I want to show, with Matlab, a temperature distribution on an object surface.
I've got a 3D data in the form of (x, y, z, V) vectors. I would like to show this object in Matlab, with the colour representing the local total "value".
I can export the object as an STL file. It can be shown easily using the STL plotting (see stldemo):
fv = stlread('file.stl');
patch(fv, 'EdgeColor', 'none', 'FaceLighting', 'gouraud', 'AmbientStrength', 0.15, 'FaceColor', [0.8 0.8 1.0]);
camlight('headlight');
material('dull');
To colour it according to (x,y,z,V), I need to attach each (x, y, z) point to a vertex in the patch (the nearest one would work). If there are many (x,y,z) points for which a single STL vertex is the nearest, I add up the corresponding V values for that vertex.
The number of vertices is thousands. The number of (x, y, z) points is also large. So doing a loop through (x, y, z) points and then an internal loop over vertices to find the nearest one (which involves calculating distances between points) is out of question. Is there any smart way to do it quickly?
Note: I cannot control the location of the data points, they are defined by an external program. The STL points are controlled by another external program. So I have to marry two different point sets.
Here is the code illustrating what I want to achieve, with 4 vertices and 3 data points:
% Create patch
figure;
p = patch;
colorbar
p.Vertices = [...
0, 0, 0; ...
1, 0, 0; ...
1, 1, 0;
0, 1, 0];
p.Faces = [ ...
1, 2, 3; ...
1, 3, 4];
% Data points
x = [0.1, 0.1, 0.25];
y = [0.01, 0.02, 0.75];
z = [0.01, 0.2, -0.01];
v = [1, 1, 1];
p.FaceVertexCData = zeros(size(p.Vertices, 1), 1);
% Point 1 (0.1, 0.01, 0.01) is closest to vertex 1 (0, 0, 0). Its value
% goes to vertex 1.
p.FaceVertexCData(1) = p.FaceVertexCData(1) + v(1);
% Point 2 (0.1, 0.02, 0.2) is also closest to vertex 1 (0, 0, 0). Its
% value also goes to vertex 1
p.FaceVertexCData(1) = p.FaceVertexCData(1) + v(2);
% Point 3 (0.25, 0.75, -0.01) is closest to vertex 4 (0, 1, 0). Its power
% goes to vertex 4.
p.FaceVertexCData(4) = p.FaceVertexCData(4) + v(3);
% Other vertices are left with 0.
p.FaceColor = 'interp';
Attaching a volume scalar value (of Temperature in your case) of a point to a neighbouring point is a tricky exercise, requires complex for loops and defining special case rules (in your case you wanted to attach the value of 2 different points to the same patch vertex, what if the 2 values to attach are different? Do you average? discard ?).
A safer approach is to re-interpolate your temperature field over your object surface. the function griddata can do that for you.
First I had to define a scalar field. Since I do not have your temperature data, I use the flow function from Matlab. I generated a scalar field the same ways than in this article: flow data.
This gave me a scalar field v (flow value but let's say it's your temperature) at for every coordinates x, y, z.
Then I created and introduced a 3D patch which will be your object. I chose a sphere but any 3D patch will work the same way.
The code to get the sphere as a patch is borrowed from surf2patch
you will have to offset and inflate the sphere to get it exactly as in the figure below
Now is the interesting bit. In the following code, v is the value of the scalar field (temperature for you) at the coordinates x, y, z.
%% // Extract patch vertices coordinates in separate variables
xp = fv.vertices(:,1) ;
yp = fv.vertices(:,2) ;
zp = fv.vertices(:,3) ;
%% // interpolate the temperature field over the patch coordinates
Tpv = griddata(x,y,z,v,xp,yp,zp) ;
%% // Set the patch color data to the new interpolated temperature
set(hp,'FaceVertexCData',Tpv) ;
And your object surface is now at the right interpolated temperature:
you can delete the slice plane if you want to observe the patch alone
Related
I tried to add relative pose measurements of 2 adjacent nodes using factorGraph's object function factorTwoPoseSE3, but no matter how much I should modify the relative angle “relT” variable(i.e. the number of quaternions), the optimized result graph remains the same, which is apparently a big problem?
If this is not the case, how do I modify the code to achieve the correct optimisation result that I want?
