I am trying to understand the TriScatteredInterp in Matlab.
I followed sample program in the help file.
x = rand(100,1)*4-2;
y = rand(100,1)*4-2;
z = x.*exp(-x.^2-y.^2);
Construct the interpolant:
F = TriScatteredInterp(x,y,z);
What I observe is F.X is same as x and y and F.V is same as z;
ti = -2:.25:2;
[qx,qy] = meshgrid(ti,ti);
qz = F(qx,qy);
I consider linear interpolation is done in the qz = F(qx,qy);. How does it do for the linear interpolation?
Thanks
Now I understand how TriScatteredInterp works in Matlab.
We have x,y,z points for N X 3 dimensions.
All these points, we need to implement Delaunay triangles in C++.
That is easy. Then, for all your desired grid points x', y', please search the triangle in which your x',y' is located. Then do Barycentric interpolation in a triangle as shown in the link. You will get z' for these x',y'. That is all what we need to do in C++ for TriScatteredInterp. Good Luck!
Related
I want to plot a surface in MATLAB using surf. I have this equation: x = y^2 +4z^2.
What I am doing is the following:
[x,y] = meshgrid(-4:.1:4, -4:.1:4);
z = sqrt((x - y.^2)./4); % Basically I'm just clearing for z
surf(x,y,z)
But with this I am getting the error: Error using surf X,Y,Z and C cannot be complex. I know there is a complex number because of the values that x and y have, plus the square root. Is there another way to plot a surface in MATLAB? because I really don't know what to do, and my skills are very basics.
Why do you feel that you need to grid x and y, and not use the form of the original equation itself?
This seems to work perfectly fine
[y,z] = meshgrid(-4:.1:4, -4:.1:4);
x = y.^2 + 4*z.^2;
surf(x,y,z)
to produce
MATLAB's surf command allows you to pass it optional X and Y data that specify non-cartesian x-y components. (they essentially change the basis vectors). I desire to pass similar arguments to a function that will draw a line.
How do I plot a line using a non-cartesian coordinate system?
My apologies if my terminology is a little off. This still might technically be a cartesian space but it wouldn't be square in the sense that one unit in the x-direction is orthogonal to one unit in the y-direction. If you can correct my terminology, I would really appreciate it!
EDIT:
Below better demonstrates what I mean:
The commands:
datA=1:10;
datB=1:10;
X=cosd(8*datA)'*datB;
Y=datA'*log10(datB*3);
Z=ones(size(datA'))*cosd(datB);
XX=X./(1+Z);
YY=Y./(1+Z);
surf(XX,YY,eye(10)); view([0 0 1])
produces the following graph:
Here, the X and Y dimensions are not orthogonal nor equi-spaced. One unit in x could correspond to 5 cm in the x direction but the next one unit in x could correspond to 2 cm in the x direction + 1 cm in the y direction. I desire to replicate this functionality but drawing a line instead of a surf For instance, I'm looking for a function where:
straightLine=[(1:10)' (1:10)'];
my_line(XX,YY,straightLine(:,1),straightLine(:,2))
would produce a line that traced the red squares on the surf graph.
I'm still not certain of what your input data are about, and what you want to plot. However, from how you want to plot it, I can help.
When you call
surf(XX,YY,eye(10)); view([0 0 1]);
and want to get only the "red parts", i.e. the maxima of the function, you are essentially selecting a subset of the XX, YY matrices using the diagonal matrix as indicator. So you could select those points manually, and use plot to plot them as a line:
Xplot = diag(XX);
Yplot = diag(YY);
plot(Xplot,Yplot,'r.-');
The call to diag(XX) will take the diagonal elements of the matrix XX, which is exactly where you'll get the red patches when you use surf with the z data according to eye().
Result:
Also, if you're just trying to do what your example states, then there's no need to use matrices just to take out the diagonal eventually. Here's the same result, using elementwise operations on your input vectors:
datA = 1:10;
datB = 1:10;
X2 = cosd(8*datA).*datB;
Y2 = datA.*log10(datB*3);
Z2 = cosd(datB);
XX2 = X2./(1+Z2);
YY2 = Y2./(1+Z2);
plot(Xplot,Yplot,'rs-',XX2,YY2,'bo--','linewidth',2,'markersize',10);
legend('original','vector')
Result:
Matlab has many built-in function to assist you.
