I am using geom3d to model 3D geometric shapes. Specifically, I am interested to calculate a volume of the ellipsoid after slicing it with an arbitrary plane. I am using the following code (here for simplicity I will use a sphere and cut it in the center).
[x, y, z] = sphere(30);
[f, v, c] = surf2patch(x, y, z, z, 'triangles');
meshVolume(v, f)
This calculates a volume of a sphere as expected (4/3*pi).
Now, I slice it in half.
P = createPlane([0, 0, 0], [0, 0, 1]);
[v1 f1] = clipConvexPolyhedronHP(v, f, P);
drawMesh(v1, f1, 'facealpha', 0.5)
Now if I calculate the volume, I get incorrect value (significantly smaller than half of the initial volume).
meshVolume(v1, f1)
What is the reason for the incorrect result?
(I have already answerd on Mathworks File Exchange page, but I think it can be useful to put it also here)
Using determinant without 'abs' is the correct way of computing volume of meshes. By doing that way you can handle non convex meshes, and even polyhedra with holes. But the faces of the mesh must be oriented consistently (usually outwards of the mesh).
A similar topic is given here
Using the absolute value of the determinant will give correct result only for convex polyhedra. So I would use it only as a quick hack for specific application.
Regarding the original question, there was a bug in the computation of the clipped mesh, resulting in faces with inconsistent orientation, and therefore in wrong resulting volume. I have updated it so it should give better results.
(note: I am the author of the geom3d package)
Actually found a bug in the implementation of the meshVolume function.
The function divides the volume into tetrahedra, which are 3-dimensional solids that have 4 vertices, 6 edges, and 4 triangular faces.
Then the total volume of the body is the sum of the volumes of all tetrahedra.
The code computes the volume by calculating the determinant of the vertices.
vols(i) = det(tetra) / 6;
It turns out that sometimes this determinant can be negative due to a different order of the rows (vertices coordinates).
The correct formula should include absolute value of the determinant.
vols(i) = abs(det(tetra) / 6);
That solves the problem. I've reported the bug on geom3d webpage.
For more details about tetrahedra see here
Related
I have a huge set of data of a timelapse of 2D laser scans of waves running up and down stairs (see fig.1fig.2fig.3).
There is a lot of noise in the scans, since the water splashes a lot.
Now I want to smoothen the scans.
I have 2 questions:
How do I apply a moving median filter (as recommended by another study dealing with a similar problem)? I can only find instructions for single e.g. (x,y) or (t,y) plots but not for x and y values that vary over time. Maybe an average filter would do it as well, but I do not have a clue on that either.
The scanner is at a fixed point (222m) so all the data spikes point towards that point at the ceiling. Is it possible or necessary to include this into the smoothing process?
This is the part of the code (I hope it's enough to get it):
% Plot data as real time profile
x1=data.x;y1=data.y;
t=data.t;
% add moving median filter here?
h1=plot(x1(1,:),y1(1,:));
axis([210 235 3 9])
ht=title('Scanner data');
for i=1:1:length(t);
set(h1,'XData',x1(i,:),'YData',y1(i,:));set(ht,'String',sprintf('t = %5.2f
s',data.t(i)));pause(.01);end
The data.x values are stored in a (mxn) matrix in which the change in time is arranged vertically and the x values i.e. "laser points" of the scanner are horizontally arranged. The data.y is stored in the same way. The data.t values are stored in a (mx1) matrix.
I hope I explained everything clearly and that somebody can help me. I am already pretty desperate about it... If there is anything missing or confusing, please let me know.
If you're trying to apply a median filter in the x-y plane, then consider using medfilt2 from the Image Processing Toolbox. Note that this function only accepts 2-D inputs, so you'll have to loop over the third dimension.
Also note that medfilt2 assumes that the x and y data are uniformly spaced, so if your x and y data don't fall onto a uniformly spaced grid you may have to manually loop over indices, extract the corresponding patches, and compute the median.
If you can/want to apply an averaging filter instead of a median filter, and if you have uniformly spaced data, then you can use convn to compute a k x k moving average by doing:
y = convn(x, ones(k,k)/(k*k), 'same');
Note that you'll get some bias on the boundaries because you're technically trying to compute an average of k^2 pixels when you have less than that number of values available.