Design a plane rectangle, its observed 4 absolute coordinate points (0,0), (1,0), (1,1), (0,1), (0,1), node (node) number 1,2,3,4, respectively, in the last node 4 observed node 1 relative coordinates for (0.8,0), that is, closed loop , you can see that the last node error is 0.2. that is, this pose graph contains 4 edges, The first 3 of these are odometers and the last one is a closed loop edge.
% Since factorTwoPoseSE2 cannot specify a starting point for the time being, se3 is used for this purpose.
abspos = [0,0,0,eul2quat([0,0,0],"ZYX"); % rotate 0 about Z axis
1,0,0,eul2quat([pi/2,0,0],"ZYX"); % rotate 90*pi/180 about Z axis
1,1,0,eul2quat([pi,0,0],"ZYX"); % rotate 180*pi/180 about Z axis
0,1,0,eul2quat([3*pi/2,0,0],"ZYX")]; % rotate 270*pi/180 about Z axis
relT = pi/2; % no matter how much I should modify this value, the result does't change???
relPose = [1,0,0,eul2quat([relT,0,0]);
1,0,0,eul2quat([relT,0,0]);
1,0,0,eul2quat([relT,0,0]);
0.8,0,0,eul2quat([relT,0,0])];
fg = factorGraph();
pf = factorPoseSE3Prior(1,Measurement=[0,0,0, 1,0,0,0]);
addFactor(fg,pf);
nodeID = [1 2];
f = factorTwoPoseSE3(nodeID,Measurement=relPose(1,:));% odometry
addFactor(fg,f);
nodeID = [2 3];
f = factorTwoPoseSE3(nodeID,Measurement=relPose(2,:));% odometry
addFactor(fg,f);
nodeID = [3 4];
f = factorTwoPoseSE3(nodeID,Measurement=relPose(3,:));% odometry
addFactor(fg,f);
nodeID = [4 1];
f = factorTwoPoseSE3(nodeID,Measurement=relPose(end,:));% loop closure
addFactor(fg,f);
% optimize
optns = factorGraphSolverOptions();
optimize(fg,optns);
newNodes = [fg.nodeState(1);
fg.nodeState(2);
fg.nodeState(3);
fg.nodeState(4)]
figure;
plot(newNodes(:,1),newNodes(:,2),'b-',Marker='.')
hold on;grid on;
plot(newNodes([4,1],1),newNodes([4,1],2),'r-',Marker='.')
text(newNodes(:,1),newNodes(:,2),string(1:4))
title('after factorGraph(3D) optimize')
Additionly, according to the factorTwoPoseSE3 official documentation "Measured relative pose, specified as a seven-element row vector of the form [dx dy dz dqw dqx dqy dqz]. dx, dy, and dz are the change in position in x, y, and z respectively. dqw, dqx, dqy, and dqz are the change in quaternion rotation in w, x, y, and z respectively."
So, I have made the following changes to the relpos and it seems that the optimisation graphs are not correct, which is very confusing!
relPose = diff(abspos);
relPose = [relPose;
abspos(1,:)-abspos(end,:)];
run in MATLAB R2022b.
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;
(Disclaimer: I thought about posting this on math.statsexchange, but found similar questions there that were moved to SO, so here I am)
The context:
I'm using fft/ifft to determine probability distributions for sums of random variables.
So e.g. I'm having two uniform probability distributions - in the simplest case two uniform distributions on the interval [0,1].
So to get the probability distribution for the sum of two random variables sampled from these two distributions, one can calculate the product of the fourier-transformed of each probabilty density.
Doing the inverse fft on this product, you get back the probability density for the sum.
An example:
function usumdist_example()
x = linspace(-1, 2, 1e5);
dx = diff(x(1:2));
NFFT = 2^nextpow2(numel(x));
% take two uniform distributions on [0,0.5]
intervals = [0, 0.5;
0, 0.5];
figure();
hold all;
for i=1:size(intervals,1)
% construct the prob. dens. function
P_x = x >= intervals(i,1) & x <= intervals(i,2);
plot(x, P_x);
% for each pdf, get the characteristic function fft(pdf,NFFT)
% and form the product of all char. functions in Y
if i==1
Y = fft(P_x,NFFT) / NFFT;
else
Y = Y .* fft(P_x,NFFT) / NFFT;
end
end
y = ifft(Y, NFFT);
x_plot = x(1) + (0:dx:(NFFT-1)*dx);
plot(x_plot, y / max(y), '.');
end
My issue is, the shape of the resulting prob. dens. function is perfect.
