Using MatLab, I know how to create a line segment connecting two points using this code:
line([0 1],[0 1])
This draws a straight line segment from the point (0,0) to the point (1,1).
What I am trying to do is continue that line to the edge of the plot. Rather than just drawing a line between these two points I want to draw a line through those two points that spans the entire figure for any set of two points.
For this particular line and a x=-10:10, y=-10:10 plot I could write:
line([-10 10], [-10 10]);
But I would need to generalize this for any set of points.
Solve the line equation going through those two points:
y = a*x + b;
for a and b:
a = (yp(2)-yp(1)) / (xp(2)-xp(1));
b = yp(1)-a*xp(1);
Find the edges of the plotting window
xlims = xlim(gca);
ylims = ylim(gca);
or take a edges far away, so you can still zoomout, later reset the x/y limits.
or if there is no plot at the moment, define your desired edges:
xlims = [-10 10];
ylims = [-10 10];
Fill in those edges into the line equation and plot the corresponding points:
y = xlims*a+b;
line( xlims, y );
And reset the edges
xlim(xlims);
ylim(ylims);
There is one special case, the vertical line, which you'll have to take care of separately.
What about
function = long_line(X,Y,sym_len)
dir = (Y-X)/norm(Y-X);
Yp = X + dir*sym_len;
Xp = X - dir*sym_len;
line(Xp,Yp);
end
being sym_len one half of the expected length of the plotted line around X?
Related
I want to make a figure in MATLAB as described in the following image
What I did is the following:
x = [1 2 3];
y = [2 2 4];
radius = [1 1.2 2.2];
theta = [-pi 0 pi];
figure;
scatter(x,y,radius)
How do I add an angle theta to the plot to represent a complex number z = radius.*exp(1j*theta) at every spacial coordinates?
Technically speaking, those are only circles if x and y axes are scaled equally. That is because scatter always plots circles, independently of the scales (and they remain circles if you zoom in nonuniformly. + you have the problem with the line, which should indicate the angle...
You can solve both issues by drawing the circles:
function plotCirc(x,y,r,theta)
% calculate "points" where you want to draw approximate a circle
ang = 0:0.01:2*pi+.01;
xp = r*cos(ang);
yp = r*sin(ang);
% calculate start and end point to indicate the angle (starting at math=0, i.e. right, horizontal)
xt = x + [0 r*sin(theta)];
yt = y + [0 r*cos(theta)];
% plot with color: b-blue
plot(x+xp,y+yp,'b', xt,yt,'b');
end
having this little function, you can call it to draw as many circles as you want
x = [1 2 3];
y = [2 2 4];
radius = [1 1.2 2.2];
theta = [-pi 0 pi];
figure
hold on
for i = 1:length(x)
plotCirc(x(i),y(i),radius(i),theta(i))
end
I went back over scatter again, and it looks like you can't get that directly from the function. Hopefully there's a clean built-in way to do this, and someone else will chime in with it, but as a backup plan, you can just add the lines yourself.
You'd want a number of lines that's the same as the length of your coordinate set, from the center point to the edge at the target angle, and fortunately 'line' does multiple lines if you feed it a matrix.
You could just tack this on to the end of your code to get the angled line:
x_lines = [x; x + radius.*cos(theta)];
y_lines = [y; y + radius.*sin(theta)];
line(x_lines, y_lines, 'Color', 'b')
I had to assign the color specifically, since otherwise 'line' makes each new line cycle through the default colors, but that also means you could easily change the line color to stand out more. There's also no center dot, but that'd just be a second scatter plot with tiny radius. Should plot most of what you're looking for, at least.
(My version of Matlab is old enough that scatter behaves differently, so I can only check the line part, but they have the right length and location.)
Edit: Other answer makes a good point on whether scatter is appropriate here. Probably better to draw the circle too.
