Ellipse around the data in MATLAB - matlab

I would like to reproduce the following figure in MATLAB:
There are two classes of points with X and Y coordinates. I'd like to surround each class with an ellipse with one parameter of standard deviation, which determine how far the ellipse will go along the axis.
The figure was created with another software and I don't exactly understand how it calculates the ellipse.
Here is the data I'm using for this figure. The 1st column is class, 2nd - X, 3rd - Y. I can use gscatter to draw the points itself.
A = [
0 0.89287 1.54987
0 0.69933 1.81970
0 0.84022 1.28598
0 0.79523 1.16012
0 0.61266 1.12835
0 0.39950 0.37942
0 0.54807 1.66173
0 0.50882 1.43175
0 0.68840 1.58589
0 0.59572 1.29311
1 1.00787 1.09905
1 1.23724 0.98834
1 1.02175 0.67245
1 0.88458 0.36003
1 0.66582 1.22097
1 1.24408 0.59735
1 1.03421 0.88595
1 1.66279 0.84183
];
gscatter(A(:,2),A(:,3),A(:,1))
FYI, here is the SO question on how to draw ellipse. So, we just need to know all the parameters to draw it.
Update:
I agree that the center can be calculated as the means of X and Y coordinates. Probably I have to use principal component analysis (PRINCOMP) for each class to determine the angle and shape. Still thinking...

Consider the code:
%# generate data
num = 50;
X = [ mvnrnd([0.5 1.5], [0.025 0.03 ; 0.03 0.16], num) ; ...
mvnrnd([1 1], [0.09 -0.01 ; -0.01 0.08], num) ];
G = [1*ones(num,1) ; 2*ones(num,1)];
gscatter(X(:,1), X(:,2), G)
axis equal, hold on
for k=1:2
%# indices of points in this group
idx = ( G == k );
%# substract mean
Mu = mean( X(idx,:) );
X0 = bsxfun(#minus, X(idx,:), Mu);
%# eigen decomposition [sorted by eigen values]
[V D] = eig( X0'*X0 ./ (sum(idx)-1) ); %#' cov(X0)
[D order] = sort(diag(D), 'descend');
D = diag(D);
V = V(:, order);
t = linspace(0,2*pi,100);
e = [cos(t) ; sin(t)]; %# unit circle
VV = V*sqrt(D); %# scale eigenvectors
e = bsxfun(#plus, VV*e, Mu'); %#' project circle back to orig space
%# plot cov and major/minor axes
plot(e(1,:), e(2,:), 'Color','k');
%#quiver(Mu(1),Mu(2), VV(1,1),VV(2,1), 'Color','k')
%#quiver(Mu(1),Mu(2), VV(1,2),VV(2,2), 'Color','k')
end
EDIT
If you want the ellipse to represent a specific level of standard deviation, the correct way of doing is by scaling the covariance matrix:
STD = 2; %# 2 standard deviations
conf = 2*normcdf(STD)-1; %# covers around 95% of population
scale = chi2inv(conf,2); %# inverse chi-squared with dof=#dimensions
Cov = cov(X0) * scale;
[V D] = eig(Cov);

I'd try the following approach:
Calculate the x-y centroid for the center of the ellipse (x,y in the linked question)
Calculate the linear regression fit line to get the orientation of the ellipse's major axis (angle)
Calculate the standard deviation in the x and y axes
Translate the x-y standard deviations so they're orthogonal to the fit line (a,b)

I'll assume there is only one set of points given in a single matrix, e.g.
B = A(1:10,2:3);
you can reproduce this procedure for each data set.
Compute the center of the ellipsoid, which is the mean of the points. Matlab function: mean
Center your data. Matlab function bsxfun
Compute the principal axis of the ellipsoid and their respective magnitude. Matlab function: eig
The successive steps are illustrated below:
Center = mean(B,1);
Centered_data = bsxfun(#minus,B,Center);
[AX,MAG] = eig(Centered_data' * Centered_data);
The columns of AX contain the vectors describing the principal axis of the ellipsoid while the diagonal of MAG contains information on their magnitude.
To plot the ellipsoid, scale each principal axis with the square root of its magnitude.

