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).
Related
I have a number of 2d probability mass functions from 2 categories. I am trying to plot the contours to visualise them (for example at their half height, but doesn't really matter).
I don't want to use contourf to plot directly because I want to control the fill colour and opacity. So I am using contourc to generate xy coordinates, and am then using fill with these xy coordinates.
The problem is that the xy coordinates from the contourc function have strange numbers in them which cause the following strange vertices to be plotted.
At first I thought it was the odd contourmatrix format, but I don't think it is this as I am only asking for one value from contourc. For example...
contourmatrix = contourc(x, y, Z, [val, val]);
h = fill(contourmatrix(1,:), contourmatrix(2,:), 'r');
Does anyone know why the contourmatrix has these odd values in them when I am only asking for one contour?
UPDATE:
My problem seems might be a failure mode of contourc when the input 2D matrix is not 'smooth'. My source data is a large set of (x,y) points. Then I create a 2D matrix with some hist2d function. But when this is noisy the problem is exaggerated...
But when I use a 2d kernel density function to result in a much smoother 2D function, the problem is lessened...
The full process is
a) I have a set of (x,y) points which form samples from a distribution
b) I convert this into a 2D pmf
c) create a contourmatrix using contourc
d) plot using fill
Your graphic glitches are because of the way you use the data from the ContourMatrix. Even if you specify only one isolevel, this can result in several distinct filled area. So the ContourMatrix may contain data for several shapes.
simple example:
isolevel = 2 ;
[X,Y,Z] = peaks ;
[C,h] = contourf(X,Y,Z,[isolevel,isolevel]);
Produces:
Note that even if you specified only one isolevel to be drawn, this will result in 2 patches (2 shapes). Each has its own definition but they are both embedded in the ContourMatrix, so you have to parse it if you want to extract each shape coordinates individually.
To prove the point, if I simply throw the full contour matrix to the patch function (the fill function will create patch objects anyway so I prefer to use the low level function when practical). I get the same glitch lines as you do:
xc = X(1,:) ;
yc = Y(:,1) ;
c = contourc(xc,yc,Z,[isolevel,isolevel]);
hold on
hp = patch(c(1,1:end),c(2,1:end),'r','LineWidth',2) ;
produces the same kind of glitches that you have:
Now if you properly extract each shape coordinates without including the definition column, you get the proper shapes. The example below is one way to extract and draw each shape for inspiration but they are many ways to do it differently. You can certainly compact the code a lot but here I detailed the operations for clarity.
The key is to read and understand how the ContourMatrix is build.
parsed = false ;
iShape = 1 ;
while ~parsed
%// get coordinates for each isolevel profile
level = c(1,1) ; %// current isolevel
nPoints = c(2,1) ; %// number of coordinate points for this shape
idx = 2:nPoints+1 ; %// prepare the column indices of this shape coordinates
xp = c(1,idx) ; %// retrieve shape x-values
yp = c(2,idx) ; %// retrieve shape y-values
hp(iShape) = patch(xp,yp,'y','FaceAlpha',0.5) ; %// generate path object and save handle for future shape control.
if size(c,2) > (nPoints+1)
%// There is another shape to draw
c(:,1:nPoints+1) = [] ; %// remove processed points from the contour matrix
iShape = iShape+1 ; %// increment shape counter
else
%// we are done => exit while loop
parsed = true ;
end
end
grid on
This will produce:
I have wrote a program using Level Set Function for curve propagation. Curve is represented by level zero in contour plot. How can I get the coordinates of these curve as I march in time?
Contour are obtained at every time step. How to get the coordinates at each step?
A zero level contour from a set of points z(x,y) can be obtained by contourc. For instance, for the peaks-example of MATLAB, we obtain the contour lines with height 0 as follows:
c = contourc(peaks, [0,0]);
If this contour is a single line, this is all you need to do. However, the contour might consists of multiple islands (which is the case for this example), which means that you need to split the set of (x,y) coordinates in c somehow. A split based on NaN values can be obtained through:
c(:,~c(1,:)) = NaN;
and the result can be plot by
plot(c(1,:),c(2,:),'k');
If this is not sufficient, you could also construct a cell array c_split of islands:
b = find(~c(1,:))+1;
e = b+c(2,b-1)-1;
c_split = arrayfun(#(idx) c(:,b(idx):e(idx)), 1:numel(b), 'uni', 0)';
and the visualisation is very similar:
idx = 1;
plot(c_split{idx}(1,:),c_split{idx}(2,:),'k');
If the contours need to be stored for multiple time steps, you combine all results into a cell:
c_all{idx} = c;
I'm trying to use Matlab for some data plotting. In particular I need to plot a series of lines, some times given two points belonging to it, some times given the orthogonal vector.
