I have two plotted lines and I want to find the least distance error between them. When I simply subtract them from each other I get the error in the x-direction. But I am looking for the error in the least distance way between the two lines.
Any help is greatly appreciated!
Best regards,
Gidi
With d = pdist2(L1, L2, 'euclidean', 'smallest', 1); you get a vector d with the distance from each point in L2 to it's closest neighbor in L1. Then, the shortest distance will be min(d).
I am assuming that both L1 and L2 are n-by-2 and m-by-2 with m and n being the number of points (n and m are allowed to be different). From your comment I would guess you had not included the x-component. To fix that, you could say L1 = [y_n, u_new] and likewise for L2 from z, assuming y_n was the x-component. If y_n is a row vector, you should transpose it as in L1 = [y_n', u_new].
If you wish to plot the minimum distance for each point and both lines, plot(y_n, [u_new, z, d]) should work. Again here, check the orientation of your vectors.
Related
I have a matrix of A=[60x60],and two coefficients a,b. Since matrix A was moved by a,b, how to multiply the coefficients into matrix A so that I could obtain A_moved? Any function to do it?
Here's part of matlab code implemented:
A=rand(60); %where it's in 2D, A(k1,k2)
a=0.5; b=0.8;
[m, n]=size(A);
[M,N] = meshgrid(1:m,1:n);
X = [M(:), N(:)];
A_moved=A(:)(X)*[a b] %I know this is not valid but you get the idea
In another word A_moved is calculated by A_moved=a*k1+b*k2.
This line of code A_moved=A(:)(X)*[a b] is to represent my idea that a,b multiply back into the original A because X represent correspond coordinates of k1 and k2. The first column represent k1, and second column represent k2. Thus it become A_moved=a*k1+b*k2. But this couldn't get me anyway.
In the end A-moved is a 60x60 matrix which have been multiplied by coefficients a,b correspondingly. To make it clearer,A is the phase of image. a,b moved it phase.
Appreciate any help. Thank you!
Reference paper: Here
EDIT:
As suggested by Noel for better understanding.
A=[2 3;5 7], a=1.5 and b=2.5.
Since A is approximated as a*k1+b*k2
Thus,
A_moved=[1.5*k1_1+2.5k2_1 1.5*k1_2+2.5k2_2; 1.5*k1_2+2.5k2_1 1.5*k1_2+2.5k2_2];
where k1 and k2, If I'm understood correctly is the coordinates of the original A matrix, as defined in X above.
On the chat we found that your problem was matrix algebra related
What you want to obtain in A_moved is the x coordinate multiplied by a contant a plus the y coordinate multiplied by a constant b.
You already have this coordinates in M and N, so you can obtain A_moved as
A_moved = (a*M) + (b*N);
And it will retain same shape as A
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 hope you will have the right answer to this question, which is rather challanging.
I have two uni-dimensional vectors Y and Z, which hold the coordinates of N grid points located on a square grid. So
Ny points along Y
Nz points along Z
N = Ny*Nz
Y = Y[N]; (Y holds N entries)
Z = Z[N]; (Z holds N entries)
Now, the goal would be generating the distance matrix D, which holds N*N entries: so each row of matrix D is defined as the distance between the i-th point on the grid and the remnant (N - i) points.
Generally, to compute the whole matrix I would call
D = squareform(pdist([Y Z]));
or,equivalently,
D = pdist2([Y Z],[Y Z]).
But, since D is a symmetric matrix, I'd like to generate only the N(N + 1)/2 independent entries and store them into a row-ordered vector Dd.
So the question is: how to generate a row-ordered array Dd whose entries are defined by the lower triangular terms of matrix D? I'd, furthermore, like storing the entries in a column-major order.
I hope the explanation is clear enough.
As woodchips commented, it is simpler and faster to compute the whole matrix and extract the elements you care about. Here's a way to do it:
ndx = find(tril(true(size(D))));
Dd = D(ndx);
If you absolutely must compute elements of Dd without computing matrix D first, you probably need a double for loop.
Given a 3-dimensional polyline P = {(x1, y1, t1), ..., (xn, yn, tn)} and another polyline Q = {(x1, y1, t1), ..., (xm, ym, tm)} (m not necessarly equal to n, so polylines could have different length), coincidence in space and time occurs when the trajectories of moving objects P and Q, have some timing and positioning in common (Point A, as seen in example figure is a coincidence point cause (xa, ya, ta)==(xb, yb, tb) obviously coincidence point could be a point outside of initial sets of point)
The concept is quite simple and visual perspective easily identify where colocation happen. Hardest part is how to realize an algorithm that efficiently compute coincidence and return the calculated (remember: point could be outside given sets of points) x, y coords and time t of where colocation happens!! This algorithm will be developed in Matlab so I have all necessary to rapidly work.
Best regards
assuming that x, y, z are functions of t for all segments of each polyline, here's a brute-force start. in 4 dimensions: P has segments p1 from (x_start(t), y_start(t), z_start(t), t) to (x_end(t), y_end(t), z_end(t), t), and similarly Q
for each segment p of P
for each segment q of Q
if p intersects q (in 4 dimensions)
output intersection point
the intersection condition is:
there exists alpha and beta in [0,1] where alpha * px_start(t) + (1 - alpha) * (px_end(t) - px_start(t)) = beta * qx_start(t) + (1 - beta) * (qx_end(t) - qx_start(t)) and 2 more similar conditions for y and z
solvability of the intersection condition depends on what are the functions x(t), y(t), z(t) -- linear? polynomials? etc.
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.