Getting right to the question: How can I make my plot go from looking like this:
to something like this:
where a right-handed Cartesian coordinate system (with axis labels at the end of the arrows, e.g., in this example the x-axis label is = $x_1^G$)
Some elaboration and prefacing: First, I am new to matlab and "matrix math". I tried searching on this site for questions similar to what I have posed above but did not see anything (but I may have missed), so hopefully this is not a duplicate.
Here is the code I have used to develop my rotated ellipsoid (since I'm new to matlab I would appreciate any comments on my code):
%Insert the components of the 3x3 matrix (i.e. the scalar values of the 2nd-rank symmetric tensor)
Pmatrix = [115.9547 12.03765 4.68235; 12.03765 116.3702 -2.47985; 4.68235 -2.47985 134.5488];
[R,D]=eig(Pmatrix); %find the eigenvectors (R) and eigenvalues (D) of the 2nd-rank tensor
[x, y, z] =sphere; %generate coordinates of a sphere, using the sphere command
%Stretch the coordinates of the sphere to form the tensor's representation ellipsoid, do this by multiplying the x, y, z coordinates by the square-roots of the eigenvalues
x1 = x*sqrt(D(1,1));
y1 = y*sqrt(D(2,2));
z1 = z*sqrt(D(3,3));
figure;
hmesh = mesh(x1,y1,z1);
set(hmesh,'FaceColor',[0.5,0.5,0.5],'FaceAlpha',0.5) %set color to gray, make mostly transparent
axis equal; %Make tick mark increments on all axes equal
xlabel('x');
ylabel('y');
zlabel('z');
theta1 = -asind(R(3,1)); %rotation around y-axis in degrees
psi1 = atan2d(R(3,2)/cos(theta1),R(3,3)/cos(theta1)); %rotation around x-axis in degrees
phi1 = atan2d(R(2,1)/cos(theta1),R(1,1)/cos(theta1)); %rotation around z-axis in degress
direction = [1 0 0]; %rotate the surface plot psi1 degrees around its x-axis
rotate(hmesh,direction,psi1)
direction = [0 1 0]; %rotate the surface plot theta1 degrees around its y-axis
rotate(hmesh,direction,theta1)
direction = [0 0 1]; %rotate the surface plot phi1 degrees around its z-axis
rotate(hmesh,direction,phi1)
view([-36 18]); %Change the camera viewpoint
To add to my question, I would like to add other (similar) items to my plot, so that the final product would look like this:
In the development of the image above, a set of axes collinear with the ellipsoid's eigenvectors are added (the red, green, blue arrows):
Then these axes are extended in the opposite direction, through the origin, where these portions of the axes are shown as a dashed lines:
Then, the angles between the global coordinate axes (black arrows) and the ellipsoid's axes (colored arrows) are notated as shown here:
To comment on this last addition, someone has built some matlab code with these features (see youtube video here). In the video's description it says the matlab code can be found here. Being new to matlab, I don't see where the code is, e.g., I don't see where on this page (screenshot below) the code is to build that matlab plot seen in the youtube video. If you can guide me on how to navigate that mathworks-fileexchage page that would be appreciated.
Related
I have two datasets. One detailing a list of angles (which I am plotting onto a rose plot):
angles
-0.8481065519
0.0367932161
2.6273740453
...
n
The other, detailing directional statistics from this group of angles:
angle,error
-0.848106563,0.8452778824
Where angle essentially defines the directional mean, and error the circular variance, essentially an error bar either side of the angle
I have thus far plotted a rose histogram using the set of angles, as such:
h = rose(angles,36)
I would like to create a plot of the directional statistic angle (it does not need a length/magnitude - just to the edge of the circle plot) with the error around it.
As an example:
I added the lines by hand in Matlab. If possible it would be good to perhaps have shading within the arc too. Alternatively, (and possibly preferred) would be to have just a sliver above the rose plot bins (so it doesn't cover the data) with a centre line (showing the angle and shading surrounding for the error.
Thanks in advance.
How about this?
%// Data
angles = 2*pi*.8*randn(1,1e4);
angle = -0.848106563;
error = 0.8452778824;
%// Plot rose
rose(angles, 36);
axis image %// make axis square
hold on
%// Plot mean
a = axis;
a = a(2); %// size of axis
plot([0 cos(angle)*a], [0 sin(angle)*a], 'r')
%// Plot error as many shaded triangles that compose a circular wedge
t = linspace(-error/2+angle,error/2+angle,100); %// increase "100" if needed
for k = 1:numel(t)-1
h = patch([0 cos(t(k))*a cos(t(k+1))*a 0], ...
[0 sin(t(k))*a sin(t(k+1))*a 0], [.5 0 0], 'edgecolor', 'none');
%// change color [.5 0 0] to something else if desired. Note also alpha
set(h,'Facealpha',.3) %// make transparent
end
%// Place rose on top by rearranging order of axis children
ch = get(gca,'children');
set(gca,'children',[ch(2:end); ch(1)]);
For this to work, you need to use a figure renderer capable of transparency. So you may need to adjust the figure's renderer property.
