I am using MatLab to solve a system of coupled PDEs, with pdenonlin.
I create a mesh for my geometry (a square box with a circular hole in the middle), and refine it until I :
[p,e,t] = initmesh('DefectGeom2');
[p,e,t] = refinemesh('DefectGeom2',p,e,t);
I solve the system
% SOLUTION:
u = pdenonlin(b_s,p,e,t,c_s,a_s,f_s);
% EXTRACT different functions from the full solutions (systems):
np = size(p,2); % number of node points
uk = reshape(u,np,[]); % each uk column has one component of u
Therefore I now have my uk (in my case 3) solutions.
Now I want to calculate integrals and derivatives of this approximate solutions. I tried by interpolating to a uniform grid using tri2grid:
x=linspace(-1,1,Npts);
y=x;
psi=tri2grid(p,t,uk(:,3),x,y);
theta=tri2grid(p,t,uk(:,1),x,y);
theta_y=derivative(theta,1,2);
psi_x=derivative(psi,1,1);
and calculate:
pressure = trapz(x,psi_x-cos(2*theta).*theta_y+sin(2*theta));
But this gives me a poor approximation, I guess because of the fact that the grid is uniform whereas the mesh is finer around the central circle and coarser elsewhere.
Is there a way I could use MatLab built-in functions to calculate derivatives of the solutions of pdenonlin without brutally interpolate with tri2grid on a uniform grid?
There is a pdeInterpolant class in the pde toolbox which might be useful. Use meshToPet to generate the input arguments from your mesh and then evaluate to query the interpolant.
You can specify a different grid when you query the interpolant so if you use a finer grid for this stage you might get better results without having to alter the original mesh.
Related
Here is the given system I want to plot and obtain the vector field and the angles they make with the x axis. I want to find the index of a closed curve.
I know how to do this theoretically by choosing convenient points and see how the vector looks like at that point. Also I can always use
to compute the angles. However I am having trouble trying to code it. Please don't mark me down if the question is unclear. I am asking it the way I understand it. I am new to matlab. Can someone point me in the right direction please?
This is a pretty hard challenge for someone new to matlab, I would recommend taking on some smaller challenges first to get you used to matlab's conventions.
That said, Matlab is all about numerical solutions so, unless you want to go down the symbolic maths route (and in that case I would probably opt for Mathematica instead), your first task is to decide on the limits and granularity of your simulated space, then define them so you can apply your system of equations to it.
There are lots of ways of doing this - some more efficient - but for ease of understanding I propose this:
Define the axes individually first
xpts = -10:0.1:10;
ypts = -10:0.1:10;
tpts = 0:0.01:10;
The a:b:c syntax gives you the lower limit (a), the upper limit (c) and the spacing (b), so you'll get 201 points for the x. You could use the linspace notation if that suits you better, look it up by typing doc linspace into the matlab console.
Now you can create a grid of your coordinate points. You actually end up with three 3d matrices, one holding the x-coords of your space and the others holding the y and t. They look redundant, but it's worth it because you can use matrix operations on them.
[XX, YY, TT] = meshgrid(xpts, ypts, tpts);
From here on you can perform whatever operations you like on those matrices. So to compute x^2.y you could do
x2y = XX.^2 .* YY;
remembering that you'll get a 3d matrix out of it and all the slices in the third dimension (corresponding to t) will be the same.
Some notes
Matlab has a good builtin help system. You can type 'help functionname' to get a quick reminder in the console or 'doc functionname' to open the help browser for details and examples. They really are very good, they'll help enormously.
I used XX and YY because that's just my preference, but I avoid single-letter variable names as a general rule. You don't have to.
Matrix multiplication is the default so if you try to do XX*YY you won't get the answer you expect! To do element-wise multiplication use the .* operator instead. This will do a11 = b11*c11, a12 = b12*c12, ...
To raise each element of the matrix to a given power use .^rather than ^ for similar reasons. Likewise division.
You have to make sure your matrices are the correct size for your operations. To do elementwise operations on matrices they have to be the same size. To do matrix operations they have to follow the matrix rules on sizing, as will the output. You will find the size() function handy for debugging.
Plotting vector fields can be done with quiver. To plot the components separately you have more options: surf, contour and others. Look up the help docs and they will link to similar types. The plot family are mainly about lines so they aren't much help for fields without creative use of the markers, colours and alpha.
To plot the curve, or any other contour, you don't have to test the values of a matrix - it won't work well anyway because of the granularity - you can use the contour plot with specific contour values.
