I have an image of a cytoskeleton. There are a lot of small objects inside and I want to calculate the length between all of them in every axis and to get a matrix with all this data. I am trying to do this in matlab.
My final aim is to figure out if there is any axis with a constant distance between the object.
I've tried bwdist and to use connected components without any luck.
Do you have any other ideas?
So, the end goal is that you want to globally stretch this image in a certain direction (linearly) so that the distances between nearest pairs end up the closest together, hopefully the same? Or may you do more complex stretching ? (note that with arbitrarily complex one you can always make it work :) )
If linear global one, distance in x' and y' is going to be a simple multiplication of the old distance in x and y, applied to every pair of points. So, the final euclidean distance will end up being sqrt((SX*x)^2 + (SY*y)^2), with SX being stretch in x and SY stretch in y; X and Y are distances in X and Y between pairs of points.
If you are interested in just "the same" part, solution is not so difficult:
Find all objects of interest and put their X and Y coordinates in a N*2 matrix.
Calculate distances between all pairs of objects in X and Y. You will end up with 2 matrices sized N*N (with 0 on the diagonal, symmetric and real, not sure what is the name for that type of matrix).
Find minimum distance (say this is between A an B).
You probably already have this. Now:
Take C. Make N-1 transformations, which all end up in C->nearestToC = A->B. It is a simple system of equations, you have X1^2*SX^2+Y1^2*SY^2 = X2^2*SX^2+Y2*SY^2.
So, first say A->B = C->A, then A->B = C->B, then A->B = C->D etc etc. Make sure transformation is normalized => SX^2 + SY^2 = 1. If it cannot be found, the only valid transformation is SX = SY = 0 which means you don't have solution here. Obviously, SX and SY need to be real.
Note that this solution is unique except in case where X1 = X2 and Y1 = Y2. In this case, grab some other point than C to find this transformation.
For each transformation check the remaining points and find all nearest neighbours of them. If distance is always the same as these 2 (to a given tolerance), great, you found your transformation. If not, this transformation does not work and you should continue with the next one.
If you want a transformation that minimizes variations between distances (but doesn't require them to be nearly equal), I would do some optimization method and search for a minimum - I don't know how to find an exact solution otherwise. I would pick this also in case you don't have linear or global stretch.
If i understand your question correctly, the first step is to obtain all of the objects center of mass points in the image as (x,y) coordinates. Then, you can easily compute all of the distances between all points. I suggest taking a look on a histogram of those distances which may provide some information as to the nature of distance distribution (for example if it is uniformly random, or are there any patterns that appear).
Obtaining the center of mass points is not an easy task, consider transforming the image into a binary one, or some sort of background subtraction with blob detection or/and edge detector.
For building a histogram you can use histogram.
Related
I am trying to randomly generate uniformly distributed vectors, which are of Euclidian length of 1. By uniformly distributed I mean that each entry (coordinate) of the vectors is uniformly distributed.
More specifically, I would like to create a set of, say, 1000 vectors (lets call them V_i, with i=1,…,1000), where each of these random vectors has unit Euclidian length and the same dimension V_i=(v_1i,…,v_ni)' (let’s say n = 5, but the algorithm should work with any dimension). If we then look on the distribution of e.g. v_1i, the first element of each V_i, then I would like that this is uniformly distributed.
In the attached MATLAB example you see that you cannot simply draw random vectors from a uniform distribution and then normalize the vectors to Euclidian length of 1, as the distribution of the elements across the vectors is then no longer uniform.
Is there a way to generate this set of vectors such, that the distribution of the single elements across the vector-set is uniform?
Thank you for any ideas.
PS: MATLAB is our Language of choice, but solutions in any languages are, of course, welcome.
clear all
rng('default')
nvar=5;
sample = 1000;
x = zeros(nvar,sample);
for ii = 1:sample
y=rand(nvar,1);
x(:,ii) = y./norm(y);
end
hist(x(1,:))
figure
hist(x(2,:))
figure
hist(x(3,:))
figure
hist(x(4,:))
figure
hist(x(5,:))
What you want cannot be accomplished.
Vectors with a length of 1 sit on a circle (or sphere or hypersphere depending on the number of dimensions). Let's focus on the 2D case, if it cannot be done there, it will be clear that it cannot be done with more dimensions either.
Because the points are on a circle, their x and y coordinates are dependent, the one can be computed based on the other. Thus, the distributions of x and y coordinates cannot be defined independently. We can define the distribution of the one, generate random values for it, but the other coordinate must be computed from the first.
Let's make points on a half circle with a uniform x coordinate (can be extended to a full circle by adding a random sign to the y coordinate):
N = 1000;
x = 2 * rand(N,1) - 1;
y = sqrt(1 - x.^2);
plot(x,y,'.')
axis equal
histogram(y)
The plot generates shows a clearly non-uniform distribution, with many more samples generated near y=1 than near y=0. If we add a random sign to the y-coordinate we'd have more samples near y=1 and y=-1 than near y=0.
