ans = 
Main Content
Results for
D.R. Kaprekar was a self taught recreational mathematician, perhaps known mostly for some numbers that bear his name.
Today, I'll focus on Kaprekar's constant (as opposed to Kaprekar numbers.)
The idea is a simple one, embodied in these 5 steps.
1. Take any 4 digit integer, reduce to its decimal digits.
2. Sort the digits in decreasing order.
3. Flip the sequence of those digits, then recompose the two sets of sorted digits into 4 digit numbers. If there were any 0 digits, they will become leading zeros on the smaller number. In this case, a leading zero is acceptable to consider a number as a 4 digit integer.
4. Subtract the two numbers, smaller from the larger. The result will always have no more than 4 decimal digits. If it is less than 1000, then presume there are leading zero digits.
5. If necessary, repeat the above operation, until the result converges to a stable result, or until you see a cycle.
Since this process is deterministic, and must always result in a new 4 digit integer, it must either terminate at either an absorbing state, or in a cycle.
For example, consider the number 6174.
7641 - 1467
We get 6174 directly back. That seems rather surprising to me. But even more interesting is you will find all 4 digit numbers (excluding the pure rep-digit nmbers) will always terminate at 6174, after at most a few steps. For example, if we start with 1234
4321 - 1234
8730 - 0378
8532 - 2358
and we see that after 3 iterations of this process, we end at 6174. Similarly, if we start with 9998, it too maps to 6174 after 5 iterations.
9998 ==> 999 ==> 8991 ==> 8082 ==> 8532 ==> 6174
Why should that happen? That is, why should 6174 always drop out in the end? Clearly, since this is a deterministic proces which always produces another 4 digit integer (Assuming we treat integers with a leading zero as 4 digit integers), we must either end in some cycle, or we must end at some absorbing state. But for all (non-pure rep-digit) starting points to end at the same place, it seems just a bit surprising.
I always like to start a problem by working on a simpler problem, and see if it gives me some intuition about the process. I'll do the same thing here, but with a pair of two digit numbers. There are 100 possible two digit numbers, since we must treat all one digit numbers as having a "tens" digit of 0.
N = (0:99)';
Next, form the Kaprekar mapping for 2 digit numbers. This is easier than you may think, since we can do it in a very few lines of code on all possible inputs.
Ndig = dec2base(N,10,2) - '0';
Nmap = sort(Ndig,2,'descend')*[10;1] - sort(Ndig,2,'ascend')*[10;1];
I'll turn it into a graph, so we can visualize what happens. It also gives me an excuse to employ a very pretty set of tools in MATLAB.
G2 = graph(N+1,Nmap+1,[],cellstr(dec2base(N,10,2)));
plot(G2)
Do you see what happens? All of the rep-digit numbers, like 11, 44, 55, etc., all map directly to 0, and they stay there, since 0 also maps into 0. We can see that in the star on the lower right.
G2cycles = cyclebasis(G2)
G2cycles{1}
All other numbers eventually end up in the cycle:
G2cycles{2}
That is
81 ==> 63 ==> 27 ==> 45 ==> 09 ==> and back to 81
looping forever.
Another way of trying to visualize what happens with 2 digit numbers is to use symbolics. Thus, if we assume any 2 digit number can be written as 10*T+U, where I'll assume T>=U, since we always sort the digits first
syms T U
(10*T + U) - (10*U+T)
So after one iteration for 2 digit numbers, the result maps ALWAYS to a new 2 digit number that is divisible by 9. And there are only 10 such 2 digit numbers that are divisible by 9. So the 2-digit case must resolve itself rather quickly.
What happens when we move to 3 digit numbers? Note that for any 3 digit number abc (without loss of generality, assume a >= b >= c) it almost looks like it reduces to the 2 digit probem, aince we have abc - cba. The middle digit will always cancel itself in the subtraction operation. Does that mean we should expect a cycle at the end, as happens with 2 digit numbers? A simple modification to our previous code will tell us the answer.
N = (0:999)';
Ndig = dec2base(N,10,3) - '0';
Nmap = sort(Ndig,2,'descend')*[100;10;1] - sort(Ndig,2,'ascend')*[100;10;1];
G3 = graph(N+1,Nmap+1,[],cellstr(dec2base(N,10,2)));
plot(G3)
This one is more difficult to visualize, since there are 1000 nodes in the graph. However, we can clearly see two disjoint groups.
We can use cyclebasis to tell us the complete story again.