Solving systems of dynamic equations is completely possible, but you will be doing a numeric simulation and your results will again be subject to the granularity of your grid. If you have closed form solutions, like your phi expression, they may be easier to work with conceptually but harder to get working in matlab.
This kind of problem is tractable in matlab but it involves some non-basic uses which are pretty hard to follow until you've got your head round Matlab's syntax. I would advise to start with a 2d grid instead
[XX, YY] = meshgrid(xpts, ypts);
and compute some functions of that like x^2.y or x^2 - y^2. Get used to plotting them using quiver or plotting the coordinates separately in intensity maps or surfaces.
I would like to plot some figures like this one:
-axis being real and imag part of some complex valued vector(usually either pure real or imag)
-have some 3D visualization like in the given case
First, define your complex function as a function of (Re(x), Im(x)). In complex analysis, you can decompose any complex function into its real parts and imaginary parts. In other words:
F(x) = Re(x) + i*Im(x)
In the case of a two-dimensional grid, you can obviously extend to defining the function in terms of (x,y). In other words:
F(x,y) = Re(x,y) + i*Im(x,y)
In your case, I'm assuming you'd want the 2D approach. As such, let's use I and J to represent the real parts and imaginary parts separately. Also, let's start off with a simple example, like cos(x) + i*sin(y) which is based on the very popular Euler exponential function. It isn't exact, but I modified it slightly as the plot looks nice.
Here are the steps you would do in MATLAB:
Define your function in terms of I and J
Make a set of points in both domains - something like meshgrid will work
Use a 3D visualization plot - You can plot the individual points, or plot it on a surface (like surf, or mesh).
NB: Because this is a complex valued function, let's plot the magnitude of the output. You were pretty ambiguous with your details, so let's assume we are plotting the magnitude.
Let's do this in code line by line:
% // Step #1
F = #(I,J) cos(I) + i*sin(J);
% // Step #2
[I,J] = meshgrid(-4:0.01:4, -4:0.01:4);
% // Step #3
K = F(I,J);
% // Let's make it look nice!
mesh(I,J,abs(K));
xlabel('Real');
ylabel('Imaginary');
zlabel('Magnitude');
colorbar;
This is the resultant plot that you get:
Let's step through this code slowly. Step #1 is an anonymous function that is defined in terms of I and J. Step #2 defines I and J as matrices where each location in I and J gives you the real and imaginary co-ordinates at their matching spatial locations to be evaluated in the complex function. I have defined both of the domains to be between [-4,4]. The first parameter spans the real axis while the second parameter spans the imaginary axis. Obviously change the limits as you see fit. Make sure the step size is small enough so that the plot is smooth. Step #3 will take each complex value and evaluate what the resultant is. After, you create a 3D mesh plot that will plot the real and imaginary axis in the first two dimensions and the magnitude of the complex number in the third dimension. abs() takes the absolute value in MATLAB. If the contents within the matrix are real, then it simply returns the positive of the number. If the contents within the matrix are complex, then it returns the magnitude / length of the complex value.
I have labeled the axes as well as placed a colorbar on the side to visualize the heights of the surface plot as colours. It also gives you an idea of how high and how long the values are in a more pleasing and visual way.
As a gentle push in your direction, let's take a slice out of this complex function. Let's make the real component equal to 0, while the imaginary components span between [-4,4]. Instead of using mesh or surf, you can use plot3 to plot your points. As such, try something like this:
F = #(I,J) cos(I) + i*sin(J);
J = -4:0.01:4;
I = zeros(1,length(J));
K = F(I,J);
plot3(I, J, abs(K));
xlabel('Real');
ylabel('Imaginary');
zlabel('Magnitude');
grid;
plot3 does not provide a grid by default, which is why the grid command is there. This is what I get:
As expected, if the function is purely imaginary, there should only be a sinusoidal contribution (i*sin(y)).
You can play around with this and add more traces if you need to.
Hope this helps!
I finished an SVM training and got data like X, Y. X is the feature matrix only with 2 dimensions, and Y is the classification labels. Because the data is only in two dimensions, so I would like to draw a decision boundary to show the surface of support vectors.
I use contouf in Matlab to do the trick, but really find it hard to understand how to use the function.
