How to be able plot eigen vector in Matlab an make it sure have the exact position as a new axes of our data??
here is my code for PCA
a= randn(100,3);
b=mean(a);
c=cov(a);
[vec,val] = eigs(c);
e1=vec(:,1);
e2=vec(:,2);
plot3(a(:,1),a(:,2),a(:,3),'ro');
hold on
%% Here is the Problem begins
plot(e1,'k--');
plot(e2,'k--');
Here is the output that I got
That two lines represent the e1&e2.
How to plot the e1 & e2 properly???
You can use the output from eig to directly plot all three eigenvectors (or a subset):
a = randn(100,3);
b = mean(a);
c = cov(a);
[vec,val] = eig(c);
plot3(a(:,1),a(:,2),a(:,3),'ro');
hold on
z = zeros(1,3);
% plot3([z(1:2);vec(1,1:2)],[z(1:2);vec(2,1:2)],[z(1:2);vec(3,1:2)],'k')
plot3([z;vec(1,:)],[z;vec(2,:)],[z;vec(3,:)],'k')
axis equal
grid on
And if you want the scaled eigenvectors:
sc_vec = vec*val;
z = zeros(1,3);
plot3([z;sc_vec(1,:)],[z;sc_vec(2,:)],[z;sc_vec(3,:)],'b')
Another option is to use quiver3 to get lines with arrowheads:
scale = 1;
z = zeros(1,3);
quiver3(z,z,z,vec(1,:),vec(2,:),vec(3,:),scale,'g')
Just don't expect the arrows to always look perfect, because Matlab isn't particularly good at this sort of thing.
Note that eig may return the eigenvalues and -vectors in a different order than eigs, but it is the function that you should be using. eigs is a specialized function that solved the eigenproblem in a different way and is much less efficient.
You can use plot3
a= randn(100,3);
b=mean(a);
c=cov(a);
[vec,val] = eigs(c);
e1=[zeros(size(vec(:,1))),vec(:,1)]'; e1 = e1/norm(e1)*4;
e2=[zeros(size(vec(:,2))),vec(:,2)]'; e2 = e2/norm(e2)*4;
e3=[zeros(size(vec(:,3))),vec(:,3)]'; e3 = e3/norm(e3)*4;
plot3(a(:,1),a(:,2),a(:,3),'ro');
hold on
%% Here is the Problem begins
plot3(e1(:,1),e1(:,2),e1(:,3),'k','LineWidth',2);
plot3(e2(:,1),e2(:,2),e2(:,3),'b','LineWidth',2);
plot3(e3(:,1),e3(:,2),e3(:,3),'g','LineWidth',2);
To represent each vector, use plot3 to plot a line from the origin to the coordinates defined by the vector, perhaps multiplied by a large number (I use 10 in the example) to make the line longer:
plot3(10*[0 e1(1)], [0 e1(2)], [0 e1(3)],'b--');
plot3(10*[0 e2(1)], [0 e2(2)], [0 e2(3)],'g--');
You may need to change the view to get a good perspective. For that you can use view function, or click the "Rotate 3D" icon (circular arrow) in the figure toolbar.
Related
Given scatter data, or a matrix, I would like to generate a nice plot such as the one shown below, with all 3 histograms and a colored matrix. I'm specifically interested in the diagonal histogram, which ideally, would correspond to the diagonals of a matrix:
Source figure: www.med.upenn.edu/mulab/jpst.html
The existing command scatterhist is not that powerful to generate this type of graph. Any ideas?
Thanks!
EDIT:
Following #Cris Luengo's hints, I came up with the following code which does some first work at the inclined histogram: WORK IN PROGRESS (HELP WELCOME)!!
b = [0 1 2 3 4 5 6 7 8 9 10];
h = [0.33477 0.40166 0.20134 0.053451 0.008112 0.000643 2.7e-05 0 0 0 0];
wid = 0.25; bb = sort([b-wid b-wid b+wid b+wid]);
kk = [zeros(numel(h),1) h(:) h(:) zeros(numel(h),1)];
kk = reshape(kk',[1,numel(kk)]);
pp=patch(bb,kk,'b');axis([-.5 5 0 .5])
set(gca,'CameraUpVector',[-1,.08,0]);axis square
EDIT 2: Using rotation
phi = pi/4;
R = [cos(phi),-sin(phi);sin(phi),cos(phi)];
rr = [bb' kk'] * R;
bb = rr(:,1); kk = rr(:,2);
patch(bb,kk,'b'); axis([-.5 3 -4 .5])
Here is a recipe to plot the diagonal histogram, if you can do that I’m sure you can figure out the rest too.
Compute the histogram, the bin counts are h, the bin centers are b.
Build a coordinate matrix, attaching the coordinates of a point on the x-axis at the left and right ends of the histogram:
coords = [b(:),h(:)];
coords = [coord;b(end),0;b(1),0];
Using patch you can now plot the histogram as follows:
patch(coords(1,:),coords(2,:));
To plot a rotated histogram you can simply multiply the coords matrix with a rotation matrix, before using patch:
phi = pi/4;
R = [cos(phi),-sin(phi);sin(phi),cos(phi)];
coords = R * coords;
You might need to shift the plot to place it at the right location w.r.t. the other elements.
I recommend that you place all these graphic elements in the same axes object; you can set the axes’ visibility to 'off' so that it works only as a canvas for the other elements.
