I want to calculate each individual barycenter (centroid) of a list of triangles. Thus far I've managed to write this much :
function Triangle_Source_Centroid(V_Epoch0, F_Epoch0)
for i = 1:length(F_Epoch0)
Centroid_X = F_Epoch0(V_Epoch0(:,1),1) + F_Epoch0(V_Epoch0(:,1),2) + F_Epoch0(V_Epoch0(:,1),3);
Centroid_Y = F_Epoch0(V_Epoch0(:,2),1) + F_Epoch0(V_Epoch0(:,2),2) + F_Epoch0(V_Epoch0(:,2),3);
Centroid_Z = F_Epoch0(V_Epoch0(:,3),1) + F_Epoch0(V_Epoch0(:,3),2) + F_Epoch0(V_Epoch0(:,3),3);
Triangle_Centroid = [Centroid_X; Centroid_Y; Centroid_Z];
end
end
it doesn't work, and only gives me an error message:
Subscript indices must either be real positive integers or logicals.
Given how the variables are named, I'm guessing that V_Epoch0 is an N-by-3 matrix of vertices (X, Y, and Z for the columns) and F_Epoch0 is an M-by-3 matrix of face indices (each row is a set of row indices into V_Epoch0 showing which points make each triangle). Assuming this is right...
You can actually avoid using a for loop in this case by making use of matrix indexing. For example, to get the X coordinates for every point in F_Epoch0, you can do this:
allX = reshape(V_Epoch0(F_Epoch0, 1), size(F_Epoch0));
Then you can take the mean across the columns to get the average X coordinate for each triangular face:
meanX = mean(allX, 2);
And meanX is now a M-by-1 column vector. You can then repeat this for Y and Z coordinates:
allY = reshape(V_Epoch0(F_Epoch0, 2), size(F_Epoch0));
meanY = mean(allY, 2);
allZ = reshape(V_Epoch0(F_Epoch0, 3), size(F_Epoch0));
meanZ = mean(allZ, 2);
centroids = [meanX meanY meanZ];
And centroids is an M-by-3 matrix of triangle centroid coordinates.
Bonus:
All of the above can actually be done with just this one line:
centroids = squeeze(mean(reshape(V_Epoch0(F_Epoch0, :), [size(F_Epoch0, 1) 3 3]), 2));
Check out the documentation for multidimensional arrays to learn more about how this works.
Related
I have two equations:
ellipseOne = '((x-1)^2)/6^2 + y^2/3^2 = 1';
and
ellipseTwo = '((x+2)^2)/2^2 + ((y-5)^2)/4^2 = 1';
and I plotted them:
ezplot(ellipseOne, [-10, 10, -10, 10])
hold on
ezplot(ellipseTwo, [-10, 10, -10, 10])
title('Ellipses')
hold off
Now I'm trying to find the intersection of the two ellipses. I tried:
intersection = solve(ellipseOne, ellipseTwo)
intersection.x
intersection.y
to find the points where they intersect, but MATLAB is giving me a matrix and an equation as an answer which I don't understand. Could anyone point me in the right direction to get the coordinates of intersection?
The solution is in symbolic form. The last step you need to take is to transform it into numeric. Simply convert the result using double. However, because this is a pair of quadratic equations, there are 4 possible solutions due to the sign ambiguity (i.e. +/-) and can possibly give imaginary roots. Therefore, isolate out the real solutions and what is left should be your answer.
Therefore:
% Your code
ellipseOne = '((x-1)^2)/6^2 + y^2/3^2 = 1';
ellipseTwo = '((x+2)^2)/2^2 + ((y-5)^2)/4^2 = 1';
intersection = solve(ellipseOne, ellipseTwo);
% Find the points of intersection
X = double(intersection.x);
Y = double(intersection.y);
mask = ~any(imag(X), 2) | ~any(imag(Y), 2);
X = X(mask); Y = Y(mask);
The first three lines of code are what you did. The next four lines extract out the roots in numeric form, then I create a logical mask that isolates out only the points that are real. This looks at the imaginary portion of both the X and Y coordinate of all of the roots and if there is such a component for any of the roots, we want to eliminate those. We finally remove the roots that are imaginary and we should be left with two real roots.
We thus get for our intersection points:
>> disp([X Y])
-3.3574 2.0623
-0.2886 2.9300
The first column is the X coordinate and the second column is the Y coordinate. This also seems to agree with the plot. I'll take your ellipse plotting code and also insert the intersection points as blue dots using the above solution:
ezplot(ellipseOne, [-10, 10, -10, 10])
hold on
ezplot(ellipseTwo, [-10, 10, -10, 10])
% New - place intersection points
plot(X, Y, 'b.');
title('Ellipses');
hold off;
I have the following code for calculating the result of a linear combination of Gaussian functions. What I'd really like to do is to vectorize this somehow so that it's far more performant in Matlab.