I'm trying to fill the intersecting area between two circles in Matlab. I've quite literally copied and pasted this piece of code from this article on Matlab Central.
t = linspace(0, 2*pi, 100);
cir = #(r,ctr) [r*cos(t)+ctr(1); r*sin(t)+ctr(2)]; % Circle Function
c1 = cir(1.0, [0; 0]);
c2 = cir(1.5, [1; 1]);
in1 = find(inpolygon(c1(1,:), c1(2,:), c2(1,:), c2(2,:))); % Circle #1 Points Inside Circle #2
in2 = find(inpolygon(c2(1,:), c2(2,:), c1(1,:), c1(2,:))); % Circle #2 Points Inside Circle #1
[fillx,ix] = sort([c1(1,in1) c2(1,in2)]); % Sort Points
filly = [c1(2,in1) (c2(2,in2))];
filly = filly(ix);
figure(1)
plot(c1(1,:), c1(2,:))
hold on
plot(c2(1,:), c2(2,:))
fill([fillx fliplr(fillx)], [filly fliplr(filly)], 'g', 'EdgeColor','none')
hold off
axis square
What I end up with is the following image:
However, it should appear as this image:
Why is the area not being filled as it is in the example article?
If you have Mapping Toolbox you can use polybool to find the intersection between to polygones, and than patch (which dosen't require Mapping Toolbox, and is better than fill) to draw it. The folowing code works even without the first 2 lines that use poly2cw, but it issues some warnings. This can be solved with the poly2cw trasfomation:
[c1(1,:), c1(2,:)] = poly2cw(c1(1,:), c1(2,:)); % clock-wise transform
[c2(1,:), c2(2,:)] = poly2cw(c2(1,:), c2(2,:)); % clock-wise transform
[xb, yb] = polybool('intersection',c1(1,:),c1(2,:),c2(1,:), c2(2,:));
plot(c1(1,:), c1(2,:))
hold on
plot(c2(1,:), c2(2,:))
patch(xb, yb, 1, 'FaceColor', 'g','EdgeColor','none')
axis equal
The code in the question does not work because it has some mistake in the order of the points in fillx and filly. We can see that if we set the 'EdgeColor' to be visible, and follow the peripheral line of the patch (see below, here reduced to 20 points, for the illustration). We can clearly see that the point of the polygon to be filled are ordered in a 'zig-zag' between the circles, so it has no area at all. I have numbered the vertices of the polygon taken from each circle to demonstrate in what order the fill function read them.
In order to fill with color all the intersection between the circles, we need to define the 'polygon' by the points on the circles that intersect (in1 and in2) in the right order. This means we want them to make a closed shape if an imaginary pencil is drawing a line between them in the same order they are given. Like this:
We start with 1 in one of the circles and continue until the numbers on that circle are ended, and then move to 1 on the other circle, and when we reach the end in the second circle we close the polygon by connecting the last point to the first one. As you can see in both figure above, the starting point and the end point of the circles are really close, so we get 2 numbers above each other.
How can we order the points correctly?
We start by getting in1 and in2 as described in the question. Let's take a look at in1:
in1 =
1 2 3 4 5 6 7 19 20
Those are the indices of the points from c1 to be taken, they appear to be in order, but contains a gap. This gap is because inpolygon checks the points by the order in c1, and the starting point of c1 is within the intersection region. So we get the first 7 points, then we are out of the intersection, and we go back in as we reach points 19 and 20. However, for our polygon, we need this points to start from the closest point to one of that places where the circle intersect and to go around the circle until we reach the second point of intersection.
To do that, we look for the 'gap' in the points order:
gap = find(diff(in1)>1);
and reorder them correctly:
X1 = [c1(1,in1(gap+1:end)) c1(1,in1(1:gap))];
Y1 = [c1(2,in1(gap+1:end)) c1(2,in1(1:gap))];
But, there may be no 'gap', as we see in in2:
in2 =
11 12 13 14
So we need to wrap this within an if to check if the points need to be reordered:
if ~isempty(gap)
X1 = [c1(1,in1(gap+1:end)) c1(1,in1(1:gap))];
Y1 = [c1(2,in1(gap+1:end)) c1(2,in1(1:gap))];
else
X1 = c1(1,in1);
Y1 = c1(2,in1);
end
Now all we need to do it to concatenate X1 and X2 (for circle 2), and the same for the Ys, and use patch (which is like fill, but better) to draw it:
patch([X1 X2],[Y1 Y2],'g','EdgeColor','none')