Alternatively, you can use nested calls to movmean since the averaging operation is separable:
y = movmean(movmean(x, k, 2), k, 1);
If your grid is separable, but not uniform, you can still use movmean, just use the SamplePoints name-value pair:
y = movmean(movmean(x, k, 2, 'SamplePoints', yv), k, 1, 'SamplePoints', xv);
You can also control the endpoint handling in movmean with the Endpoints name-value pair.
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.
I have 2 sets of DICOM image data for 1 subject, consisting of a PET scan and CT scan which were taken at the same time. The Frame of Reference UIDs are different, which I think means that their reference origins are different. So that the 'Image Position Patient' tag can't be compared.
What I want to do is resample both images such that their spatial dimensions are equal and their pixel dimensions are equal. The task seems relatively straightforward, but for the fact that their origins are different.
Download link for data
For any two images A and B deemed to represent the same object, registration is the act of identifying for each pixel / landmark in A the equivalent pixel / landmark in B.
Assuming each pixel in both A and B can be embedded in a coordinate system, registration usually entails transforming A such that after the transformation, the coordinates of each pixel in A coincide with those of the equivalent pixel in B (i.e. the objective is for the two objects overlap in that coordinate space)
An isometric transformation is one where the distance between any two pixels in A, and the distance between the equivalent two pixels in B does not change after the transformation has been applied. For instance, rotation in space, reflection (i.e. mirror image), and translation (i.e. shifting the object in a particular direction) are all isometric transformations. A registration algorithm applying only isometric transformations is said to be rigid.
An affine transformation is similar to an isometric one, except scaling may also be involved (i.e. the object can also grow or shrink in size).
In medical imaging If A and B were obtained at different times, it is highly unlikely that the transformation is a simple affine or isometric one. For instance, say during scan A the patient had their arms down by their side, and in scan B the patient had their arms over their head. There is no rigid registration of A that would result in perfect overlap with B, since distances between equivalent points have changed (e.g. the distance between head-to-hand, and hand-to-foot in each case). Therefore more elaborate non-rigid registration algorithms would need to be used.
The fact that in your case A and B were obtained during the same scanning session in the same machine means that it's a reasonable assumption that the transformation will be a simple affine one. I.e. you will probably only need to rotate and translate the object a bit; if the coordinate system of A is 'denser' than B, you might also need to grow / shrink it a bit. But that's it, no weird 'warping' will be necessary to compensate for 'movement' occurring between scans A and B being obtained, since they happened at the same time.
A 3D vector, denoting a 'magnitude and direction' in 3D space can be transformed to another 3D vector using a 3x3 transformation matrix T. For example, if you apply transformation to vector (using matrix multiplication), the resulting vector u is . In other words, the 'new' x-coordinate depends on the old x, y, and z coordinates in a manner specified by the transformation matrix, and similarly for the new y and new z coordinates.
If you apply a 3x3 transformation T to three vectors at the same time, you'll get three transformed vectors out. e.g. for v = [v1, v2, v3] where v1 = [1; 2; 3], v2 = [2; 3; 4], v3 = [3; 4; 5], then T*v will
give you a 3x3 matrix u, where each column corresponds to a
transformed vector of x,y,z coordinates.
Now, consider the transformation matrix T is unknown and we want to discover it. Say we have a known point and we know that after the transformation it becomes a known point . We have:
Consider the top row; even if you know p and p', it should be clear that you cannot determine a, b, and c from a single point. You have three unknowns and only one equation. Therefore to solve for a, b, and c, you need at least a system of three equations. The same applies for the other two rows. Therefore, to find the transformation matrix T you need three known points (before and after transformation).
In matlab, you can solve such a system of equations where T*v = u, by typing T = u/v. For a 3x3 transformation matrix T, u and v need to contain at least 3 vectors, but they can contain more (i.e. the system of equations is overrepresented). The more vectors you pass in, the more accurate the transformation matrix from a numerical point of view. But in theory you only need three.
If your transformation also involves a translation element, then you need to do the trick described in the image you posted. I.e. you represent a 3D vector [x,y,z] as a homogeneous-coordinates vector [x,y,z,1]. This enables you to add a 4th column in your transformation matrix, which results in a 'translation' for each point, i.e. adding an extra value in the new x', y' and z' coefficients, which is independent of the input vector. Since the translation coefficients are also unknown, you now have 12 instead of 9 unknowns, and therefore you need 4 points to solve this system. i.e.