It will be a bit of work to get everything placed as in the plot you show, but none of it is difficult. Use the low-level image, line,patch and text to place those types of elements, don’t try to use the higher-level plotting functions such as plot, they don’t provide any benefits over the low-level ones in this case.
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
Im just trying to draw a line through the following points in matlab. Currently the line extends only to the points. I need to to extend and intercept the x axis. The code is below
A = [209.45 198.066 162.759];
B = [1.805 1.637 1.115];
plot(A,B,'*');
axis([0 210 0 2]);
hold on
line(A,B)
hold off
If you want to augment your points with a corresponding y==0 point, I suggest using interp1 to obtain the x-intercept:
A = [209.45 198.066 162.759];
B = [1.805 1.637 1.115];
x0 = interp1(B,A,0,'linear','extrap'); %extrapolate (y,x) at y==0 to get x0
[newA, inds] = sort([x0 A]); %insert x0 where it belongs
newB = [0 B];
newB = newB(inds); %keep the same order with B
plot(A,B,'b*',newA,newB,'b-');
This will use interp1 to perform a linear interpolant, with extrapolation switched on. By interpolating (B,A) pairs, we in effect invert your linear function.
Next we add the (x0,0) point to the data, but since matlab draws lines in the order of the points, we have to sort the vector according to x component. The sorting order is then used to keep the same order in the extended B vector.
Finally the line is plotted. I made use of plot with a linespec of '-' to draw the line in the same command as the points themselves. If it doesn't bother you that the (x0,0) point is also indicated, you can plot both markers and lines together using plot(newA,newB,'*-'); which ensures that the colors match up (in the above code I manually set the same blue colour on both plots).
I would like to customize the color of my rootlocus plot.
I use a for cycle to plot 10 rootlocus (with slightly different systems in the loop) and I would like every of them to be of a different shade of grey. I thought to use the gray command to obtain a matrix to store the RGB data and then use this matrix in the rlocus(sys,K,'style') command (choosing the i-th line at the i-th iteration of my cycle). Unfortunately the command requires the style to be a cell (for example 'g' or 'b') and not a vector of numbers.
This is a sample of my code:
figure()
hold on
L = [sys1, sys2, ..., sys10];
colors = gray(10);
for i = 0:9
rlocus (L(i+1), 'Color', colors(i+1, :));
end
The rlocus() function is not as powerful as the plot() function and only has limited support for setting colours with rlocus(sys, 'b') as you've noticed. However, we can combine it with the plot() function to make use of its power.
Here I use [R, K] = rlocus(sys) to return the values of the root locus, R. Each row of R represents a different trajectory. We can plot 1 trajectory of the root locus with plot(R(m, :)) and utilise the strength of plot() to change the colour however we wish.
L = [sys1, sys2, sys3, sys4, sys5, sys6, sys7, sys8, sys9, sys10];
C = gray(numel(L) + 1); % Extra 1 because the last value will be
% white and plotting white on white does
% not look well :P
figure;
hold on
for n = 1:numel(L)
[R, K] = rlocus(L(n));
for m = 1:numel(R)/length(R)
plot(R(m, :), 'Color', C(n, :));
end
end
hold off
Generate a plot showing the graphs of
y=(2*a+1)*exp(-x)-(a+1)*exp(2*x)
in the range x ∈ <-2, 4> for all integer values of a between -3 and 3
I know how to make typical plot for 2 values and set a range on the axes, but how to draw the graph dependent on the parameter a?
To elaborate on Ben Voigt's comment: A more advanced technique would be to replace the for-loop with a call to bsxfun to generate a matrix of evaluations of M(i,j) = f(x(i),a(j)) and call plot with this matrix. Matlab will then use the columns of the matrix and plot each column with individual colors.
%%// Create a function handle of your function
f = #(x,a) (2*a+1)*exp(-x)-(a+1)*exp(2*x);
%%// Plot the data
x = linspace(-2, 4);
as = -3:3;
plot(x, bsxfun(f,x(:),as));
%%// Add a legend
legendTexts = arrayfun(#(a) sprintf('a == %d', a), as, 'uni', 0);
legend(legendTexts, 'Location', 'best');
You could also create the evaluation matrix using ndgrid, which explicitly returns all combinations of the values of x and as. Here you have to pay closer attention on properly vectorizing the code. (We were lucky that the bsxfun approach worked without having to change the original f.)
f = #(x,a) (2*a+1).*exp(-x)-(a+1).*exp(2*x); %// Note the added dots.
[X,As] = ndgrid(x,as);
plot(x, f(X,As))
However for starters, you should get familiar with loops.
You can do it using a simple for loop as follows. You basically loop through each value of a and plot the corresponding y function.
clear
clc
close all
x = -2:4;
%// Define a
a = -3:3;
%// Counter for legend
p = 1;
LegendText = cell(1,numel(a));
figure;
hold on %// Important to keep all the lines on the same plot.
for k = a
CurrColor = rand(1,3);
y= (2*k+1).*exp(-x)-(k+1).*exp(2.*x);
plot(x,y,'Color',CurrColor);
%// Text for legend
LegendText{p} = sprintf('a equals %d',k);
p = p+1;
end
legend(LegendText,'Location','best')
Which gives something like this:
You can customize the graph as you like. Hope that helps get you started!