In 2D the easiest way to do this is polar that allows you to make a graph using theta and rho vectors:
theta = linspace(0,2*pi,100);
r = sin(2*theta);
figure(1)
polar(theta, r), grid on
So, you would get this.
There also is pol2cart function that would convert your data into x and y format:
[x,y] = pol2cart(theta,r);
figure(2)
plot(x, y), grid on
This would look slightly different
Then, if we extend this to 3D, you are only left with plot3. So, If you have data like:
theta = linspace(0,10*pi,500);
r = ones(size(theta));
z = linspace(-10,10,500);
you need to use pol2cart with 3 arguments to produce this:
[x,y,z] = pol2cart(theta,r,z);
figure(3)
plot3(x,y,z),grid on
Finally, if you have spherical data, you have sph2cart:
theta = linspace(0,2*pi,100);
phi = linspace(-pi/2,pi/2,100);
rho = sin(2*theta - phi);
[x,y,z] = sph2cart(theta, phi, rho);
figure(4)
plot3(x,y,z),grid on
view([-150 70])
That would look this way
It seems to be very basic question, but I wonder when I plot x values against y values, what interpolation technique is used behind the scene to show me the discrete data as continuous? Consider the following example:
x = 0:pi/100:2*pi;
y = sin(x);
plot(x,y)
My guess is it is a Lagrangian interpolation?
No, it's just a linear interpolation. Your example uses a quite long dataset, so you can't tell the difference. Try plotting a short dataset and you'll see it.
MATLAB's plot performs simple linear interpolation. For finer resolution you'd have to supply more sample points or interpolate between the given x values.
For example taking the sinus from the answer of FamousBlueRaincoat, one can just create an x vector with more equidistant values. Note, that the linear interpolated values coincide with the original plot lines, as the original does use linear interpolation as well. Note also, that the x_ip vector does not include (all) of the original points. This is why the do not coincide at point (~0.8, ~0.7).
Code
x = 0:pi/4:2*pi;
y = sin(x);
x_ip = linspace(x(1),x(end),5*numel(x));
y_lin = interp1(x,y,x_ip,'linear');
y_pch = interp1(x,y,x_ip,'pchip');
y_v5c = interp1(x,y,x_ip,'v5cubic');
y_spl = interp1(x,y,x_ip,'spline');
plot(x,y,x_ip,y_lin,x_ip,y_pch,x_ip,y_v5c,x_ip,y_spl,'LineWidth',1.2)
set(gca,'xlim',[pi/5 pi/2],'ylim',[0.5 1],'FontSize',16)
hLeg = legend(...
'No Interpolation','Linear Interpolation',...
'PChip Interpolation','v5cubic Interpolation',...
'Spline Interpolation');
set(hLeg,'Location','south','Fontsize',16);
By the way..this does also apply to mesh and others
[X,Y] = meshgrid(-8:2:8);
R = sqrt(X.^2 + Y.^2) + eps;
Z = sin(R)./R;
figure
mesh(Z)
No, Lagrangian interpolation with 200 equally spaced points would be an incredibly bad idea. (See: Runge's phenomenon).
The plot command simply connects the given (x,y) points by straight lines, in the order given. To see this for yourself, use fewer points:
x = 0:pi/4:2*pi;
y = sin(x);
plot(x,y)
I'm not very familiar with matlab, but I'm trying to plot the intersection of a plane (x + y + z = 1) with a surface. the surface is defined implicitly (x, y, and z as functions of alpha, beta). this is my code:
alpha = linspace(0,pi);
beta = linspace(0,pi);
[alpha,beta]=meshgrid(alpha,beta);
L= 4*exp(-.6*beta).*sin(alpha);
%converting to x,y,z coordinates:
x = L.*sin(alpha).*cos(beta);
y = L.*cos(alpha);
z= L.*sin(alpha).*sin(beta);
to plot this i generally would use surf(x,y,z). but in this case i want to plot its intersection with a plane, for example the one defined by z2 = 1-x+y. (im not sure whether it would be better to define separate matrices for new x, y values, or whether it is better to use the existing ones.) i hope this question isn't too confusing. if you have any advice, please help.