I wrote like:
#1 try:
contourf(X);
#2 try:
contourf([X(:,1) X(:,2) Y]);
#3 try:
Z(:,:,1)=X(Y==1,:);
Z(:,:,2)=X(Y==2,:);
contourf(Z);
all these things do not correctly. And I checked the Matlab help files, most of them make Z as a function, so I really do not know how to form the correct Z matrix.
If you're using the svmtrain and svmclassify commands from Bioinformatics Toolbox, you can just use the additional input argument (...'showplot', true), and it will display a scatter plot with a decision boundary and the support vectors highlighted.
If you're using your own SVM, or a third-party tool such as libSVM, what you probably need to do is to:
Create a grid of points in your 2D input feature space using the meshgrid command
Classify those points using your trained SVM
Plot the grid of points and the classifications using contourf.
For example, in kind-of-MATLAB-but-pseudocode, assuming your input features are called X1 and X2:
numPtsInGrid = 100;
x1Range = linspace(x1lower, x1upper, numPtsInGrid);
x2Range = linspace(x2lower, x2upper, numPtsInGrid);
[X1, X2] = meshgrid(x1Range, x2Range);
Z = classifyWithMySVMSomehow([X1(:), X2(:)]);
contourf(X1(:), X2(:), Z(:))
Hope that helps.
I know it's been a while but I will give it a try in case someone else will come up with that issue.
Assume we have a 2D training set so as to train an SVM model, in other words the feature space is a 2D space. We know that a kernel SVM model leads to a score (or decision) function of the form:
f(x) = sumi=1 to N(aiyik(x,xi)) + b
Where N is the number of support vectors, xi is the i -th support vector, ai is the estimated Lagrange multiplier and yi the associated class label. Values(scores) of decision function in way depict the distance of the observation x frοm the decision boundary.
Now assume that for every point (X,Y) in the 2D feature space we can find the corresponding score of the decision function. We can plot the results in the 3D euclidean space, where X corresponds to values of first feature vector f1, Y to values of second feature f2, and Z to the the return of decision function for every point (X,Y). The intersection of this 3D figure with the Z=0 plane gives us the decision boundary into the two-dimensional feature space. In other words, imagine that the decision boundary is formed by the (X,Y) points that have scores equal to 0. Seems logical right?
Now in MATLAB you can easily do that, by first creating a grid in X,Y space:
d = 0.02;
[x1Grid,x2Grid] = meshgrid(minimum_X:d:maximum_X,minimum_Y:d:maximum_Y);
d is selected according to the desired resolution of the grid.
Then for a trained model SVMModel find the scores of every grid's point:
xGrid = [x1Grid(:),x2Grid(:)];
[~,scores] = predict(SVMModel,xGrid);
Finally plot the decision boundary
figure;
contour(x1Grid,x2Grid,reshape(scores(:,2),size(x1Grid)),[0 0],'k');
Contour gives us a 2D graph where information about the 3rd dimension is depicted as solid lines in the 2D plane. These lines implie iso-response values, in other words (X,Y) points with same Z value. In our occasion contour gives us the decision boundary.
Hope I helped to make all that more clear. You can find very useful information and examples in the following links:
MATLAB's example
Representation of decision function in 3D space
In matlab, how to generate two clusters of random points like the following graph. Can you show me the scripts/code?
If you want to generate such data points, you will need to have their probability distribution to be able to generate the points.
For your point, I do not have the real distributions, so I can only give an approximation. From your figure I see that both lay approximately on a circle, with a random radius and a limited span for the angle. I assume those angles and radii are uniformly distributed over certain ranges, which seems like a pretty good starting point.
Therefore it also makes sense to generate the random data in polar coordinates (i.e. angle and radius) instead of the cartesian ones (i.e. horizontal and vertical), and transform them to allow plotting.
C1 = [0 0]; % center of the circle
C2 = [-5 7.5];
R1 = [8 10]; % range of radii
R2 = [8 10];
A1 = [1 3]*pi/2; % [rad] range of allowed angles
A2 = [-1 1]*pi/2;
nPoints = 500;
urand = #(nPoints,limits)(limits(1) + rand(nPoints,1)*diff(limits));
randomCircle = #(n,r,a)(pol2cart(urand(n,a),urand(n,r)));
[P1x,P1y] = randomCircle(nPoints,R1,A1);
P1x = P1x + C1(1);
P1y = P1y + C1(2);
[P2x,P2y] = randomCircle(nPoints,R2,A2);
P2x = P2x + C2(1);
P2y = P2y + C2(2);
figure
plot(P1x,P1y,'or'); hold on;
plot(P2x,P2y,'sb'); hold on;
axis square
This yields:
This method works relatively well when you deal with distributions that you can transform easily and when you can easily describe the possible locations of the points. If you cannot, there are other methods such as the inverse transforming sampling method which offer algorithms to generate the data instead of manual variable transformations as I did here.