G3cycles = cyclebasis(G3)
G3cycles{:}
And we see that all 3 digit numbers must either terminate at 000, or 495. For example, if we start with 181, we would see:
811 - 118
963 - 369
954 - 459
It will terminate there, forever trapped at 495. And cyclebasis tells us there are no other cycles besides the boring one at 000.
What is the maximum length of any such path to get to 495?
D3 = distances(G3,496) % Remember, MATLAB uses an index origin of 1
D3(isinf(D3)) = -inf; % some nodes can never reach 495, so they have an infinite distance
plot(D3)
The maximum number of steps to get to 495 is 6 steps.
find(D3 == 6) - 1
So the 3 digit number 100 required 6 iterations to eventually reach 495.
shortestpath(G3,101,496) - 1
I think I've rather exhausted the 3 digit case. It is time now to move to the 4 digit problem, but we've already done all the hard work. The same scheme will apply to compute a graph. And the graph theory tools do all the hard work for us.
N = (0:9999)';
Ndig = dec2base(N,10,4) - '0';
Nmap = sort(Ndig,2,'descend')*[1000;100;10;1] - sort(Ndig,2,'ascend')*[1000;100;10;1];
G4 = graph(N+1,Nmap+1,[],cellstr(dec2base(N,10,2)));
plot(G4)
cyclebasis(G4)
ans{:}
And here we see the behavior, with one stable final point, 6174 as the only non-zero ending state. There are no circular cycles as we had for the 2-digit case.
How many iterations were necessary at most before termination?
D4 = distances(G4,6175);
D4(isinf(D4)) = -inf;
plot(D4)
The plot tells the story here. The maximum number of iterations before termination is 7 for the 4 digit case.
find(D4 == 7,1,'last') - 1
shortestpath(G4,9986,6175) - 1
Can you go further? Are there 5 or 6 digit Kaprekar constants? Sadly, I have read that for more than 4 digits, things break down a bit, there is no 5 digit (or higher) Kaprekar constant.
We can verify that fact, at least for 5 digit numbers.
N = (0:99999)';
Ndig = dec2base(N,10,5) - '0';
Nmap = sort(Ndig,2,'descend')*[10000;1000;100;10;1] - sort(Ndig,2,'ascend')*[10000;1000;100;10;1];
G5 = graph(N+1,Nmap+1,[],cellstr(dec2base(N,10,2)));
plot(G5)
cyclebasis(G5)
ans{:}
The result here are 4 disjoint cycles. Of course the rep-digit cycle must always be on its own, but the other three cycles are also fully disjoint, and are of respective length 2, 4, and 4.
Following on from my previous post The Non-Chaotic Duffing Equation, now we will study the chaotic behaviour of the Duffing Equation
P.s:Any comments/advice on improving the code is welcome.
The Original Duffing Equation is the following:
Let 
. This implies that
. This implies thatThen we rewrite it as a System of First-Order Equations
Using the substitution 
for 
, the second-order equation can be transformed into the following system of first-order equations:
for , the second-order equation can be transformed into the following system of first-order equations:
Exploring the Effect of γ.
% Define parameters
gamma = 0.1;
alpha = -1;
beta = 1;
delta = 0.1;
omega = 1.4;
% Define the system of equations
odeSystem = @(t, y) [y(2);
-delta*y(2) - alpha*y(1) - beta*y(1)^3 + gamma*cos(omega*t)];
% Initial conditions
y0 = [0; 0]; % x(0) = 0, v(0) = 0
% Time span
tspan = [0 200];
% Solve the system
[t, y] = ode45(odeSystem, tspan, y0);
% Plot the results
figure;
plot(t, y(:, 1));
xlabel('Time');
ylabel('x(t)');
title('Solution of the nonlinear system');
grid on;
% Plot the phase portrait
figure;
plot(y(:, 1), y(:, 2));
xlabel('x(t)');
ylabel('v(t)');
title('Phase Portrait');
grid on;
% Define the tail (e.g., last 10% of the time interval)
tail_start = floor(0.9 * length(t)); % Starting index for the tail
tail_end = length(t); % Ending index for the tail
% Plot the tail of the solution