What if you try the following: (I based my code on this answer to a previous question Plotting Implicit Algebraic equations in MATLAB) and it is not perfect but maybe it can help you:
clear
clc
alpha = linspace(0,pi);
beta = linspace(0,pi);
[alpha,beta]=meshgrid(alpha,beta);
L= 4*exp(-.6*beta).*sin(alpha);
x = L.*sin(alpha).*cos(beta);
y = L.*cos(alpha);
z= L.*sin(alpha).*sin(beta);
% Use an anonymous function to define the plane you want to plot
[X,Y] = meshgrid(min(x(:)):.5:max(x(:)),min(y(:)):0.5:max(y(:))); % x and y limits
f = #(X,Y) -X+Y+1; % sorry for the choice of capital letters; it was the most intuitive I thought.
hold on
surf(x,y,z);
contour3(X,Y,f(X,Y),200); % the 200 is arbitrary; play with it to change the # of lines making up the plane
hold off
rotate3d on
The result looks like this:
As I said I'm not 100% confident this is the most robust way but it looks fine to me :)
I have a function z = f(x, y), where z is the value at point (x, y). How may I integrate z over the x-y plane in MATLAB?
By function above, I actually mean I have something similar to a hash table. That is, given a (x, y) pair, I can look up the table to find the corresponding z value.
The problem would be rather simple, if the points were uniformly distributed over x-y plane, in which case I can simply sum up all the z values, multiply it with the bottom area, and finally divide it by the number of points I have. However, the distribution is not uniform as shown below. So I am actually asking for the computation method that minimises the error.
The currently accepted answer will only work for gridded data. If your data is scattered you can use the following approach instead:
scatteredInterpolant + integral2:
f = scatteredInterpolant(x(:), y(:), z(:), 'linear');
int = integral2(#(x,y) f(x,y), xmin, xmax, ymin, ymax);
This defines the linear interpolant f of the data z(i) = f(x(i),y(i)) and uses it as an argument to integral2. Note that ymin and ymax, instead of doubles, can be function handles depending on x. So usually you will be integrating rectangles, but this could be used for integration regions a bit more complicated.
If your integration area is rather complicated or has holes, you should consider triangulating your data.
DIY using triangulation:
Let's say your integration area is given by the triangulation trep, which for example could be obtained by trep = delaunayTriangulation(x(:), y(:)). If you have your values z corresponding to z(i) = f(trep.Points(i,1), trep.Points(i,2)), you can use the following integration routine. It computes the exact integral of the linear interpolant. This is done by evaluating the areas of all the triangles and then using these areas as weights for the midpoint(mean)-value on each triangle.
function int = integrateTriangulation(trep, z)
P = trep.Points; T = trep.ConnectivityList;
d21 = P(T(:,2),:)-P(T(:,1),:);
d31 = P(T(:,3),:)-P(T(:,1),:);
areas = abs(1/2*(d21(:,1).*d31(:,2)-d21(:,2).*d31(:,1)));
int = areas'*mean(z(T),2);
If you have a discrete dataset for which you have all the x and y values over which z is defined, then just obtain the Zdata matrix corresponding to those (x,y) pairs. Save this matrix, and then you can make it a continuous function using interp2:
function z_interp = fun(x,y)
z_interp = interp2(Xdata,Ydata,Zdata,x,y);
end
Then you can use integral2 to find the integral:
q = integral2(#fun,xmin,xmax,ymin,ymax)
where #fun is your function handle that takes in two inputs.
I had to integrate a biavariate normal distribution recently in MatLab. The idea is very simple. Matlab defines a surface through a meshgrid, so from x, y you need to do this:
x = -10:0.05:10;
y = x;
[X,Y] = meshgrid(x',y');
...for example. Then, let's call FX the function that defines the value at each point of the surface. To calculate the integral you just need to do this:
surfint = zeros(length(X),1);
for a = 1:length(X)
surfint(a,1) = trapz(x,FX(:,a));
end
trapz(x, surfint)
For me, this is the simplest way.