I have a point-cloud, for which i want to calculate the distance between all individual points in Matlab (preferably without duplicates).
The matrix with the coordinates is formatted as: points [ p x n x d ]. Where p = 1 (for now), n is as large as the number of points and d as large as the number of dimensions (3 in this case).
This is the data i have:
points(:,:,1) = 1 2 3
points(:,:,2) = 4 5 6
points(:,:,3) = 7 8 9
So i have three points in three dimensions. Now using pdist, i have tried to calculate the euclidian distance between each point using distances = pdist(points(:,1:3)); and distances = pdist(X(:,:,1:3)); But both just return an empty matrix.
Does anyone know how to use pdist to calculate these distances? It should return 3 distances, instead of 0, but i must be doing something wrong.
Btw. this question is a follow up to this one. I asked it in the comments at first, but thought it deserved a new question because it is a considerable expansion of the original question.
Try
distances = pdist(squeeze(X(:,:,1:3)));
or the transpose of squeeze(...) if that isn't correct.
Related
I have declared 2 matrixes like this:
a = [ 1 2;
11 12];
[m, n] = size(a);
b = a(2,:);
dist( b , a ); % the first column is not interesting
This works, I get a vector
[ 10.0499 9.0000 ]
However if I want to add a column or a line to my matrix a:
a = [ 1 2 3 ;
11 12 13];
then apply the same algorithm, than above, ignoring or not the first column, I get this error:
Error using -
Matrix dimensions must agree
I have no idea why it does not work, can someone explain to me please?
Actually I don't even know how to retrieve the way this euclidian distance is computed, I failed at trying to retrieve those values [ 10.0499 9.0000 ] by hand.
The Matlab mathworks manual says he algorithm used is the following:
d = sum((x-y).^2).^0.5
Any help
It does not work because the dist function when called with two arguments works like this:
Z = dist(W,P) takes an SxR weight matrix and RxQ input matrix and
returns the SxQ matrix of distances between W's rows and P's columns.
dist(P',P) returns the same result as dist(P).
That is, if you do this:
a = [ 1 2 3 ;
11 12 13]
b = a(2,:) % Then b = [11 12 13]
...and call:
dist(b, a)
It will try to compute the distance between b's rows (in this case, only a row with three numbers, that is, a 3D point) and a's columns (each column has two numbers, that is, a 2D point). Measuring a distance between them makes no sense.
The reason it worked on your first example was because the matrix was square (2x2). Therefore, you're computing distances between a row (2D) to the other columns (also 2D).
A distance by definition is a single number, not a vector. The fact that you've getting a vector for matrices of the same size already indicates that something is wrong. In particular, it gives you distance between each of the corresponding column vectors to each other. So it doesn't work when your matrices are of different sizes. A distance between matrices is not defined in any one particular way. I don't know of a notion of Euclidean distance between two matrices.
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
How can I calculate the norm of quaternion in matlab?
I tried this example
a = [1 4 4 -4];
norm = quatnorm(a)
My expected output is 7 but matlab returns 49.
What am I doing wrong?
As #Dan points out, using the native implementation, you are probably getting the square of the formal norm definition. For some reason quatnorm returns the square, after estimating the Euclidean norm (square root of sum of squares).
q = [1 4 4 -4];
MATLABquatnorm:
for index = size(q, 1):-1:1
qnorm(index,:) = norm(q(index,:), 2);
end
qout = qnorm.*qnorm;
Alternative (for vectors):
sqrt(q*q')
This is equivalent to getting sqrt(quatnorm(q)). As you will note above, quatnorm is also adapted to estimate norms for quaternions stored in successive matrix rows (estimates the norm of each row and then squares)
Alternative (for matrices N x 4):
Q = [q; 2*q]; % example
sqrt(diag(Q*Q'))
You can either take square root of the returned number, or you can use function quatmod(q) instead, which calculates the proper Euclidean norm (modulus) of the complex number by not taking square of it.
Let a circle of known radius be plotted in MATLAB.
Assume a pair of random points whose location has to be determined in terms of coordinates (x1,y1) (x2,y2)..(xn,yn). Pairs should be close to each other. For example T1 and R1 should be near.
As shown in figure, there are four random pairs (T1,R1)..(T4,R4).
There coordinates need to be determined wrt to center (0,0).
How can I generate this in MATLAB?
The simplest approach to pick a point from a uniform distribution over a circle with reduce R is using Gibbs sampling. Here is the code:
function [x y] = circular uniform (R)
while true
x = 2*R*rand() - R
y = 2*R*rand() - R
if (x*x + y*y) > R*R
return
end
end
The loop runs 4/π times on average.
(Complete edit after the question was edited).
To complete this task, I think that you need to combine the different approaches that have been mentioned before your edit:
To generate the centers T1,T2,T3,... in the green torus, use the polar coordinates. (Edit: this turned out to be wrong, rejection sampling must also be used here, otherwise, the distribution is not uniform!)
To generate the points R1,R2,R3,... in the circle around T1,T2,T3,... but still in the torus, use the rejection sampling.