I've used the following to obtain the plot of the line:
Line given two points A = [A(1), A(2)] B = [B(1), B(2)]:
plot([A(1),B(1)],[A(2),B(2)])
Line given the vector W = [W(1), W(2)]':
if( W(1) == 0 )
plot( [W(1), rand(1)] ,[W(2), W(2)])
else
plot([W(1), W(1) + (W(2)^2 / W(1))],[W(2),0])
end
where I'm calculating the intersection between the x-axis and the line using the second theorem of Euclid on the triangle rectangle formed by the vector W and the line.
My problem as you can see from the picture above is that the line will only be plotted between the two points and not on all the range of my axis.
I have 2 questions:
How can I have a line going across the whole axis range?
Is there a more easy and direct way (maybe a function?) to plot the line perpendicular to a vector? (An easier and more clean way to solve point 2 above.)
Thanks in advance.
Do you know the bounds of your axis for displaying the plot? If so, you can specify the range of the plot with the axis([xmin, xmax, ymin, ymax]) function.
So, from your question, if you know the slope m and intercept b, you can make sure your function plots the line across the whole window by specifying:
plot([xmin, xmax], [m*xmin + b, m*xmax + b]);
axis([xmin, xmax, min(m*xmin+b, m*xmax+b), max(m*xmin+b, m*xmax+b)]);
where xmin and xmax are values you specify as the range of your x-axis. This will make your line go from the corner of your plot to the other corner. If you want a buffer in the y-direction, then add one like so:
buffer = 5; % for example, you set this to something that looks good.
axis([xmin, xmax, min(m*xmin+b, m*xmax+b)-buffer, max(m*xmin+b, m*xmax+b)+buffer]);
I have a 3D plot and two points coordinates A(0,0,0) and B(13,-11,19). I just want to plot a visible line connecting this two points ... I tried plot3(0,0,0, 13,-11,19) and other stuff but everything i tried failed miserably.
Here's how:
% Your two points
P1 = [0,0,0];
P2 = [13,-11,19];
% Their vertial concatenation is what you want
pts = [P1; P2];
% Because that's what line() wants to see
line(pts(:,1), pts(:,2), pts(:,3))
% Alternatively, you could use plot3:
plot3(pts(:,1), pts(:,2), pts(:,3))
Admittedly, this might seem a bit counter-intuitive at first, but in the long run it'll make sense.
If you read doc plot or doc line, you'll see that each expects sets of x, y and z data, respectively. That is, using
plot3(X,Y,Z)
with X, Y and Z some matrices, plot3 will draw a line from the first triplet (X(1) Y(1) Z(1)) to the second triplet (X(2) Y(2) Z(2)) and so on -- same for line.
for an implicit equation(name it "y") of lambda and beta-bar which is plotted with "ezplot" command, i know it is possible that by a root finding algorithm like "bisection method", i can find solutions of beta-bar for each increment of lambda. but how to build such an algorithm to obtain the lines correctly.
(i think solutions of beta-bar should lie in an n*m matrix)
would you in general show the methods of plotting such problem? thanks.
one of my reasons is discontinuity of "ezplot" command for my equation.
ok here is my pic:
alt text http://www.mojoimage.com/free-image-hosting-view-05.php?id=5039TE-beta-bar-L-n2-.png
or
http://www.mojoimage.com/free-image-hosting-05/5039TE-beta-bar-L-n2-.pngFree Image Hosting
and my code (in short):
h=ezplot('f1',[0.8,1.8,0.7,1.0]);
and in another m.file
function y=f1(lambda,betab)
n1=1.5; n2=1; z0=120*pi;
d1=1; d2=1; a=1;
k0=2*pi/lambda;
u= sqrt(n1^2-betab^2);
wb= sqrt(n2^2-betab^2);
uu=k0*u*d1;
wwb=k0*wb*d2 ;
z1=z0/u; z1_b=z1/z0;
a0_b=tan(wwb)/u+tan(uu)/wb;
b0_b=(1/u^2-1/wb^2)*tan(uu)*tan(wwb);
c0_b=1/(u*wb)*(tan(uu)/u+tan(wwb)/wb);
uu0= k0*u*a; m=0;
y=(a0_b*z1_b^2+c0_b)+(a0_b*z1_b^2-c0_b)*...
cos(2*uu0+m*pi)+b0_b*z1_b*sin(2*uu0+m*pi);
end
fzero cant find roots; it says "Function value must be real and finite".
anyway, is it possible to eliminate discontinuity and only plot real zeros of y?
heretofore,for another function (namely fTE), which is :
function y=fTE(lambda,betab,s)
m=s;
n1=1.5; n2=1;
d1=1; d2=1; a=1;
z0=120*pi;
k0=2*pi/lambda;
u = sqrt(n1^2-betab^2);
w = sqrt(betab^2-n2^2);
U = k0*u*d1;
W = k0*w*d2 ;
z1 = z0/u; z1_b = z1/z0;
a0_b = tanh(W)/u-tan(U)/w;
b0_b = (1/u^2+1/w^2)*tan(U)*tanh(W);
c0_b = -(tan(U)/u+tanh(W)/w)/(u*w);
U0 = k0*u*a;
y = (a0_b*z1_b^2+c0_b)+(a0_b*z1_b^2-c0_b)*cos(2*U0+m*pi)...