With 20 points the circle is not really a circle and the intersection is partially colored, so here is the full code and the result with 200 points:
t = linspace(0, 2*pi, 200);
cir = #(r,ctr) [r*cos(t)+ctr(1); r*sin(t)+ctr(2)]; % Circle Function
c1 = cir(1.0, [0; 0]);
c2 = cir(1.5, [1; 1]);
plot(c1(1,:), c1(2,:))
hold on
plot(c2(1,:), c2(2,:))
axis equal
in1 = find(inpolygon(c1(1,:), c1(2,:), c2(1,:), c2(2,:)));
in2 = find(inpolygon(c2(1,:), c2(2,:), c1(1,:), c1(2,:)));
gap = find(diff(in1)>1);
if ~isempty(gap)
X1 = [c1(1,in1(gap+1:end)) c1(1,in1(1:gap))];
Y1 = [c1(2,in1(gap+1:end)) c1(2,in1(1:gap))];
else
X1 = c1(1,in1);
Y1 = c1(2,in1);
end
gap = find(diff(in2)>1);
if ~isempty(gap)
X2 = [c2(1,in2(gap+1:end)) c2(1,in2(1:gap))];
Y2 = [c2(2,in2(gap+1:end)) c2(2,in2(1:gap))];
else
X2 = c2(1,in2);
Y2 = c2(2,in2);
end
patch([X1 X2],[Y1 Y2],'g','EdgeColor','none')
hold off
All mentioned above could be replace with the use of convhull on the vertices that intersect and give the same result:
x = [c1(1,in1) c2(1,in2)]; % all x's for intersecting vertices
y = [c1(2,in1) c2(2,in2)]; % all y's for intersecting vertices
k = convhull(x,y); % calculate the convex polygon
patch(x(k),y(k),'g','EdgeColor','none')
I have two equations:
ellipseOne = '((x-1)^2)/6^2 + y^2/3^2 = 1';
and
ellipseTwo = '((x+2)^2)/2^2 + ((y-5)^2)/4^2 = 1';
and I plotted them:
ezplot(ellipseOne, [-10, 10, -10, 10])
hold on
ezplot(ellipseTwo, [-10, 10, -10, 10])
title('Ellipses')
hold off
Now I'm trying to find the intersection of the two ellipses. I tried:
intersection = solve(ellipseOne, ellipseTwo)
intersection.x
intersection.y
to find the points where they intersect, but MATLAB is giving me a matrix and an equation as an answer which I don't understand. Could anyone point me in the right direction to get the coordinates of intersection?
The solution is in symbolic form. The last step you need to take is to transform it into numeric. Simply convert the result using double. However, because this is a pair of quadratic equations, there are 4 possible solutions due to the sign ambiguity (i.e. +/-) and can possibly give imaginary roots. Therefore, isolate out the real solutions and what is left should be your answer.
Therefore:
% Your code
ellipseOne = '((x-1)^2)/6^2 + y^2/3^2 = 1';
ellipseTwo = '((x+2)^2)/2^2 + ((y-5)^2)/4^2 = 1';
intersection = solve(ellipseOne, ellipseTwo);
% Find the points of intersection
X = double(intersection.x);
Y = double(intersection.y);
mask = ~any(imag(X), 2) | ~any(imag(Y), 2);
X = X(mask); Y = Y(mask);
The first three lines of code are what you did. The next four lines extract out the roots in numeric form, then I create a logical mask that isolates out only the points that are real. This looks at the imaginary portion of both the X and Y coordinate of all of the roots and if there is such a component for any of the roots, we want to eliminate those. We finally remove the roots that are imaginary and we should be left with two real roots.
We thus get for our intersection points:
>> disp([X Y])
-3.3574 2.0623
-0.2886 2.9300
The first column is the X coordinate and the second column is the Y coordinate. This also seems to agree with the plot. I'll take your ellipse plotting code and also insert the intersection points as blue dots using the above solution:
ezplot(ellipseOne, [-10, 10, -10, 10])
hold on
ezplot(ellipseTwo, [-10, 10, -10, 10])
% New - place intersection points
plot(X, Y, 'b.');
title('Ellipses');
hold off;
I want to plot a line at a certain "x" (being a scalar) value so every second, a vertical line with certain values will be drawn. I know that one way to plot a vertical line (without considering time) is to declare a vector like:
y=0:0.01:5;
or something like:
y=3:0.01:6;
and write plot(x,y);
Also, I've done "animated" plots using the "pause" parameter but I don't know how to do that in this case. Thanks for your help.
Another way I can suggest is to use plot in combination with hold on. With plot, you'd only specify two points where each point has the same x value, but the y value can change to whatever you want. When you draw points using plot, the default behaviour is that a line is drawn in between the points. If we specify two points having the same x coordinate, but different y coordinates, we would essentially draw a vertical line in between these points.