However, the x-axis does not fit to the x I create in the beginning, but is shifted.
In the example, the peak is at 1.5, while it should be 0.5.
The shift changes if I e.g. add a third random variable or if I modify the range of x.
But I can't get figure how.
I'm afraid it might have to do with the fact that I'm having negative x values, while fourier transforms usually work in a time/frequency domain, where frequencies < 0 don't make sense.
I'm aware I could find e.g. the peak and shift it to its proper place, but seems nasty and error prone...
Glad about any ideas!
The problem is that your x origin is -1, not 0. You expect the center of the triangular pdf to be at .5, because that's twice the value of the center of the uniform pdf. However, the correct reasoning is: the center of the uniform pdf is 1.25 above your minimum x, and you get the center of the triangle at 2*1.25 = 2.5 above the minimum x (that is, at 1.5).
In other words: although your original x axis is (-1, 2), the convolution (or the FFT) behave as if it were (0, 3). In fact, the FFT knows nothing about your x axis; it only uses the y samples. Since your uniform is zero for the first samples, that zero interval of width 1 is amplified to twice its width when you do the convolution (or the FFT). I suggest drawing the convolution on paper to see this (draw original signal, reflected signal about y axis, displace the latter and see when both begin to overlap). So you need a correction in the x_plot line to compensate for this increased width of the zero interval: use
x_plot = 2*x(1) + (0:dx:(NFFT-1)*dx);
and then plot(x_plot, y / max(y), '.') will give the correct graph:
I have a matrix with x and y coordinates as well as the temperature values for each of my data points. When I plot this in a scatter plot, some of the data points will obscure others and therefore, the plot will not give a true representation of how the temperature varies in my data set.
To fix this, I would like to decrease the resolution of my graph and create pixels which represent the average temperature for all data points within the area of the pixel. Another way to think about the problem that I need to put a grid over the current plot and average the values within each segment of the grid.
I have found this thread - Generate a heatmap in MatPlotLib using a scatter data set - which shows how to use python to achieve the end result that I want. However, my current code is in MATLAB and even though I have tried different suggestions such as heatmap, contourf and imagesc, I can't get the result I want.
You can "reduce the resolution" of your data using accumarray, where you specify which output "bin" each point should go in and specify that you wish to take a mean over all points in that bin.
Some example data:
% make points that overlap a lot
n = 10000
% NOTE: your points do not need to be sorted.
% I only sorted so we can visually see if the code worked,
% see the below plot
Xs = sort(rand(n, 1));
Ys = rand(n, 1);
temps = sort(rand(n, 1));
% plot
colormap("hot")
scatter(Xs, Ys, 8, temps)
(I only sorted by Xs and temps in order to get the stripy pattern above so that we can visually verify if the "reduced resolution" worked)
Now, suppose I want to decrease the resolution of my data by getting just one point per 0.05 units in the X and Y direction, being the average of all points in that square (so since my X and Y go from 0 to 1, I'll get 20*20 points total).
% group into bins of 0.05
binsize = 0.05;
% create the bins
xbins = 0:binsize:1;
ybins = 0:binsize:1;
I use histc to work out which bin each X and Y is in (note - in this case since the bins are regular I could also do idxx = floor((Xs - xbins(1))/binsize) + 1)
% work out which bin each X and Y is in (idxx, idxy)
[nx, idxx] = histc(Xs, xbins);
[ny, idxy] = histc(Ys, ybins);
Then I use accumarray to do a mean of temps within each bin:
% calculate mean in each direction
out = accumarray([idxy idxx], temps', [], #mean);
(Note - this means that the point in temps(i) belongs to the "pixel" (of our output matrix) at row idxy(1) column idxx(1). I did [idxy idxx] as opposed to [idxx idxy] so that the resulting matrix has Y == rows and X == columns))
You can plot like this:
% PLOT
imagesc(xbins, ybins, out)
set(gca, 'YDir', 'normal') % flip Y axis back to normal
Or as a scatter plot like this (I plot each point in the midpoint of the 'pixel', and drew the original data points on too for comparison):
xx = xbins(1:(end - 1)) + binsize/2;
yy = ybins(1:(end - 1)) + binsize/2;
[xx, yy] = meshgrid(xx, yy);
scatter(Xs, Ys, 2, temps);
hold on;
scatter(xx(:), yy(:), 20, out(:));
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.