With these ingredients, you should be able to do everything you need. Here is the code I wrote:
d=859.23;
D=1432.05;
R=100;
N=400;
% Generate the angle
theta = 2*pi*rand(N,1);
% Generate the radius
r = d + (D-d)*rand(N,1);
% Get the centers of the circles
Tx = r.*cos(theta);
Ty = r.*sin(theta);
% Generate the R points
Rx=zeros(N,1);
Ry=zeros(N,1);
for i=1:N
while true
% Try
alpha = 2*pi*rand();
rr = R*rand();
Rx(i) = Tx(i) + rr*cos(alpha);
Ry(i) = Ty(i) + rr*sin(alpha);
% Check if in the correct zone
if ( (Rx(i)*Rx(i) + Ry(i)*Ry(i) > d*d) && (Rx(i)*Rx(i) + Ry(i)*Ry(i) < D*D) )
break
end
end
end
% Display
figure(1);
clf;
angle=linspace(0,2*pi,1000);
plot( d*cos(angle), d*sin(angle),'-b');
hold on;
plot( D*cos(angle), D*sin(angle),'-b');
for i=1:N
plot(Tx(i),Ty(i),'gs');
plot(Rx(i),Ry(i),'rx');
plot([Tx(i) Rx(i)],[Ty(i) Ry(i)],'-k');
end
hold off;
let R be radious of (0;0) centered circle.
(x,y) : x^2+y^2<=R^2 (LE) to be inside the circle
x = rand()*2*R - R;
y should be in interval (-sqrt(R^2 - x^2);+sqrt(R^2 - x^2))
so, let it be
y = rand()*sqrt(R^2 - x^2)*2-sqrt(R^2 - x^2);
Hope, that's right, i have no matlab to test.
Hope, you'll manage to find close pairs your self.
Ok, i'll spend a bit more time for a hint.
To find a random number k in interval [a,b] use
k = rand()*(b-a)+a
Now it should really help if i still remember the matlab syntaxis. Good luck.
Here is a low quality solution that is very easy to use with uniformly distributed points. Assuming the number of points is small efficiency should not be a concern, if you want better quality you can use something more powerfull than nearest neighbor:
While you have less than n points: Generate a random point
If it is in the circle, store it else go to step 1
While there are unpaired points: check which point is nearest to the first unpaired point, make them a pair
As a result most pairs should be good, but some can be really really bad. I would recommend you to try it and perhaps add a step 4 with k-opt or some other local search if required. And if you really have little points (e.g. less than 20) you can of course just calculate all distances and find the optimum matching.
If you don't really care about the uniform distribution, here is an even easier solution:
While you have less than n points: Generate a random point
If it is in the circle, store it else go to step 1
For each of these points, generate a point near it
If it is in the circle, store it else go to step 3
Generate the random points as #PheuVerg suggested (with a slight vectorized tweak)
n = 8; %must be even!
x = rand(n, 1)*2*R - R;
y = rand(n, 1).*sqrt(R^2 - x.^2).*2-sqrt(R^2 - x.^2);
Then use kmeans clustering to get n/2 centers
[~ c] = kmeans([x y], n/2);
now you have to loop through each center and find it's distance to each point
dists = zeros(n, n/2);
for cc = 1:n/2
for pp = 1:n
dists(pp, cc) = sqrt((c(cc,1) - x(pp))^2 + (c(cc,2) - y(pp))^2);
end
end
now you must find the smallest 2 values for each columns of dists
[sorted, idx] = sort(dists);
so now the top two rows of each column are the two nearest points. But there could be clashes! i.e. points that are nearest to two different centers. So for repeated values you have to loop through and choose swap for the point that will give you the smallest extra distance.
Example data:
x =
0.7894
-0.7176
-0.5814
0.0708
0.5198
-0.2299
0.2245
-0.8941
y =
-0.0800
-0.3339
0.0012
0.9765
-0.4135
0.5733
-0.1867
0.2094
sorted =
0.1870 0 0 0.1555
0.2895 0.5030 0.5030 0.2931
0.3145 1.1733 0.6715 0.2989
1.0905 1.1733 0.7574 0.7929
1.1161 1.2326 0.8854 0.9666
1.2335 1.2778 1.0300 1.2955
1.2814 1.4608 1.2106 1.3051
1.4715 1.5293 1.2393 1.5209
idx =
5 4 6 3
7 6 4 2
1 3 3 8
6 7 8 6
3 8 7 7
2 1 2 4
4 5 1 5
8 2 5 1
So now it's clear that 5 and 7 are pairs, and that 3 and 2 are pairs. But 4 and 6 are both repeated. (in this case it is clear that they are pairs too I guess!) but what I would suggest is to leave point 4 with center 2 and point 6 with center 3. Then we start at column 2 and see the next available point is 8 with a distance of 1.2326. This would leave point 1 paired with point 6 but then it's distance from the center is 1.2106. Had we paired point 6 with 8 and point 4 with 1 we would have got distances of 0.7574 and 1.2778 respectively which is actually less total distance. So finding 'close' pairs is easy but finding the set of pairs with the globally smallest minimum is hard! This solutions gets you something decent quite easily but fi you need the global best then I'm afraid you have quite a bit of work to do still :(
Finally let me add some visualisation. First lets (manually) create a vector that shows which points are paired:
I = [1 2 2 1 3 4 3 4];
Remember that that will depend on your data! Now you can plot is nicely like this:
gscatter(x, y, I)
Hope this gets you close and that you can eliminate the manual pairing of mine at the end by yourself. It shouldn't be too hard to get a crude solution.
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.