
end
end