Note that y is a column vector (output), x is a matrix where each column corresponds to a data point and each row corresponds to a dimension (i.e. 2 rows = 2D), variance is a double, gaussians is a matrix where each column is a vector corresponding to the mean point of the gaussian and weights is a row vector of the weights in front of each gaussian. Note that the length of weights is 1 bigger than gaussians as weights(1) is the 0th order weight.
function [ y ] = CalcPrediction( gaussians, variance, weights, x )
basisFunctions = size(gaussians, 2);
xvalues = size(x, 2);
if length(weights) ~= basisFunctions + 1
ME = MException('TRAIN:CALC', 'The number of weights should be equal to the number of basis functions plus one');
throw(ME);
end
y = weights(1) * ones(xvalues, 1);
for xIdx = 1:xvalues
for i = 1:basisFunctions
diff = x(:, xIdx) - gaussians(:, i);
y(xIdx) = y(xIdx) + weights(i+1) * exp(-(diff')*diff/(2*variance));
end
end
end
You can see that at the moment I simply iterate over the x vectors and then the gaussians inside 2 for loops. I'm hoping that this can be improved - I've looked at meshgrid but that seems to only apply to vectors (and I have matrices)
Thanks.
Try this
diffx = bsxfun(#minus,x,permute(gaussians,[1,3,2])); % binary operation with singleton expansion
diffx2 = squeeze(sum(diffx.^2,1)); % dot product, shape is now [XVALUES,BASISFUNCTIONS]
weight_col = weights(:); % make sure weights is a column vector
y = exp(-diffx2/2/variance)*weight_col(2:end); % a column vector of length XVALUES
Note, I changed diff to diffx since diff is a builtin. I'm not sure this will improve performance as allocating arrays will offset increase by vectorization.
I'm trying to generate a cloud of 2D points (uniformly) distributed within a triangle. So far, I've achieved the following:
The code I've used is this:
N = 1000;
X = -10:0.1:10;
for i=1:N
j = ceil(rand() * length(X));
x_i = X(j);
y_i = (10 - abs(x_i)) * rand;
E(:, i) = [x_i y_i];
end
However, the points are not uniformly distributed, as clearly seen in the left and right corners. How can I improve that result? I've been trying to search for the different shapes too, with no luck.
You should first ask yourself what would make the points within a triangle distributed uniformly.
To make a long story short, given all three vertices of the triangle, you need to transform two uniformly distributed random values like so:
N = 1000; % # Number of points
V = [-10, 0; 0, 10; 10, 0]; % # Triangle vertices, pairs of (x, y)
t = sqrt(rand(N, 1));
s = rand(N, 1);
P = (1 - t) * V(1, :) + bsxfun(#times, ((1 - s) * V(2, :) + s * V(3, :)), t);
This will produce a set of points which are uniformly distributed inside the specified triangle:
scatter(P(:, 1), P(:, 2), '.')
Note that this solution does not involve repeated conditional manipulation of random numbers, so it cannot potentially fall into an endless loop.
For further reading, have a look at this article.
That concentration of points would be expected from the way you are building the points. Your points are equally distributed along the X axis. At the extremes of the triangle there is approximately the same amount of points present at the center of the triangle, but they are distributed along a much smaller region.
The first and best approach I can think of: brute force. Distribute the points equally around a bigger region, and then delete the ones that are outside the region you are interested in.
N = 1000;
points = zeros(N,2);
n = 0;
while (n < N)
n = n + 1;
x_i = 20*rand-10; % generate a number between -10 and 10
y_i = 10*rand; % generate a number between 0 and 10
if (y_i > 10 - abs(x_i)) % if the points are outside the triangle
n = n - 1; % decrease the counter to try to generate one more point
else % if the point is inside the triangle
points(n,:) = [x_i y_i]; % add it to a list of points
end
end
% plot the points generated
plot(points(:,1), points(:,2), '.');
title ('1000 points randomly distributed inside a triangle');
The result of the code I've posted:
one important disclaimer: Randomly distributed does not mean "uniformly" distributed! If you generate data randomly from an Uniform Distribution, that does not mean that it will be "evenly distributed" along the triangle. You will see, in fact, some clusters of points.
You can imagine that the triangle is split vertically into two halves, and move one half so that together with the other it makes a rectangle. Now you sample uniformly in the rectangle, which is easy, and then move the half triangle back.