To summarise:
To transform your image A to occupy the same space as B, interpret the coordinates of A as if they were in the same coordinate system as B, find four equivalent landmarks in both, and obtain a suitable transformation matrix as above by solving this system of equations using the / right matrix division operator. You can then use this transformation matrix T you found, to transform all the coordinates in A (expressed as homogeneous coordinates) to the new ones.
I want to evaluate the grid quality where all coordinates differ in the real case.
Signal is of a ECG signal where average life-time is 75 years.
My task is to evaluate its age at the moment of measurement, which is an inverse problem.
I think 2D approximation of the 3D case is hard (done here by Abo-Zahhad) with with 3-leads (2 on chest and one at left leg - MIT-BIT arrhythmia database):
where f is a piecewise continuous function in R^2, \epsilon is the error matrix and A is a 2D matrix.
Now, I evaluate the average grid distance in x-axis (time) and average grid distance in y-axis (energy).
I think this can be done by Matlab's Image Analysis toolbox.
However, I am not sure how complete the toolbox's approaches are.
I think a transform approach must be used in the setting of uneven and noncontinuous grids. One approach is exact linear time euclidean distance transforms of grid line sampled shapes by Joakim Lindblad et all.
The method presents a distance transform (DT) which assigns to each image point its smallest distance to a selected subset of image points.
This kind of approach is often a basis of algorithms for many methods in image analysis.
I tested unsuccessfully the case with bwdist (Distance transform of binary image) with chessboard (returns empty square matrix), cityblock, euclidean and quasi-euclidean where the last three options return full matrix.
Another pseudocode
% https://stackoverflow.com/a/29956008/54964
%// retrieve picture
imgRGB = imread('dummy.png');
%// detect lines
imgHSV = rgb2hsv(imgRGB);
BW = (imgHSV(:,:,3) < 1);
BW = imclose(imclose(BW, strel('line',40,0)), strel('line',10,90));
%// clear those masked pixels by setting them to background white color
imgRGB2 = imgRGB;
imgRGB2(repmat(BW,[1 1 3])) = 255;
%// show extracted signal
imshow(imgRGB2)
where I think the approach will not work here because the grids are not necessarily continuous and not necessary ideal.
pdist based on the Lumbreras' answer
In the real examples, all coordinates differ such that pdist hamming and jaccard are always 1 with real data.
The options euclidean, cytoblock, minkowski, chebychev, mahalanobis, cosine, correlation, and spearman offer some descriptions of the data.
However, these options make me now little sense in such full matrices.
I want to estimate how long the signal can live.
Sources
J. Müller, and S. Siltanen. Linear and nonlinear inverse problems with practical applications.
EIT with the D-bar method: discontinuous heart-and-lungs phantom. http://wiki.helsinki.fi/display/mathstatHenkilokunta/EIT+with+the+D-bar+method%3A+discontinuous+heart-and-lungs+phantom Visited 29-Feb 2016.
There is a function in Matlab defined as pdist which computes the pairwisedistance between all row elements in a matrix and enables you to choose the type of distance you want to use (Euclidean, cityblock, correlation). Are you after something like this? Not sure I understood your question!
cheers!
Simply, do not do it in the post-processing. Those artifacts of the body can be about about raster images, about the viewer and/or ... Do quality assurance in the signal generation/processing step.
It is much easier to evaluate the original signal than its views.
i am having 3D matrix in which most of the values are zeros but there are some nonzeros values.
when I am plotting this 3D matrix in matlab I am getting plot like as below
here u can see there are two groups of points are nearer to each other(that's why the color became dark) and two individual group of points is far away....
so my objective is to cluster that two nearer group of points and make it as one cluster1 and other two will be called as cluster2 and cluster3 ....
I tried kmeans clustering, BIC clustering...but as kmeans clustering is basically build up for 2D data input, I faced hurdle there ...then I reshape 3D matrix into 2D matrix but still I am getting another error Subscripted assignment dimension mismatch
so could u plz come out with some fruitful idea to do this......
Based on your comment that you used vol3d I assume that your data has to interpreted this way. If your data-matrix is called M, try
[A,B,C] = ind2sub(size(M),find(M));
points = [A,B,C];
idx = kmeans(points,3);
Here, I assumed that M(i,j,k) = 1 means that you have measured a point with properties i,j and k, which in your case would be velocity, angle and range.