Related

How to interpolate using in polar coordinate

I have polar coordinates, radius 0.05 <= r <= 1 and 0 ≤ θ ≤ 2π. The radius r is 50 values between 0.05 to 1, and polar angle θ is 24 values between 0 to 2π.
How do I interpolate r = 0.075 and theta = pi/8?
I dunno what you have tried, but interp2 works just as well on polar data as it does on Cartesian. Here is some evidence:
% Coordinates
r = linspace(0.05, 1, 50);
t = linspace(0, 2*pi, 24);
% Some synthetic data
z = sort(rand(50, 24));
% Values of interest
ri = 0.075;
ti = pi/8;
% Manually interpolate
rp = find(ri <= r, 1, 'first');
rm = find(ri >= r, 1, 'last');
tp = find(ti <= t, 1, 'first');
tm = find(ti >= t, 1, 'last');
drdt = (r(rp) - r(rm)) * (t(tp) - t(tm));
dr = [r(rp)-ri ri-r(rm)];
dt = [t(tp)-ti ti-t(tm)];
fZ = [z(rm, tm) z(rm, tp)
z(rp, tm) z(rp, tp)];
ZI_manual = (dr * fZ * dt.') / drdt
% Interpolate with MATLAB
ZI_MATLAB = interp2(r, t, z', ri, ti, 'linear')
Result:
ZI_manual =
2.737907208525297e-002
ZI_MATLAB =
2.737907208525298e-002
Based on comments you have the following information
%the test point
ri=0.53224;
ti = pi/8;
%formula fo generation of Z
g=9.81
z0=#(r)0.01*(g^2)*((2*pi)^-4)*(r.^-5).*exp(-1.25*(r/0.3).^-4);
D=#(t)(2/pi)*cos(t).^2;
z2=#(r,t)z0(r).*D(t) ;
%range of vlaues of r and theta
r=[0.05,0.071175,0.10132,0.14422,0.2053, 0.29225,0.41602,0.5922,0.84299,1.2];
t=[0,0.62832,1.2566,1.885, 2.5133,3.1416,3.7699,4.3982,5.0265,5.6549,6.2832];
and you want interplation of the test point.
When you sample some data to use them for interpolation you should consider how to sample data according to your requirements.
So when you are sampling a regular grid of polar coordinates ,those coordinates when converted to rectangular will form a circular shape that
most of the points are concentrated in the center of the cricle and when we move from the center to outer regions distance between the points increased.
%regular grid generated for r and t
[THETA R] = meshgrid(t ,r);
% Z for polar grid
Z=z2(R,THETA);
%convert coordinate from polar to cartesian(rectangular):
[X, Y] = pol2cart (THETA, R);
%plot points
plot(X, Y, 'k.');
axis equal
So when you use those point for interpolation the accuracy of the interpolation is greater in the center and lower in the outer regions where the distance between points increased.
In the other word with this sampling method you place more importance on the center region related to outer ones.
To increase accuracy density of grid points (r and theta) should be increased so if length of r and theta is 11 you can create r and theta with size 20 to increase accuracy.
In the other hand if you create a regular grid in rectangular coordinates an equal importance is given to each region . So accuracy of the interpolation will be the same in all regions.
For it first you create a regular grid in the polar coordinates then convert the grid to rectangular coordinates so you can calculate the extents (min max) of the sampling points in the rectangular coordinates. Based on this extents you can create a regular grid in the rectangular coordinates
Regular grid of rectangular coordinates then converted to polar coordinated to get z for grid points using z2 formula.
%get the extent of points
extentX = [min(X(:)) max(X(:))];
extentY = [min(Y(:)) max(Y(:))];
%sample 100 points(or more or less) inside a region specified be the extents
X_samples = linspace(extentX(1),extentX(2),100);
Y_samples = linspace(extentY(1),extentY(2),100);
%create regular grid in rectangular coordinates
[XX YY] = meshgrid(X_samples, Y_samples);
[TT RR] = cart2pol(XX,YY);
Z_rect = z2(RR,TT);
For interpolation of a test point say [ri ti] first it converted to rectangular then using XX ,YY value of z is interpolated
[xi yi] = pol2cart (ti, ri);
z=interp2(XX,YY,Z_rect,xi,yi);
If you have no choice to change how you sample the data and only have a grid of polar points as discussed with #RodyOldenhuis you can do the following:
Interpolate polar coordinates with interp2 (interpolation for gridded data)
this approach is straightforward but has the shortcoming that r and theta are not of the same scale and this may affect the accuracy of the interpolation.
z = interp2(THETA, R, Z, ti, ri)
convert polar coordinates to rectangular and then apply an interpolation method that is for scattered data.