figure;
plot(y(tail_start:tail_end, 1), y(tail_start:tail_end, 2), 'r', 'LineWidth', 1.5);
xlabel('x(t)');
ylabel('v(t)');
title('Phase Portrait - Tail of the Solution');
grid on;
% Define parameters
gamma = 0.318;
alpha = -1;
beta = 1;
delta = 0.1;
omega = 1.4;
% Define the system of equations
odeSystem = @(t, y) [y(2);
-delta*y(2) - alpha*y(1) - beta*y(1)^3 + gamma*cos(omega*t)];
% Initial conditions
y0 = [0; 0]; % x(0) = 0, v(0) = 0
% Time span
tspan = [0 800];
% Solve the system
[t, y] = ode45(odeSystem, tspan, y0);
% Plot the results
figure;
plot(t, y(:, 1));
xlabel('Time');
ylabel('x(t)');
title('Solution of the nonlinear system');
grid on;
% Plot the phase portrait
figure;
plot(y(:, 1), y(:, 2));
xlabel('x(t)');
ylabel('v(t)');
title('Phase Portrait');
grid on;
% Define the tail (e.g., last 10% of the time interval)
tail_start = floor(0.9 * length(t)); % Starting index for the tail
tail_end = length(t); % Ending index for the tail
% Plot the tail of the solution
figure;
plot(y(tail_start:tail_end, 1), y(tail_start:tail_end, 2), 'r', 'LineWidth', 1.5);
xlabel('x(t)');
ylabel('v(t)');
title('Phase Portrait - Tail of the Solution');
grid on;
% Define parameters
gamma = 0.338;
alpha = -1;
beta = 1;
delta = 0.1;
omega = 1.4;
% Define the system of equations
odeSystem = @(t, y) [y(2);
-delta*y(2) - alpha*y(1) - beta*y(1)^3 + gamma*cos(omega*t)];
% Initial conditions
y0 = [0; 0]; % x(0) = 0, v(0) = 0
% Time span with more points for better resolution
tspan = linspace(0, 200,2000); % Increase the number of points
% Solve the system
[t, y] = ode45(odeSystem, tspan, y0);
% Plot the results
figure;
plot(t, y(:, 1));
xlabel('Time');
ylabel('x(t)');
title('Solution of the nonlinear system');
grid on;
% Plot the phase portrait
figure;
plot(y(:, 1), y(:, 2));
xlabel('x(t)');
ylabel('v(t)');
title('Phase Portrait');
grid on;
% Define the tail (e.g., last 10% of the time interval)
tail_start = floor(0.9 * length(t)); % Starting index for the tail
tail_end = length(t); % Ending index for the tail
% Plot the tail of the solution
figure;
plot(y(tail_start:tail_end, 1), y(tail_start:tail_end, 2), 'r', 'LineWidth', 1.5);
xlabel('x(t)');
ylabel('v(t)');
title('Phase Portrait - Tail of the Solution');
grid on;
ax = gca;
chart = ax.Children(1);
datatip(chart,0.5581,-0.1126);
% Define parameters
gamma = 0.35;
alpha = -1;
beta = 1;
delta = 0.1;
omega = 1.4;
% Define the system of equations
odeSystem = @(t, y) [y(2);
-delta*y(2) - alpha*y(1) - beta*y(1)^3 + gamma*cos(omega*t)];
% Initial conditions
y0 = [0; 0]; % x(0) = 0, v(0) = 0
% Time span with more points for better resolution
tspan = linspace(0, 400,3000); % Increase the number of points
% Solve the system
[t, y] = ode45(odeSystem, tspan, y0);
% Plot the results
figure;
plot(t, y(:, 1));
xlabel('Time');
ylabel('x(t)');
title('Solution of the nonlinear system');
grid on;
% Plot the phase portrait
figure;
plot(y(:, 1), y(:, 2));
xlabel('x(t)');
ylabel('v(t)');
title('Phase Portrait');
grid on;
% Define the tail (e.g., last 10% of the time interval)
tail_start = floor(0.9 * length(t)); % Starting index for the tail
tail_end = length(t); % Ending index for the tail
% Plot the tail of the solution
figure;
plot(y(tail_start:tail_end, 1), y(tail_start:tail_end, 2), 'r', 'LineWidth', 1.5);
xlabel('x(t)');
ylabel('v(t)');
title('Phase Portrait - Tail of the Solution');
grid on;
Studying the attached document Duffing Equation from the University of Colorado, I noticed that there is an analysis of The Non-Chaotic Duffing Equation and all the graphs were created with Matlab. And since the code is not given I took the initiative to try to create the same graphs with the following code.