I have a matrix M of size NxP. Every P columns are orthogonal (M is a basis). I also have a vector V of size N.
My objective is to transform the first vector of M into V and to update the others in order to conservate their orthogonality. I know that the origins of V and M are the same, so it is basically a rotation from a certain angle. I assume we can find a matrix T such that T*M = M'. However, I can't figure out the details of how to do it (with MATLAB).
Also, I know there might be an infinite number of transforms doing that, but I'd like to get the simplest one (in which others vectors of M approximately remain the same, i.e no rotation around the first vector).
A small picture to illustrate. In my actual case, N and P can be large integers (not necessarily 3):
Thanks in advance for your help!
[EDIT] Alternative solution to Gram-Schmidt (accepted answer)
I managed to get a correct solution by retrieving a rotation matrix R by solving an optimization problem minimizing the 2-norm between M and R*M, under the constraints:
V is orthogonal to R*M[1] ... R*M[P-1] (i.e V'*(R*M[i]) = 0)
R*M[0] = V
Due to the solver constraints, I couldn't indicate that R*M[0] ... R*M[P-1] are all pairwise orthogonal (i.e (R*M)' * (R*M) = I).
Luckily, it seems that with this problem and with my solver (CVX using SDPT3), the resulting R*M[0] ... R*M[P-1] are also pairwise orthogonal.
I believe you want to use the Gram-Schmidt process here, which finds an orthogonal basis for a set of vectors. If V is not orthogonal to M[0], you can simply change M[0] to V and run Gram-Schmidt, to arrive at an orthogonal basis. If it is orthogonal to M[0], instead change another, non-orthogonal vector such as M[1] to V and swap the columns to make it first.
Mind you, the vector V needs to be in the column space of M, or you will always have a different basis than you had before.
Matlab doesn't have a built-in Gram-Schmidt command, although you can use the qr command to get an orthogonal basis. However, this won't work if you need V to be one of the vectors.
Option # 1 : if you have some vector and after some changes you want to rotate matrix to restore its orthogonality then, I believe, this method should work for you in Matlab
http://www.mathworks.com/help/symbolic/mupad_ref/numeric-rotationmatrix.html
(edit by another user: above link is broken, possible redirect: Matrix Rotations and Transformations)
If it does not, then ...
Option # 2 : I did not do this in Matlab but a part of another task was to find Eigenvalues and Eigenvectors of the matrix. To achieve this I used SVD. Part of SVD algorithm was Jacobi Rotation. It says to rotate the matrix until it is almost diagonalizable with some precision and invertible.
https://math.stackexchange.com/questions/222171/what-is-the-difference-between-diagonalization-and-orthogonal-diagonalization
Approximate algorithm of Jacobi rotation in your case should be similar to this one. I may be wrong at some point so you will need to double check this in relevant docs :
1) change values in existing vector
2) compute angle between actual and new vector
3) create rotation matrix and ...
put Cosine(angle) to diagonal of rotation matrix
put Sin(angle) to the top left corner of the matric
put minus -Sin(angle) to the right bottom corner of the matrix
4) multiple vector or matrix of vectors by rotation matrix in a loop until your vector matrix is invertible and diagonalizable, ability to invert can be calculated by determinant (check for singularity) and orthogonality (matrix is diagonalized) can be tested with this check - if Max value in LU matrix is less then some constant then stop rotation, at this point new matrix should contain only orthogonal vectors.
Unfortunately, I am not able to find exact pseudo code that I was referring to in the past but these links may help you to understand Jacobi Rotation :
http://www.physik.uni-freiburg.de/~severin/fulltext.pdf
http://web.stanford.edu/class/cme335/lecture7.pdf
https://www.nada.kth.se/utbildning/grukth/exjobb/rapportlistor/2003/rapporter03/maleko_mercy_03003.pdf
Okay, so I'm working on a problem related to quantum chaos and one of the things I need to do is to map the unit cube in n-dimensions to a parallelepiped in n-dimensions and find all integer points in the interior of this parallelepiped. I have been trying to do this using the following scheme:
Given the linear map B and the dimension of the cube n, we find the coordinates of the corners of the unit hypercube by converting numbers j from 0 to (2^n -1) into their binary representation and turning them into vectors that describe the vertices of the cube.
The next step was to apply the map B to each of these vectors, which gives me a set of 2^n vectors describing the coordinates of the vertices of the parallelepiped in n dimensions
Now, we take the maximum and minimum value attained by any of these vertices in each coordinate direction, i.e the first element of my vectors might have a maximum value of 4 across all of the vertices and a minimum value of -3 etc. This gives me an n-dimensional rectangular prism that contains my parallelepiped and some extra unwanted space.
I now find all points with integer coordinates in this bounding rectangular prism described as vectors in n dimensions
Finally, I apply the inverse of the map B to each of the points and throw away any points that have any coefficients greater than 1 as they must originally have lain outside my unit hypercube.