K-means is not going to give you what you want.
For K-means, vectors are classified based on their nearest cluster center. I can only think of two ways you could get the non-convex assignment shown in the picture:
Your input data is actually higher-dimensional, and your sample image is just a 2-d projection.
You're using a distance metric with different scaling across the dimensions.
To achieve your aim:
Use a non-linear clustering algorithm.
Apply a non-linear transform to your input data. (Probably not feasible).
You can find a list on non-linear clustering algorithms here. Specifically, look at this reference on the MST clustering page. Your exact shape appears on the fourth page of the PDF together with a comparison of what happens with K-Means.
For existing MATLAB code, you could try this Kernel K-Means implementation. Also, check out the Clustering Toolbox.
Assuming that you really want to do the clustering operation on existing data, as opposed to generating the data itself. Since you have a plot of some data, it seems logical that you already know how to do that! If I am wrong in this assumption, then you should word your questions more carefully in the future.
The human brain is quite good at seeing patterns in things like this, that writing a code for on a computer will often take some serious effort.
As has been said already, traditional clustering tools such as k-means will fail. Luckily, the image processing toolbox has good tools for these purposes already written. I might suggest converting the plot into an image, using filled in dots to plot the points. Make sure the dots are large enough that they touch each other within a cluster, with some overlap. Then use dilation/erosion tools if necessary to make sure that any small cracks are filled in, but don't go so far as to cause the clusters to merge. Finally, use region segmentation tools to pick out the clusters. Once done, transform back from pixel units in the image into your spatial units, and you have accomplished your task.
For the image processing approach to work, you will need sufficient separation between the clusters compared to the coarseness within a cluster. But that seems obvious for any method to succeed.
Starting from the plot of one curve, it is possible to obtain the parametric equation of that curve?
In particular, say x={1 2 3 4 5 6....} the x axis, and y = {a b c d e f....} the corresponding y axis. I have the plot(x,y).
Now, how i can obtain the equation that describe the plotted curve? it is possible to display the parametric equation starting from the spline interpolation?
Thank you
If you want to display a polynomial fit function alongside your graph, the following example should help:
x=-3:.1:3;
y=4*x.^3-5*x.^2-7.*x+2+10*rand(1,61);
p=polyfit(x,y,3); %# third order polynomial fit, p=[a,b,c,d] of ax^3+bx^2+cx+d
yfit=polyval(p,x); %# evaluate the curve fit over x
plot(x,y,'.')
hold on
plot(x,yfit,'-g')
equation=sprintf('y=%2.2gx^3+%2.2gx^2+%2.2gx+%2.2g',p); %# format string for equation
equation=strrep(equation,'+-','-'); %# replace any redundant signs
text(-1,-80,equation) %# place equation string on graph
legend('Data','Fit','Location','northwest')
Last year, I wrote up a set of three blogs for Loren, on the topic of modeling/interpolationg a curve. They may cover some of your questions, although I never did find the time to add another 3 blogs to finish the topic to my satisfaction. Perhaps one day I will get that done.
The problem is to recognize there are infinitely many curves that will interpolate a set of data points. A spline is a nice choice, because it can be made well behaved. However, that spline has no simple "equation" to write down. Instead, it has many polynomial segments, pieced together to be well behaved.
You're asking for the function/mapping between two data sets. Knowing the physics involved, the function can be derived by modeling the system. Write down the differential equations and solve it.
Left alone with just two data series, an input and an output with a 'black box' in between you may approximate the series with an arbitrary function. You may start with a polynomial function
y = a*x^2 + b*x + c
Given your input vector x and your output vector y, parameters a,b,c must be determined applying a fitness function.
There is an example of Polynomial Curve Fitting in the MathWorks documentation.
Curve Fitting Tool provides a flexible graphical user interfacewhere you can interactively fit curves and surfaces to data and viewplots. You can:
Create, plot, and compare multiple fits
Use linear or nonlinear regression, interpolation,local smoothing regression, or custom equations
View goodness-of-fit statistics, display confidenceintervals and residuals, remove outliers and assess fits with validationdata
Automatically generate code for fitting and plottingsurfaces, or export fits to workspace for further analysis