I am working on a joint pdf problem in which the random variable
U = sqrt(X^2+Y^2)
X and Y are uniformly distributed over (-2,2). I want to plot joint pdf of X and Y. Then compute pdf of U and plot it as well. I am using matlab R2011a, and so far, I have come up with the following code. On running the code I got an error message
Undefined function or method 'makedist' for input arguement type 'char'.
I found out that makedist is not on 2011 version. So I tried using
a=-2;
b=2;
X=a+(b-a)*rand(-10,10);
Y= a+(b-a)*rand(-10,10).
However, I am not sure how to compute pdfs of X and Y, and then joint pdf of XY from this. Any help, partial or holistic, is appreciated.
Here is the matlab code for the problem
%% Create distribution objects for X~U(-2,2) and Y~U(-2,2)
pdx=makedist('Uniform','lower',-2,'upper',2);
pdy=makedist('Uniform','lower',-2,'upper',2);
%Compute the pfs
x_ref=-10:1:10;
y_ref=-10:1:10;
pdf_x=pdf(pdx,x_ref);
pdf_y=pdf(pdy,y_ref);
% Plot the pdfs
figure 1;
stairs(x_ref,pdf_x,'g','Linewidth',2);
hold on;
stairs(y_ref,pdf_y,'r','Linewidth',2);
ylim([0 1.5]);
hold off;
% Joint pdf of x and Y
pdfXY=pdf_x*pdf_y;
figure 2;
plot(pdfXY);
%CDF and PDF of U
U=sqrt(X^2+Y^2);
Umin=0;
Umax=sqrt(b^2+b^2);
a=lower;
b=upper;
x=sqrt(U^2-Y^2);
xmin=0;
xmax=x;
ymin=0;
ymax=U;
Ucdf=integral2(pdfXY,xmin,xmax,ymin,ymax);
% plot CDF of U
figure 3;
plot(Ucdf)
I am just looking to plot the regions than for any specific sample set. X and Y are continuous independent uniform random variables.
As your x and y are independent at random, the theoretical joint distribution is just a product of the two
P(x,y) = P(x)*P(y)
In terms of MATLAB code, you may think of x and y running along two different dimensions:
N = 10; %// think of a probability mass function over N points
x = linspace(-2,2, N);
y = linspace(-2,2, N)';
Px = ones(N,1)./N;
Py = ones(1,10)./N;
%// Then the joint will be:
Jxy = bsxfun(#times, Px , Py);
figure
pcolor(x,y,Jxy)
You can now plug whatever distribution you like, if they are independent for Px and Py, and it will work
Related
I have started to learn Machine Learning, and programming in matlab.
I want to plot a matrix sized m*d where d=3 and m are the number of points.
with y binary vector I'd like to color each point with blue/red.
and plot a plane which is described with the vertical vector to it w.
The problem I trying to solve is to give some kind of visual representation of the data and the linear predictor.
All I know is how to single points with plot3, but no any number of points.
Thanks.
Plot the points using scatter3()
scatter3(X(y,1),X(y,2),X(y,3),'filled','fillcolor','red');
hold on;
scatter3(X(~y,1),X(~y,2),X(~y,3),'filled','fillcolor','blue');
or using plot3()
plot(X(y,1),X(y,2),X(y,3),' o','MarkerEdgeColor','red','MarkerFaceColor','red');
hold on;
plot(X(~y,1),X(~y,2),X(~y,3),' o','MarkerEdgeColor','blue','MarkerFaceColor','blue');
There are a few ways to plot a plane. As long as w(3) isn't very close to 0 then the following will work okay. I'm assuming your plane is defined by x'*w+b=0 where b is a scalar and w and x are column vectors.
x1min = min(X(:,1)); x2min = min(X(:,2));
x1max = max(X(:,1)); x2max = max(X(:,2));
[x1,x2] = meshgrid(linspace(x1min,x1max,20), linspace(x2min, x2max, 20));
x3 = -(w(1)*x1 + w(2)*x2 + b)/w(3);
surf(x1,x2,x3,'FaceColor',[0.6,0.6,0.6],'FaceAlpha',0.7,'EdgeColor',[0.4,0.4,0.4],'EdgeAlpha',0.4);
xlabel('x_1'); ylabel('x_2'); zlabel('x_3'); axis('vis3d');
Resulting plot
I have the following equations:
FindBurrow = .3*log(x+2)-.2
DistEffect= normpdf(y,0,6) + 0.92
VegEffect = -.006*z+1
Detection = FindBurrow*DistEffect*VegEffect
I would like to visualize the Detection function as part of a 4-D visualization. The fourth dimension would be color to represent Detection. I see that it is possible to do this if I have matrices with real x,y,z values but I currently don't.
I have tried to convert my symbolic x,y,z values to real 100 by 100 matrices using the following but it doesn't seem to work properly.