this approach requires more computations but result of it is more reliable.
MATLAB has griddata function that given scattered points first generates a triangulation of points and then creates a regular grid on top of the triangles and interpolates values of grid points.
So if you want to interpolate value of point [ri ti] you should then apply a second interpolation to get value of the point from the interpolated grid.
With the help of some information from spatialanalysisonline and Wikipedia linear interpolation based on triangulation calculated this way (tested in Octave. In newer versions of MATLAB use of triangulation and pointLocation recommended instead of delaunay and tsearch ):
ri=0.53224;
ti = pi/8;
[THETA R] = meshgrid(t ,r);
[X, Y] = pol2cart (THETA, R);
[xi yi] = pol2cart (ti, ri);
%generate triangulation
tri = delaunay (X, Y);
%find the triangle that contains the test point
idx = tsearch (X, Y, tri, xi, yi);
pts= tri(idx,:);
%create a matrix that repesents equation of a plane (triangle) given its 3 points
m=[X(pts);Y(pts);Z(pts);ones(1,3)].';
%calculate z based on det(m)=0;
z= (-xi*det(m(:,2:end)) + yi*det([m(:,1) m(:,3:end)]) + det(m(:,1:end-1)))/det([m(:,1:2) m(:,end)]);
More refinement:
Since it is known that the search point is surrounded by 4 points we can use only those point for triangulation. these points form a trapezoid. Each diagonal of trapezoid forms two triangles so using vertices of the trapezoid we can form 4 triangles, also a point inside a trapezoid can lie in at least 2 triangles.
the previous method based on triangulation only uses information from one triangle but here z of the test point can be interpolated two times from data of two triangles and the calculated z values can be averaged to get a better approximation.
%find 4 points surrounding the test point
ft= find(t<=ti,1,'last');
fr= find(cos(abs(diff(t(ft+(0:1))))/2) .* r < ri,1,'last');
[T4 R4] = meshgrid(t(ft+(0:1)), r(fr+(0:1)));
[X4, Y4] = pol2cart (T4, R4);
Z4 = Z(fr+(0:1),ft+(0:1));
%form 4 triangles
tri2= nchoosek(1:4,3);
%empty vector of z values that will be interpolated from 4 triangles
zv = NaN(4,1);
for h = 1:4
pts = tri2(h,:);
% test if the point lies in the triangle
if ~isnan(tsearch(X4(:),Y4(:),pts,xi,yi))
m=[X4(pts) ;Y4(pts) ;Z4(pts); [1 1 1]].';
zv(h)= (-xi*det(m(:,2:end)) + yi*det([m(:,1) m(:,3:end)]) + det(m(:,1:end-1)))/det([m(:,1:2) m(:,end)]);
end
end
z= mean(zv(~isnan(zv)))
Result:
True z:
(0.0069246)
Linear Interpolation of (Gridded) Polar Coordinates :
(0.0085741)
Linear Interpolation with Triangulation of Rectangular Coordinates:
(0.0073774 or 0.0060992) based on triangulation
Linear Interpolation with Triangulation of Rectangular Coordinates(average):
(0.0067383)
Conclusion:
Result of interpolation related to structure of original data and the sampling method. If the sampling method matches pattern of original data result of interpolation is more accurate, so in cases that grid points of polar coordinates follow pattern of data result of interpolation of regular polar coordinate can be more reliable. But if regular polar coordinates do not match the structure of data or structure of data is such as an irregular terrain, method of interpolation based on triangulation can better represent the data.
please check this example, i used two for loops, inside for loop i used condition statement, if u comment this condition statement and run the program, u'll get correct answer, after u uncomment this condition statement and run the program, u'll get wrong answer. please check it.
% Coordinates
r = linspace(0.05, 1, 10);
t = linspace(0, 2*pi, 8);
% Some synthetic data
%z = sort(rand(50, 24));
z=zeros();
for i=1:10
for j=1:8
if r(i)<0.5||r(i)>1
z(i,j)=0;
else
z(i,j) = r(i).^3'*cos(t(j)/2);
end
end
end
% Values of interest
ri = 0.55;
ti = pi/8;
% Manually interpolate
rp = find(ri <= r, 1, 'first');
rm = find(ri >= r, 1, 'last');
tp = find(ti <= t, 1, 'first');
tm = find(ti >= t, 1, 'last');
drdt = (r(rp) - r(rm)) * (t(tp) - t(tm));
dr = [r(rp)-ri ri-r(rm)];
dt = [t(tp)-ti ti-t(tm)];
fZ = [z(rm, tm) z(rm, tp)
z(rp, tm) z(rp, tp)];
ZI_manual = (dr * fZ * dt.') / drdt
% Interpolate with MATLAB
ZI_MATLAB = interp2(r, t, z', ri, ti, 'linear')
Result:
z1 =
0.1632
ZI_manual =
0.1543
ZI_MATLAB =
0.1582