Also, it's easier to work with unit lengths (the rectangle becomes a square) and then stretch the triangle to the desired dimensions.
x = [-10 10]; % //triangle base
y = [0 10]; % //triangle height
N = 1000; %// number of points
points = rand(N,2); %// sample uniformly in unit square
ind = points(:,2)>points(:,1); %// points to be unfolded
points(ind,:) = [2-points(ind,2) points(ind,1)]; %// unfold them
points(:,1) = x(1) + (x(2)-x(1))/2*points(:,1); %// stretch x as needed
points(:,2) = y(1) + (y(2)-y(1))*points(:,2); %// stretch y as needed
plot(points(:,1),points(:,2),'.')
We can generalize this case. If you want to sample points from some (n - 1)-dimensional simplex in Euclidean space UNIFORMLY (not necessarily a triangle - it can be any convex polytope), just sample a vector from a symmetric n-dimensional Dirichlet distribution with parameter 1 - these are the convex (or barycentric) coordinates relative to the vertices of the polytope.
I am trying to find the sum of certain coordinates in a matrix.
I have a N x M matrix. I have a vector which contains 2xM values. Every pair of values in the vector is a coordinate in the matrix. Hence their are M number of coordinates. I want to find the sum of all of the coordinates without using a for loop.
Is there a matrix operation I can use to get this?
Thanks
As I understand it the vector contains the (row,column) coordinates to the matrix elements. You can just transform them into matrix element number indices. This example shows how to do it. I'm assuming that your coordinate vector looks like this:
[n-coordinate1 m-coordinate1 n-coordinate2 m-coordinate2 ...]
n = 5; % number of rows
m = 5; % number of columns
matrix = round(10*rand(n,m)); % An n by m example matrix
% A vector with 2*m elements. Element 1 is the n coordinate,
% Element 2 the m coordinate, and so on. Indexes into the matrix:
vector = ceil(rand(1,2*m)*5);
% turn the (n,m) coordinates into the element number index:
matrixIndices = vector(1:2:end) + (vector(2:2:end)-1)*n);
sumOfMatrixElements = sum(matrix(matrixIndices)); % sums the values of the indexed matrix elements
If you want to find the centroid of your 2xM array coords, then you can simply write
centroid = mean(coords,2)
If you want to find the weighted centroid, where each coordinate pair is weighted by the corresponding entry in the MxN array A, you can use sub2ind like this:
idx = sub2ind(size(A),coords(1,:)',coords(2,:)');
weights = A(idx);
weightedCentroid = sum( bsxfun( #times, coords', weights), 1 ) / sum(weights);
If all you want is the sum of all the entries to which the coordinates point, you can do the above and simply sum the weights:
idx = sub2ind(size(A),coords(1,:)',coords(2,:)');
weights = A(idx);
sumOfValues = sum(weights);
I have a vector of coordinates called x. I want to get the element(s) with the min y coordinate:
a = find(x(:,2)==min(x(:,2))); % Contains indices
This returns the indexes of the elements with the smallest y coordinates. I say element*s* because sometimes this would return more than 1 value (e.g. (10,2) and (24,2) both have 2 as y coordinate and if 2 is the min y coordinate...).
Anyway, my next step is to sort (ascending) the elements with the min y coordinates according to their x coordinates. First I do:
b = sort(x(a,1));
The above operation might rearrange the elements with min y coordinates so I want to apply this rearrangement to a as well. So I do:
[v i] = ismember(b, x(:, 1));
Unfortunately, if there are elements with the same x value but different y values and one of these elements turns out to be a member of a (i.e. b) then the above matrix chooses it. For example if (10,2) and (24,2) are the elements with smallest y coordinates and there is a 3rd element (24, 13) then it will mess up the above operation. Is there a better way? I wrote my script using loops and everything was fine but in line with Matlab's methodology I rewrote it and I fear my unfamiliarity with matlab is causing this error.
Sorry, I might have misunderstood your question but lemme rephrase what I think you want here:
You have a set of 2D coordinates:
x = [24,2; 10,2; 24,13];
You want the pairs of coordinates to stay together (24,2) (10,2) and (24,13). And you want to find the pairs of coordinates that has the min y-coordinate and if there are multiples, then you want to sort them by x-coordinate. And you want the row indices of what those coordinate pairs were in the original matrix x. So in other words, you want a final answer of:
v = [10,2; 24,2];
i = [2,1];
If I understood correctly, then this is how you can do it:
(Note: I changed x to have one more pair (40,13) to illustrate the difference between idx(i) and i)
>> x = [40,13; 24,2; 10,2; 24,13];
>> idx = find(x(:,2)==min(x(:,2))) %Same as what you've done before.
idx =
2
3
>> [v,i] = sortrows(x(idx,:)) %Use sortrows to sort by x-coord while preserving pairings
v =
10 2
24 2
i = % The indices in x(idx,:)
2
1
>> idx(i) %The row indices in the original matrix x
ans =
3
2
And finally, if this is not what you wanted, can you indicate what you think your answer [v,i] should be in the example you gave?