%% Conversion to "Real" Numbers
syms f(x,y,z)
f(x,y,z) = Detection;
[x,y] = ndgrid(linspace(1/10,50), linspace(1/10, 50))
z] = ndgrid(linspace(1/10,50), linspace(1/10, 50))
g = matlabFunction(Detection);
DetKernel = real(g(x,y,z))
%% Graphing the IPM
figure
surf(x,y,z,DetKernel) % surface map
clear title xlabel ylabel %Clears old runs index
xlabel ('Burrow Size') %x-axis label
ylabel('Distance From Line') %y-axis label
zlabel('Vegetation Effect') %z-axis label
cb = colorbar;
cb.Label.String = 'Probability of Detection';
I suspect this is due to a lack of knowledge of how to define the x,y,z values to make them act like grid points that Matlab will then use to make a surface or mesh. The graphing part seems to work okay, just doesn't make the shape I would expect based on experimental plotting of the XY, YZ, Z*X equations.
In short I have two questions: 1. Can I graph 4D using symbolic variables? 2. If not, how do I convert to real numbers without messing up my data?
For reference, x will be between 0 and 55, y will be between 0 and 25 and z will be between 0 and 160.
I have a two columns of data. X = Model values of NOx concentrations and Y = Observations of NOx concentrations. Now, I want to scatter plot X, Y (markers varying with colors) as well as the colourbar which would show me the counts (i.e. number of data points in that range). X and Y are daily data for a year, i.e. 365 rows.
Please help me. Any help is greatly appreciated.
I have attached a sample image.
If I understand you correctly, the real problem is creating the color information, which is, creating a bivariate histogram. Luckily, MATLAB has a function, hist3, for that in the Statistics & Machine Learning Toolbox. The syntax is
[N,C] = hist3(X,nbins)
where X is a m-by-2 matrix containing the data, and nbins is a 1-by-2 vector containing the number of bins in each dimension. The return value N is a matrix of size nbins(1)-by-nbins(2), and contains the histogram data. C is a 1-by-2 cell array, containing the bin centers in both dimensions.
% Generate sample data
X = randn(10000, 1);
Y = X + rand(10000, 1);
% Generate histogram
[N,C] = hist3([X,Y], [100,100]);
% Plot
imagesc(C{1},C{2},N);
set(gca,'YDir','normal');
colormap(flipud(pink));
colorbar;
Result:
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
I am trying to simulate a network of mobile robots that uses artificial potential fields for movement planning to a shared destination xd. This is done by generating a series of m-files (one for each robot) from a symbolic expression, as this seems to be the best way in terms of computational time and accuracy. However, I can't figure out what is going wrong with my gradient computation: the analytical gradient that is being computed seems to be faulty, while the numerical gradient is calculated correctly (see the image posted below). I have written a MWE listed below, which also exhibits this problem. I have checked the file generating part of the code, and it does return a correct function file with a correct gradient. But I can't figure out why the analytic and numerical gradient are so different.
(A larger version of the image below can be found here)
% create symbolic variables
xd = sym('xd',[1 2]);
x = sym('x',[2 2]);
% create a potential function and a gradient function for both (x,y) pairs
% in x
for i=1:size(x,1)
phi = norm(x(i,:)-xd)/norm(x(1,:)-x(2,:)); % potential field function
xvector = reshape(x.',1,size(x,1)*size(x,2)); % reshape x to allow for gradient computation
grad = gradient(phi,xvector(2*i-1:2*i)); % compute the gradient
gradx = grad(1);grady=grad(2); % split the gradient in two components
% create function file names
gradfun = strcat('GradTester',int2str(i),'.m');
phifun = strcat('PotTester',int2str(i),'.m');
% generate two output files
matlabFunction(gradx, grady,'file',gradfun,'outputs',{'gradx','grady'},'vars',{xvector, xd});
matlabFunction(phi,'file',phifun,'vars',{xvector, xd});
end
clear all % make sure the workspace is empty: the functions are in the files
pause(0.1) % ensure the function file has been generated before it is called
% these are later overwritten by a specific case, but they can be used for
% debugging
x = 0.5*rand(2);
xd = 0.5*rand(1,2);
% values for the Stackoverflow case
x = [0.0533 0.0023;
0.4809 0.3875];
xd = [0.4087 0.4343];
xp = x; % dummy variable to keep x intact
% compute potential field and gradient for both (x,y) pairs
for i=1:size(x,1)
% create a grid centered on the selected (x,y) pair
xGrid = (x(i,1)-0.1):0.005:(x(i,1)+0.1);
yGrid = (x(i,2)-0.1):0.005:(x(i,2)+0.1);
% preallocate the gradient and potential matrices
gradx = zeros(length(xGrid),length(yGrid));