- Plotting the Potential Energy and Identifying Extrema
% Define the range of x values
x = linspace(-2, 2, 1000);
% Define the potential function V(x)
V = -x.^2 / 2 + x.^4 / 4;
% Plot the potential function
figure;
plot(x, V, 'LineWidth', 2);
hold on;
% Mark the minima at x = ±1
plot([-1, 1], [-1/4, -1/4], 'ro', 'MarkerSize', 5, 'MarkerFaceColor', 'g');
% Add LaTeX title and labels
title('Duffing Potential Energy: $$V(x) = -\frac{x^2}{2} + \frac{x^4}{4}$$', 'Interpreter', 'latex');
xlabel('$$x$$', 'Interpreter', 'latex');
ylabel('$$V(x)$$','Interpreter', 'latex');
grid on;
hold off;
- Solving and Plotting the Duffing Equation
% Define the system of ODEs for the non-chaotic Duffing equation
duffing_ode = @(t, X) [X(2);
X(1) - X(1).^3];
% Time span for the simulation
tspan = [0 10];
% Initial conditions [x(0), v(0)]
initial_conditions = [1; 1];
% Solve the ODE using ode45
[t, X] = ode45(duffing_ode, tspan, initial_conditions);
% Extract displacement (x) and velocity (v)
x = X(:, 1);
v = X(:, 2);
% Plot both x(t) and v(t) in the same figure
figure;
plot(t, x, 'b-', 'LineWidth', 2); % Plot x(t) with blue line
hold on;
plot(t, v, 'r--', 'LineWidth', 2); % Plot v(t) with red dashed line
% Add title, labels, and legend
title(' Component curve solutions to $$\ddot{x}-x+x^3=0$$','Interpreter', 'latex');
xlabel('t','Interpreter', 'latex');
ylabel('$$x(t) $$ and $$v(t) $$','Interpreter', 'latex');
legend('$$x(t)$$', ' $$v(t)$$','Interpreter', 'latex');
grid on;
hold off;
% Phase portrait with nullclines, equilibria, and vector field
figure;
hold on;
% Plot phase portrait
plot(x, v,'r', 'LineWidth', 2);
% Plot equilibrium points
plot([0 1 -1], [0 0 0], 'ro', 'MarkerSize', 5, 'MarkerFaceColor', 'g');
% Create a grid of points for the vector field
[x_vals, v_vals] = meshgrid(linspace(-2, 2, 20), linspace(-1, 1, 20));
% Compute the vector field components
dxdt = v_vals;
dvdt = x_vals - x_vals.^3;
% Plot the vector field
quiver(x_vals, v_vals, dxdt, dvdt, 'b');
% Set axis limits to [-1, 1]
xlim([-1.7 1.7]);
ylim([-1 1]);
% Labels and title
title('Phase-Plane solutions to $$\ddot{x}-x+x^3=0$$','Interpreter', 'latex');
xlabel('$$ (x)$$','Interpreter', 'latex');
ylabel('$$v(v)$$','Interpreter', 'latex');
grid on;
hold off;
An attractor is called strange if it has a fractal structure, that is if it has non-integer Hausdorff dimension. This is often the case when the dynamics on it are chaotic, but strange nonchaotic attractors also exist. If a strange attractor is chaotic, exhibiting sensitive dependence on initial conditions, then any two arbitrarily close alternative initial points on the attractor, after any of various numbers of iterations, will lead to points that are arbitrarily far apart (subject to the confines of the attractor), and after any of various other numbers of iterations will lead to points that are arbitrarily close together. Thus a dynamic system with a chaotic attractor is locally unstable yet globally stable: once some sequences have entered the attractor, nearby points diverge from one another but never depart from the attractor.
The term strange attractor was coined by David Ruelle and Floris Takens to describe the attractor resulting from a series of bifurcations of a system describing fluid flow. Strange attractors are often differentiable in a few directions, but some are like a Cantor dust, and therefore not differentiable. Strange attractors may also be found in the presence of noise, where they may be shown to support invariant random probability measures of Sinai–Ruelle–Bowen type.
Lorenz
% Lorenz Attractor Parameters
sigma = 10;
beta = 8/3;
rho = 28;
% Lorenz system of differential equations
f = @(t, a) [-sigma*a(1) + sigma*a(2);
rho*a(1) - a(2) - a(1)*a(3);
-beta*a(3) + a(1)*a(2)];
% Time span
tspan = [0 100];
% Initial conditions
a0 = [1 1 1];
% Solve the system using ode45
[t, a] = ode45(f, tspan, a0);
% Plot using scatter3 with time-based color mapping
figure;
scatter3(a(:,1), a(:,2), a(:,3), 5, t, 'filled'); % 5 is the marker size
title('Lorenz Attractor');
xlabel('x(t)');
ylabel('y(t)');
zlabel('z(t)');
grid on;
colorbar; % Add a colorbar to indicate the time mapping
view(3); % Set the view to 3D
Sprott
% Define the parameters
a = 2.07;
b = 1.79;
% Define the system of differential equations
dynamics = @(t, X) [ ...