I have trajectory information in 3 dimensions in matlab. These are of a gesture someone is making. When I connect the points in matlab by using plot3, I can see the trajectory nicely.
However, the trajectory is a line in the plot, but I don't know in which direction the gesture has been made as the time is not visualized. Is it possible to visualize this in a 3d plot (where the dimensions are x, y and z)? For example, the colour at the start is bright red and the colour at the end is black.
Thanks for your help,
Héctor
You need the comet3 plot (if you don't mind animations).
If you do mind animations, and you're looking for a static figure, I'd use a quiver.
Example:
% value of the parameter in the parametric equation
t = 0:0.5:2*pi;
% modified coordinate axes
u = [1 0 0].';
v = [0 2 0].';
% coordinates of the ellipse
Ell = bsxfun(#plus, bsxfun(#times, u, cos(t)), bsxfun(#times, v, sin(t)));
% difference vectors between all data points will be used as "velocities"
dEll = diff(Ell, 1,2);
% Quiver the ellipse
quiver3(...
Ell(1,1:end-1), Ell(2,1:end-1), Ell(3,1:end-1), ...
dEll(1,:), dEll(2,:), dEll(3,:), ...
2, 'r') % = scale, LineSpec
axis tight equal
Result:
i just started with my master thesis and i already am in trouble with my capability/understanding of matlab.
The thing is, i have a trajectory on a surface of a planet/moon whatever (a .mat with the time, and the coordinates. Then i have some .mat with time and the measurement at that time.
I am able to plot this as a color coded trajectory (using the measurement and the coordinates) in scatter(). This works awesomely nice.
However my problem is that i need something more sophisticated.
I now need to take the trajectory and instead of color-coding it, i am supposed to add the graph (value) of the measurement (which is given for each point) to the trajectory (which is not always a straight line). I will added a little sketch to explain what i want. The red arrow shows what i want to add to my plot and the green shows what i have.
You can always transform your data yourself: (using the same notation as #Shai)
x = 0:0.1:10;
y = x;
m = 10*sin(x);
So what you need is the vector normal to the curve at each datapoint:
dx = diff(x); % backward finite differences for 2:end points
dx = [dx(1) dx]; % forward finite difference for 1th point
dy = diff(y);
dy = [dy(1) dy];
curve_tang = [dx ; dy];
% rotate tangential vectors 90° counterclockwise
curve_norm = [-dy; dx];
% normalize the vectors:
nrm_cn = sqrt(sum(abs(curve_norm).^2,1));
curve_norm = curve_norm ./ repmat(sqrt(sum(abs(curve_norm).^2,1)),2,1);
Multiply that vector with the measurement (m), offset it with the datapoint coordinates and you're done:
mx = x + curve_norm(1,:).*m;
my = y + curve_norm(2,:).*m;
plot it with:
figure; hold on
axis equal;
scatter(x,y,[],m);
plot(mx,my)
which is imo exactly what you want. This example has just a straight line as coordinates, but this code can handle any curve just fine:
x=0:0.1:10;y=x.^2;m=sin(x);
t=0:pi/50:2*pi;x=5*cos(t);y=5*sin(t);m=sin(5*t);
If I understand your question correctly, what you need is to rotate your actual data around an origin point at a certain angle. This is pretty simple, as you only need to multiply the coordinates by a rotation matrix. You can then use hold on and plot to overlay your plot with the rotated points, as suggested in the comments.
Example
First, let's generate some data that resembles yours and create a scatter plot:
% # Generate some data
t = -20:0.1:20;
idx = (t ~= 0);
y = ones(size(t));
y(idx) = abs(sin(t(idx)) ./ t(idx)) .^ 0.25;
% # Create a scatter plot
x = 1:numel(y);
figure
scatter(x, x, 10, y, 'filled')
Now let's rotate the points (specified by the values of x and y) around (0, 0) at a 45° angle:
P = [x(:) * sqrt(2), y(:) * 100] * [1, 1; -1, 1] / sqrt(2);
and then plot them on top of the scatter plot:
hold on
axis square
plot(P(:, 1), P(:, 2))
Note the additional things have been done here for visualization purposes:
The final x-coordinates have been stretched (by sqrt(2)) to the appropriate length.
The final y-coordinates have been magnified (by 100) so that the rotated plot stands out.
The axes have been squared to avoid distortion.
This is what you should get:
It seems like you are interested in 3D plotting.
If I understand your question correctly, you have a 2D curve represented as [x(t), y(t)].
Additionally, you have some value m(t) for each point.
Thus we are looking at the plot of a 3D curve [x(t) y(t) m(t)].
you can easily achieve this using
plot3( x, y, m ); % assuming x,y, and m are sorted w.r.t t
alternatively, you can use the 3D version of scatter
scatter3( x, y, m );
pick your choice.