grady = zeros(length(xGrid),length(yGrid));
phi = zeros(length(xGrid),length(yGrid));
% generate appropriate function handles
fun = str2func(strcat('GradTester',int2str(i)));
fun2 = str2func(strcat('PotTester',int2str(i)));
% compute analytic gradient and potential for each position in the xGrid and
% yGrid vectors
for ii = 1:length(yGrid)
for jj = 1:length(xGrid)
xp(i,:) = [xGrid(ii) yGrid(jj)]; % select the position
Xvec = reshape(xp.',1,size(x,1)*size(x,2)); % turn the input into a vector
[gradx(ii,jj),grady(ii,jj)] = fun(Xvec,xd); % compute gradients
phi(jj,ii) = fun2(Xvec,xd); % compute potential value
end
end
[FX,FY] = gradient(phi); % compute the NUMERICAL gradient for comparison
%scale the numerical gradient
FX = FX/0.005;
FY = FY/0.005;
% plot analytic result
subplot(2,2,2*i-1)
hold all
xlim([xGrid(1) xGrid(end)]);
ylim([yGrid(1) yGrid(end)]);
quiver(xGrid,yGrid,-gradx,-grady)
contour(xGrid,yGrid,phi)
title(strcat('Analytic result for position ',int2str(i)));
xlabel('x');
ylabel('y');
subplot(2,2,2*i)
hold all
xlim([xGrid(1) xGrid(end)]);
ylim([yGrid(1) yGrid(end)]);
quiver(xGrid,yGrid,-FX,-FY)
contour(xGrid,yGrid,phi)
title(strcat('Numerical result for position ',int2str(i)));
xlabel('x');
ylabel('y');
end
The potential field I am trying to generate is defined by an (x,y) position, in my code called xd. x is the position matrix of dimension N x 2, where the first column represents x1, x2, and so on, and the second column represents y1, y2, and so on. Xvec is simply a reshaping of this vector to x1,y1,x2,y2,x3,y3 and so on, as the matlabfunction I am generating only accepts vector inputs.
The gradient for robot i is being calculated by taking the derivative w.r.t. x_i and y_i, these two components together yield a single derivative 'vector' shown in the quiver plots. The derivative should look like this, and I checked that the symbolic expression for [gradx,grady] indeed looks like that before an m-file is generated.
To fix the particular problem given in the question, you were actually calculating phi in such a way that meant you doing gradient(phi) was not giving the correct results compared to the symbolic gradient. I'll try and explain. Here is how you created xGrid and yGrid:
% create a grid centered on the selected (x,y) pair
xGrid = (x(i,1)-0.1):0.005:(x(i,1)+0.1);
yGrid = (x(i,2)-0.1):0.005:(x(i,2)+0.1);
But then in the for loop, ii and jj were used like phi(jj,ii) or gradx(ii,jj), but corresponding to the same physical position. This is why your results were different. Another problem you had was you used gradient incorrectly. Matlab assumes that [FX,FY]=gradient(phi) means that phi is calculated from phi=f(x,y) where x and y are matrices created using meshgrid. You effectively had the elements of phi arranged differently to that, an so gradient(phi) gave the wrong answer. Between reversing the jj and ii, and the incorrect gradient, the errors cancelled out (I suspect you tried doing phi(jj,ii) after trying phi(ii,jj) first and finding it didn't work).
Anyway, to sort it all out, on the line after you create xGrid and yGrid, put this in:
[X,Y]=meshgrid(xGrid,yGrid);
Then change the code after you load fun and fun2 to:
for ii = 1:length(xGrid) %// x loop
for jj = 1:length(yGrid) %// y loop
xp(i,:) = [X(ii,jj);Y(ii,jj)]; %// using X and Y not xGrid and yGrid
Xvec = reshape(xp.',1,size(x,1)*size(x,2));
[gradx(ii,jj),grady(ii,jj)] = fun(Xvec,xd);
phi(ii,jj) = fun2(Xvec,xd);
end
end
[FX,FY] = gradient(phi,0.005); %// use the second argument of gradient to set spacing
subplot(2,2,2*i-1)
hold all
axis([min(X(:)) max(X(:)) min(Y(:)) max(Y(:))]) %// use axis rather than xlim/ylim
quiver(X,Y,gradx,grady)
contour(X,Y,phi)
title(strcat('Analytic result for position ',int2str(i)));
xlabel('x');
ylabel('y');
subplot(2,2,2*i)
hold all
axis([min(X(:)) max(X(:)) min(Y(:)) max(Y(:))])
quiver(X,Y,FX,FY)
contour(X,Y,phi)
title(strcat('Numerical result for position ',int2str(i)));
xlabel('x');
ylabel('y');
I have some other comments about your code. I think your potential function is ill-defined, which is causing all sorts of problems. You say in the question that x is an Nx2 matrix, but you potential function is defined as
norm(x(i,:)-xd)/norm(x(1,:)-x(2,:));
which means if N was three, you'd have the following three potentials:
norm(x(1,:)-xd)/norm(x(1,:)-x(2,:));
norm(x(2,:)-xd)/norm(x(1,:)-x(2,:));
norm(x(3,:)-xd)/norm(x(1,:)-x(2,:));
and I don't think the third one makes sense. I think this could be causing some confusion with the gradients.
Also, I'm not sure if there is a reason to create the .m file functions in your real code, but they are not necessary for the code you posted.