My issue arises in step 4, I'm struggling to come up with a way of generating all vectors with integer coordinates in my rectangular hyper-prism such that I can change the number of dimensions n on the fly. Ideally, i'd like to be able to increase n at will until it becomes too computationally heavy to do so, but every method of finding all integer points in the prism i've tried so far has relied on n for loops to permute each element and thus I need to rewrite the code every time.
So I guess my question is this, is there any way to code this up so that I can change n on the fly? Also, any thoughts on the idea of the algorithm itself would be appreciated :) It wouldn't surprise me if i've overcomplicated things massively...
EDIT:
Of course as soon as I post the question I see a lovely little link in the side-bar where a clever method has been given already for how to do this: Generate a matrix containing all combinations of elements taken from n vectors
I'll leave this up for the moment just in case anyone has any comments on the method in general, but otherwise (since I can't upvote yet I'll just say it here) Luis Mendo, you are a hero!
I am trying to create an orthogonal coordinate system based on two "almost" perpendicular vectors, which are deduced from medical images. I have two vectors, for example:
Z=[-1.02,1.53,-1.63];
Y=[2.39,-1.39,-2.8];
that are almost perpendicular, since their inner product is equal to 5e-4.
Then I find their cross product to create my 3rd basis:
X=cross(Y,Z);
Even this third vector is not completely orthogonal to Z and Y, as their inner products are in the order of -15 and -16, but I guess that is almost zero. In order to use this set of vectors as my orthogonal basis for a local coordinate system, I assume they should be almost completely perpendicular. I first thought I can do this by rounding my vectors to less decimal figures, but did not help. I guess I need to find a way to alter my initial vectors a little to make them more perpendicular, but I don't know how to do that.
I would appreciate any suggestions.
Gram-Schmidt is right as pointed out above.
Basically, you want to subtract the component of Y that is in the direction of Z from Y (Note: you can alternatively operate on Z instead of Y).
The component of Y in the Z direction is given by:
dot(Y,Z)*Z/(norm(Z)^2)
(projection of Y onto Z)
Note that if Y is orthogonal to Z, then this is 0.
So:
Y = Y - dot(Y,Z)*Z/(norm(Z)^2)
and Z stays unchanged.
let V=Y+aZ
Z dot V = 0 so you can solve a and get V
Now use V and Z as you basis
You may need to normalize the vectors and use double type to get the desired precision.
I wanted to generate a set of coordinates distributed uniformly at random within a ball of radius R. Is there any way to do this in Matlab without for loops, in a matrix-like form?
Thanks
UPDATE:
I'm sorry for the confusion. I only need to generate n points uniformly at random over a circle of radius R, not a sphere.
the correct answer is here http://mathworld.wolfram.com/DiskPointPicking.html. The distribution is known as "Disk point picking"
I was about to mark this as a duplicate of a previous question on generating uniform distribution of points in a sphere, but I think you deserve the benefit of doubt here, as although there's a matlab script in the question, most of that thread is python.
This little function given in the question (and I'm pasting it directly from there), is what you need.
function X = randsphere(m,n,r)
% This function returns an m by n array, X, in which
% each of the m rows has the n Cartesian coordinates
% of a random point uniformly-distributed over the
% interior of an n-dimensional hypersphere with
% radius r and center at the origin. The function
% 'randn' is initially used to generate m sets of n
% random variables with independent multivariate
% normal distribution, with mean 0 and variance 1.
% Then the incomplete gamma function, 'gammainc',
% is used to map these points radially to fit in the
% hypersphere of finite radius r with a uniform % spatial distribution.
% Roger Stafford - 12/23/05
X = randn(m,n);
s2 = sum(X.^2,2);
X = X.*repmat(r*(gammainc(s2/2,n/2).^(1/n))./sqrt(s2),1,n);
To learn why you can't just use uniform random variable for all three co-ordinates as one might think is the correct way, give this article a read.
For the sake of completeness, here is some MATLAB code for a point-culling solution. It generates a set of random points within a unit cube, removes points that are outside a unit sphere, and scales the coordinate points up to fill a sphere of radius R:
XYZ = rand(1000,3)-0.5; %# 1000 random 3-D coordinates
index = (sum(XYZ.^2,2) <= 0.25); %# Find the points inside the unit sphere
XYZ = 2*R.*XYZ(index,:); %# Remove points and scale the coordinates
One key drawback to this point-culling method is that it makes it difficult to generate a specific number of points. For example, if you want to generate 1000 points within your sphere, how many do you have to create in the cube before culling them? If you scale up the number of points generated in the cube by a factor of 6/pi (i.e. the ratio of the volume of a unit cube to a unit sphere), then you can get close to the number of desired points in the sphere. However, since we're dealing with (pseudo)random numbers after all, we can never be absolutely certain we will generate enough points that fall in the sphere.
In short, if you want to generate a specific number of points, I'd try out one of the other solutions suggested. Otherwise, the point-culling solution is nice and simple.
Not sure if I understand your question correctly, but can't you just generate any random number inside a sphere by setting φ, θ and r, assigned to random numbers?