+ b0_b*z1_b*sin(2*U0+m*pi);
end
i'd plotted real zeros of "y" by these codes:
s=0; % s=0 for even modes and s=1 for odd modes.
lmin=0.8; lmax=1.8;
bmin=1; bmax=1.5;
lam=linspace(lmin,lmax,1000);
for n=1:length(lam)
increment=0.001; tolerence=1e-14; xstart=bmax-increment;
x=xstart;
dx=increment;
m=0;
while x > bmin
while dx/x >= tolerence
if fTE(lam(n),x,s)*fTE(lam(n),x-dx,s)<0
dx=dx/2;
else
x=x-dx;
end
end
if abs(real(fTE(lam(n),x,s))) < 1e-6 %because of discontinuity some answers are not correct.%
m=m+1;
r(n,m)=x;
end
dx=increment;
x=0.99*x;
end
end
figure
hold on,plot(lam,r(:,1),'k'),plot(lam,r(:,2),'c'),plot(lam,r(:,3),'m'),
xlim([lmin,lmax]);ylim([1,1.5]),
xlabel('\lambda(\mum)'),ylabel('\beta-bar')
you see i use matrix to save data for this plot.
![alt text][2]
because here lines start from left(axis) to rigth. but if the first line(upper) starts someplace from up to rigth(for the first figure and f1 function), then i dont know how to use matrix. lets improve this method.
[2]: http://www.mojoimage.com/free-image-hosting-05/2812untitled.pngFree Image Hosting
Sometimes EZPLOT will display discontinuities because there really are discontinuities or some form of complicated behavior of the function occurring there. You can see this by generating your plot in an alternative way using the CONTOUR function.
You should first modify your f1 function by replacing the arithmetic operators (*, /, and ^) with their element-wise equivalents (.*, ./, and .^) so that f1 can accept matrix inputs for lambda and betab. Then, run the code below:
lambda = linspace(0.8,1.8,500); %# Create a vector of 500 lambda values
betab = linspace(0.7,1,500); %# Create a vector of 500 betab values
[L,B] = meshgrid(lambda,betab); %# Create 2-D grids of values
y = f1(L,B); %# Evaluate f1 at every point in the grid
[c,h] = contour(L,B,y,[0 0]); %# Plot contour lines for the value 0
set(h,'Color','b'); %# Change the lines to blue
xlabel('\lambda'); %# Add an x label
ylabel('$\overline{\beta}$','Interpreter','latex'); %# Add a y label
title('y = 0'); %# Add a title
And you should see the following plot:
Notice that there are now additional lines in the plot that did not appear when using EZPLOT, and these lines are very jagged. You can zoom in on the crossing at the top left and make a plot using SURF to get an idea of what's going on:
lambda = linspace(0.85,0.95,100); %# Some new lambda values
betab = linspace(0.95,1,100); %# Some new betab values
[L,B] = meshgrid(lambda,betab); %# Create 2-D grids of values
y = f1(L,B); %# Evaluate f1 at every point in the grid
surf(L,B,y); %# Make a 3-D surface plot of y
axis([0.85 0.95 0.95 1 -5000 5000]); %# Change the axes limits
xlabel('\lambda'); %# Add an x label
ylabel('$\overline{\beta}$','Interpreter','latex'); %# Add a y label
zlabel('y'); %# Add a z label
Notice that there is a lot of high-frequency periodic activity going on along those additional lines, which is why they look so jagged in the contour plot. This is also why a very general utility like EZPLOT was displaying a break in the lines there, since it really isn't designed to handle specific cases of complicated and poorly behaved functions.
EDIT: (response to comments)
These additional lines may not be true zero crossings, although it is difficult to tell from the SURF plot. There may be a discontinuity at those lines, where the function shoots off to -Inf on one side of the line and Inf on the other side of the line. When rendering the surface or computing the contour, these points on either side of the line may be mistakenly connected, giving the false appearance of a zero crossing along the line.
If you want to find a zero crossing given a value of lambda, you can try using the function FZERO along with an anonymous function to turn your function of two variables f1 into a function of one variable fcn:
lambda_zero = 1.5; %# The value of lambda at the zero crossing
fcn = #(x) f1(lambda_zero,x); %# A function of one variable (lambda is fixed)
betab_zero = fzero(fcn,0.94); %# Find the value of betab at the zero crossing,
%# using 0.94 as an initial guess