Plotting a zplot using Matlab

I am trying to plot a zplot in Matlab that displays a unit circle, centered at 0 along with the poles and zeros of the plot. I am not allowed to use any other matlab function such as zplane or pzplot to do this. So far I am able to plot a unit circle just fine but I am having trouble getting my plot to display more of the axis without warping my circle. I also am having a heard time finding the poles and zeros of my function and also how to display the poles as little x's and the zeros as little o's on my plot. Any help would be greatly appreciated! My assignment looks like this and must correctly handle cases such as
zplot([0 1 1], [0 1]);
zplot([0 1 1], [0 0 1]);
function zplot(b, a)
% ZPLOT Plot a zero-pole plot.
-1 -nb
B(z) b(1) + b(2)z + .... + b(nb+1)z
H(z) = ---- = ---------------------------------
-1 -na
A(z) a(1) + a(2)z + .... + a(na+1)z
% zplot(b, a) plots the zeros and poles which determined by vectors b and a
% The plot includes the unit circle and axes for reference, plotted in black.
% Each zero is represented with a blue 'o' and each pole with a red 'x' on the
%plot.
xmin;
xmax;
ymin;
ymax;
% vector of angles at which points are drawn
angle = 0:2*pi/100:2*pi;
% Unit radius
R = 1;
% Coordinates of the circle
x = R*cos(angle);
y = R*sin(angle);
% Plot the circle
plot(x,y);
axis ([xmin, xmax, ymin, ymax]);
grid on;
end
If you can't use pzplot() it is not hard. Here is a hint:
num = [1 4 1];%numerator coefficients of transfer function
den = [1 2 1];%denominator coefficients
z = roots(num)%zeros
p = roots(den)%poles
angle = 0:2*pi/100:2*pi;
xp = cos(angle);
yp = sin(angle);
figure(1)
scatter(z,zeros(length(z),1),'o');
hold on
scatter(p,zeros(length(p),1),'x');
plot(xp,yp);
axis equal
The output
Note that I haven't dealt with imaginary poles/zeros in this example. You'll need to calculate the proper x,y coordinates for a given imaginary pole or zero. (all the poles/zeros in this example are real, not imaginary)
Given a transfer function G, you can use the pzplot() command and add a circle to it.
G = tf([1 4 1],[1 2 1]);
angle = [0:0.1:2*pi+0.1];
xp = cos(angle);
yp = sin(angle);
figure(1)
hold on
pzplot(G);
plot(xp,yp);
axis equal;
This should give you the pole-zero plot with x's for poles, o's for zeros, and the unit circle.
Here is the result.