Nice plot BTW.
Good luck with your thesis.
With the following code I'm able to draw the plot of a single 2D-Gaussian function:
x=linspace(-3,3,1000);
y=x';
[X,Y]=meshgrid(x,y);
z=exp(-(X.^2+Y.^2)/2);
surf(x,y,z);shading interp
This is the produced plot:
However, I'd like to plot a grid having a specified number x of these 2D-Gaussians.
Think of the following picture as an above view of the plot I'd like to produce (where in particular the grid is made of 5x5 2D-Gaussians). Each Gaussian should be weighed by a coefficient such that if it's negative the Gaussian is pointing towards negative values of the z axis (black points in the grid below) and if it's positive it's as in the above image (white points in the grid below).
Let me provide some mathematical details. The grid corresponds to a mixture of 2D-Gaussians summed as in the following equation:
In which each Gaussian has its own mean and deviation.
Note that each Gaussian of the mixture should be put in a determined (X,Y) coordinate, in such a way that they are equally distant from each other. e.g think of the central Gaussian in (0,0) then the other ones should be in (-1,1) (0,1) (1,1) (-1,0) (1,0) (-1,-1) (0,-1) (1,-1) in the case of a grid with dimension 3x3.
Can you provide me (and explain to me) how can I do such a plot?
Thanks in advance for the help.
Indeed you said yourself, put (as an example just for the means)
[X,Y]=meshgrid(x,y); % //mesh
g_centers = -3:3;
[x_g,y_g] = meshgrid(g_centers,g_centers); % //grid of centers (coarser)
mu = [x_g(:) , y_g(:)]; % // mesh of centers in column
z = zeros(size(X));
for i = 1:size(mu,1)
z= z + exp(-((X-mu(i,1)).^2+(Y-mu(i,2)).^2)/( 2* .001) );
end
surf(X,Y,z);shading interp
I am working in MATLAB and I'm stuck on a very simple problem: I've got an object defined by its position (x,y) and theta (an angle, in degrees). I would like to plot the point and add an arrow, starting from the point and pointing toward the direction defined by the angle. It actually doesn't even have to be an arrow, anything graphically showing the value of the angle will do!
Here's a picture showing the kind of thing I'm trying to draw:
removed dead ImageShack link
The quiver() plotting function plots arrows like this. Take your theta value and convert it to (x,y) cartesian coordinates representing the vector you want to plot as an arrow and use those as the (u,v) parameters to quiver().
theta = pi/9;
r = 3; % magnitude (length) of arrow to plot
x = 4; y = 5;
u = r * cos(theta); % convert polar (theta,r) to cartesian
v = r * sin(theta);
h = quiver(x,y,u,v);
set(gca, 'XLim', [1 10], 'YLim', [1 10]);
Take a look through online the Matlab documentation to see other plot types; there's a lot, including several radial plots. They're in the MATLAB > Functions > Graphics > Specialized Plotting section. Do "doc quiver" at the command line and browse around.
If you want to try and make something that looks like the image you linked to, here's some code to help you do it (NOTE: you would first have to download the submission arrow.m by Erik Johnson on the MathWorks File Exchange, which I always like to use for generating arrows of any shape and size):
x = 1; % X coordinate of arrow start
y = 2; % Y coordinate of arrow start
theta = pi/4; % Angle of arrow, from x-axis
L = 2; % Length of arrow
xEnd = x+L*cos(theta); % X coordinate of arrow end
yEnd = y+L*sin(theta); % Y coordinate of arrow end
points = linspace(0, theta); % 100 points from 0 to theta
xCurve = x+(L/2).*cos(points); % X coordinates of curve
yCurve = y+(L/2).*sin(points); % Y coordinates of curve
plot(x+[-L L], [y y], '--k'); % Plot dashed line
hold on; % Add subsequent plots to the current axes
axis([x+[-L L] y+[-L L]]); % Set axis limits
axis equal; % Make tick increments of each axis equal
arrow([x y], [xEnd yEnd]); % Plot arrow
plot(xCurve, yCurve, '-k'); % Plot curve
plot(x, y, 'o', 'MarkerEdgeColor', 'k', 'MarkerFaceColor', 'w'); % Plot point
And here's what it would look like:
You can then add text to the plot (for the angle and the coordinate values) using the text function.
Here's a partial answer, I expect you can figure out the rest. I fired up the Figures editor and opened the plot tools. I dragged an arrow from the palette onto my figure. Then I generated an m-file. This included the line:
annotation(figure1,'arrow',[0.1489 0.2945],[0.5793 0.6481]);
So, the first pair of coordinates is the start of the arrow. You're going to have to figure out the pointy end (second pair of coordinates) using a little bit of trigonometry. You might even be able to get the little arc if you do some more fiddling around with plot tools.
Let us know if the trig defeats you. Oh, and I forgot to plot the point, but I guess you can figure that out ?