For example, given that you want to have every "second" have a vertical line, we can do something like this:
ystart = [-1 -2 -3 -1 -2 -3];
yend = [1 2 3 1 2 3];
figure; hold on;
for idx = 1 : numel(ystart)
plot([idx idx], [ystart(idx) yend(idx)]);
end
This is what we get:
We define two arrays of 6 elements where ystart denote the starting y point and yend denote the ending y point. We spawn a new figure, use hold on to plot multiple lines on the same graph, then use a for loop with plot so that we draw a line in between two points: (x,y) = (idx, ystart(idx)) and (x,y) = (idx, yend(idx)). idx goes from 1 to 6. Obviously, you can change the location of where the x values are being plotted by specifying another array... call it x:
x = 0:2:10; %// Time values
ystart = [-1 -2 -3 -1 -2 -3];
yend = [1 2 3 1 2 3];
figure; hold on;
for idx = 1 : numel(ystart)
plot([x(idx) x(idx)], [ystart(idx) yend(idx)]);
end
Here, we will draw vertical lines starting from x = 0 up to x = 10 in steps of 2. The result would basically be the same, but the x values would be different, as well as where the vertical lines are being drawn.
There are several ways to plot vertical lines in Matlab. The easiest recommendation is the line function:
line(X,Y) adds the line defined in vectors X and Y to the current
axes. If X and Y are matrices of the same size, line draws one line
per column.
Call this as many times as you want.
h=line(X,Y)
will give you properties of the line
Another way is to
X= X0*(Y./Y)
then
plot(X,Y)
will plot a vertical line at the point X0. Another thing you can do is draw a line using the above line function from the point (X0, min(Y)) to the point (X0, max(Y)) which is the most elegant solution. If you are trying to create a movie, you will need to access the properties of this line using h=line(X,Y). To move the line to a new position, you will have to set the properties of this line by calling set(h, Property, value). For example in your movie, you need to move it to a new position so you will set that property. This way, by minimum change of data, you can move a line or show its accelerating.
I'm looking to create a "web" between a set of points where the data tells whether there is a link between any two points.
The way I thought of would be by plotting every couple points, and overlaying each couple on top of eachother.
However, if there is a way to just simple draw a line between two points that would be much easier.
Any help would be appreciated!
If you can organize the x and y coordinates of your line segments into 2-by-N arrays, you can use the function PLOT to plot each column of the matrices as a line. Here's a simple example to draw the four lines of a unit square:
x = [0 1 1 0; ...
1 1 0 0];
y = [0 0 1 1; ...
0 1 1 0];
plot(x,y);
This will plot each line in a different color. To plot all of the lines as black, do this:
plot(x,y,'k');
Use plot. Suppose your two points are a = [x1 y1] and b = [x2 y2], then:
plot([x1 x2],[y1 y2]);
If you meant by I'm looking to create a "web" between a set of points where the data tells whether there is a link between any two points actually some kind of graph represented by its adjacency matrix (opposite to other answers simple means to connect points), then:
this gplot function may indeed be the proper tool for you. It's the basic visualization tool to plot nodes and links of a graph represented as a adjacency matrix.
use this function:
function [] = drawline(p1, p2 ,color)
%enter code here
theta = atan2( p2(2) - p1(2), p2(1) - p1(1));
r = sqrt( (p2(1) - p1(1))^2 + (p2(2) - p1(2))^2);
line = 0:0.01: r;
x = p1(1) + line*cos(theta);
y = p1(2) + line*sin(theta);
plot(x, y , color)
call it like:
drawline([fx(i) fy(i)] ,[y(i,1) y(i,2)],'red')
Credit: http://www.mathworks.com/matlabcentral/answers/108652-draw-lines-between-points#answer_139175
Lets say you want a line with coordinates (x1,y1) and (x2,y2). Then you make a vector with the x and y coordinates: x = [x1 x2] and y=[y1 y2].
Matlab has a function called 'Line', this is used in this way:
line(x,y)
If you want to see the effect of drawing lines, you can use plot inside for loop note that data is a n*2 matrix containing the 'x,y' of 'n' points
clf(figure(3))
for i = 1 : length(data)-1
plot([data(i,1),data(i+1,1)], [data(i,2),data(i+1,2)], '-*');
hold on
end
hold off
Or can use this statement to draw it in one step
plot(data(:,1), data(:,2), '-*');