MatLab--How would I generate an n-sided shape wher n >= 4

I'm new to Matlab but I know a bit about programming.
For class, we have been asked to generate a matrix that gives the vertices of a two dimensional n-sided shape where n>=4. Then, generate the vectors to connect the vertices. We were also given a hint: a vector for each segment can be found by adding the vectors drawn from the origin to each of two adjacent vertices.
I know how to create a matrix using A = [1 1; 1 2; 2 2; 2 1] but I'm not sure how to draw the vectors given this or any other matrix.
The plot() function looks promising, but I'm unsure how to use it with the matrix.
Thank you for any suggestions.
Btw, I'm using Matlab 2011a
I'm not exactly sure how your matrix represents your shape but you might for example let the x-coordinates of the shape be the first column of your array, then let the y-coordinates be the 2nd column, like:
A = [1 1; 1 2; 2 2; 2 1];
x = A(:,1);
y = A(:,2);
fill(x,y,'g');
axis([0 3 0 3]);
axis square;
Which in your case plots a square from the matrix A:
Or construct something a little more complicated like a pentagon:
theta = [0:pi/2.5:2*pi];
x = sin(theta);
y = cos(theta);
% your matrix is then:
B(:,1) = x;
B(:,2) = y;
B
figure;fill(x,y,'g');
axis square;
Which gives:
If you just want to plot the outline with plot (not fill the interior with fill), just remember you have to repeat the initial point at the end so that the polygonal line is closed:
A = [1 1; 1 2; 2 2; 2 1];
B = [A; A(1,:) ]; %// repeat first row at the end
plot(B(:,1),B(:,2))
axis equal %// same scale on both axes
axis([min(x)-.5 max(x)+.5 min(y)-.5 max(y)+.5]) %// larger axes for better display

Matlab irregular shape surf-like plot

I intend to use Matlab to plot the probability distribution from stochastic process on its state space. The state space can be represented by the lower triangle of a 150x150 matrix. Please see the figure (a surf plot without mesh) for a probability distribution at a certain time point.
As we can see, there is a high degree of symmetry in the graph, but because it is plotted as a square matrix, it looks kind of weird. It we could transform the rectangle the plot would look perfect. My question is, how can I use Matlab to plot/transform the lower triangle portion as/to an equal-lateral triangle?
This function should do the job for you. If not, please let me know.
function matrix_lower_tri_to_surf(A)
%Displays lower triangle portion of matrix as an equilateral triangle
%Martin Stålberg, Uppsala University, 2013-07-12
%mast4461 at gmail
siz = size(A);
N = siz(1);
if ~(ndims(A)==2) || ~(N == siz(2))
error('Matrix must be square');
end
zeds = #(N) zeros(N*(N+1)/2,1); %for initializing coordinate vectors
x = zeds(N); %x coordinates
y = zeds(N); %y coordinates
z = zeds(N); %z coordinates, will remain zero
r = zeds(N); %row indices
c = zeds(N); %column indices
l = 0; %base index
xt = 1:N; %temporary x coordinates
yt = 1; %temporary y coordinates
for k = N:-1:1
ind = (1:k)+l; %coordinate indices
l = l+k; %update base index
x(ind) = xt; %save temporary x coordinates
%next temporary x coordinates are the k-1 middle pairwise averages
%calculated by linear interpolation through convolution
xt = conv(xt,[.5,.5]);
xt = xt(2:end-1);
y(ind) = yt; %save temporary y coordinates
yt = yt+1; %update temporary y coordinates
r(ind) = N-k+1; %save row indices
c(ind) = 1:k; % save column indices
end
v = A(sub2ind(size(A),r,c)); %extract values from matrix A
tri = delaunay(x,y); %create triangular mesh
h = trisurf(tri,x,y,z,v,'edgecolor','none','facecolor','interp'); %plot surface
axis vis3d; view(2); %adjust axes projection and proportions
daspect([sqrt(3)*.5,1,1]); %adjust aspect ratio to display equilateral triangle
end %end of function

Set surf minimum for matlab

I have a function which takes a voxel representation of a 3D landscape and can plot a X-Y section to show the middle of the landscape. The voxel representation is stored in a 3 dimensional matrix with a number that represents something important. Obviously the matrix is
1,1,1
2,2,2
in terms of accessing the elements but the actual 3D locations are found in the following method:
(index-1)*resolution+0.5*resolution+minPos;
where resolution is the grid size :
resolution
<-->
__ __ __
|__|__|__|
<- Min pos
and minPos is where the grid starts.
Now in terms of the actual question, i would like to extract a single X-Y section of this voxel representation and display it as a surf. This can be done by just doing this:
surf(voxel(:, :, section))
however then you get this:
The obvious problem is that the grid will start at 0 because that is how the matrix representation is. How can i set the minimum and cell size for surf, ie so that the grid will start at the minimum (shown above) and will have the grid spacing of resolution (shown above).
Read the documentation of surf, you can also provide x and y coordinates corresponding to your data points.
surf(X,Y,Z)
X and Y can be either vectors or matrices:
surf(X,Y,Z) uses Z for the color data and surface height. X and Y are vectors or matrices defining the x and y components of a surface. If X and Y are vectors, length(X) = n and length(Y) = m, where [m,n] = size(Z). In this case, the vertices of the surface faces are (X(j), Y(i), Z(i,j)) triples. To create X and Y matrices for arbitrary domains, use the meshgrid function
Example
Z=[ 0 1 2 3;
7 6 5 4;
8 9 10 11];
x=[-1 0 1 2];
y=[-2 0 2];
surf(x,y,Z);
Of course you have to match Z, x and y matrices/vectors as clearly described in the doc^^
Just remember that elements in columns of Z are surf'ed as values along the y-axis, elements in rows of Z are surf'ed as values along the x-axis. This is clearly to be seen in the example picture.
Solution
I think you switched the x and y-axis around, which you can fix by just transposing z:
s = size(voxel);
xi = (minPosX:resolution:(minPosX+resolution*s(1)-1));
yi = (minPosY:resolution:(minPosY+resolution*s(2)-1));
z = (voxel(:,:,section));
surf(xi, yi, z');
or that you're picking the wrong numbers for constructing xi and yi and it should be this instead:
xi = (minPosX:resolution:(minPosX+resolution*s(2)-1));
yi = (minPosY:resolution:(minPosY+resolution*s(1)-1));
z = (voxel(:,:,section));
surf(xi, yi, z);
So it was easy enough to do:
lets say we have a 3D matrix "voxel";
s = size(voxel);
xi = (minPosX:resolution:(minPosX+resolution*s(1)-1));
yi = (minPosY:resolution:(minPosY+resolution*s(2)-1));
z = (voxel(:,:,section));
[x y] = meshgrid(xi, yi);
x = x';
y = y';
surf(x, y, z);
Provides the following plot:
This is rotated which is annoying, I cant seem to get it to rotate back (I could just visualise around the other way but that's ok)