X(2) + a * X(1) * X(2) + X(1) * X(3); % dx/dt
1 - b * X(1)^2 + X(2) * X(3); % dy/dt
X(1) - X(1)^2 - X(2)^2 % dz/dt
];
% Initial conditions
X0 = [0.63; 0.47; -0.54];
% Time span
tspan = [0 100];
% Solve the system using ode45
[t, X] = ode45(dynamics, tspan, X0);
% Plot the results with color gradient
figure;
colormap(jet); % Set the colormap
c = linspace(1, 10, length(t)); % Color data based on time
% Create a 3D line plot with color based on time
for i = 1:length(t)-1
plot3(X(i:i+1,1), X(i:i+1,2), X(i:i+1,3), 'Color', [0 0.5 0.9]*c(i)/10, 'LineWidth', 1.5);
hold on;
end
% Set plot properties
title('Sprott Attractor');
xlabel('x(t)');
ylabel('y(t)');
zlabel('z(t)');
grid on;
colorbar; % Add a colorbar to indicate the time mapping
view(3); % Set the view to 3D
hold off;
Rössler
% Define the parameters
a = 0.2;
b = 0.2;
c = 5.7;
% Define the system of differential equations
dynamics = @(t, X) [ ...
-(X(2) + X(3)); % dx/dt
X(1) + a * X(2); % dy/dt
b + X(3) * (X(1) - c) % dz/dt
];
% Initial conditions
X0 = [10.0; 0.00; 10.0];
% Time span
tspan = [0 100];
% Solve the system using ode45
[t, X] = ode45(dynamics, tspan, X0);
% Plot the results
figure;
scatter3(X(:,1), X(:,2), X(:,3), 5, t, 'filled');
title('Rössler Attractor');
xlabel('x(t)');
ylabel('y(t)');
zlabel('z(t)');
grid on;
colorbar; % Add a colorbar to indicate the time mapping
view(3); % Set the view to 3D
Rabinovich-Fabrikant
%% Parameters for Rabinovich-Fabrikant Attractor
alpha = 0.14;
gamma = 0.10;
dt = 0.01;
num_steps = 5000;
% Initial conditions
x0 = -1;
y0 = 0;
z0 = 0.5;
% Preallocate arrays for performance
x = zeros(1, num_steps);
y = zeros(1, num_steps);
z = zeros(1, num_steps);
% Set initial values
x(1) = x0;
y(1) = y0;
z(1) = z0;
% Generate the attractor
for i = 1:num_steps-1
x(i+1) = x(i) + dt * (y(i)*(z(i) - 1 + x(i)^2) + gamma*x(i));
y(i+1) = y(i) + dt * (x(i)*(3*z(i) + 1 - x(i)^2) + gamma*y(i));
z(i+1) = z(i) + dt * (-2*z(i)*(alpha + x(i)*y(i)));
end
% Create a time vector for color mapping
t = linspace(0, 100, num_steps);
% Plot using scatter3
figure;
scatter3(x, y, z, 5, t, 'filled'); % 5 is the marker size
title('Rabinovich-Fabrikant Attractor');
xlabel('x(t)');
ylabel('y(t)');
zlabel('z(t)');
grid on;
colorbar; % Add a colorbar to indicate the time mapping
view(3); % Set the view to 3D
References
This project discusses predator-prey system, particularly the Lotka-Volterra equations,which model the interaction between two sprecies: prey and predators. Let's solve the Lotka-Volterra equations numerically and visualize the results.% Define parameters
% Define parameters
alpha = 1.0; % Prey birth rate
beta = 0.1; % Predator success rate
gamma = 1.5; % Predator death rate
delta = 0.075; % Predator reproduction rate
% Define the symbolic variables
syms R W
% Define the equations
eq1 = alpha * R - beta * R * W == 0;
eq2 = delta * R * W - gamma * W == 0;
% Solve the equations
equilibriumPoints = solve([eq1, eq2], [R, W]);
% Extract the equilibrium point values
Req = double(equilibriumPoints.R);
Weq = double(equilibriumPoints.W);
% Display the equilibrium points
equilibriumPointsValues = [Req, Weq]
% Solve the differential equations using ode45
lotkaVolterra = @(t,Y)[alpha*Y(1)-beta*Y(1)*Y(2);
delta*Y(1)*Y(2)-gamma*Y(2)];
% Initial conditions
R0 = 40;
W0 = 9;
Y0 = [R0, W0];
tspan = [0, 100];
% Solve the differential equations
[t, Y] = ode45(lotkaVolterra, tspan, Y0);
% Extract the populations
R = Y(:, 1);
W = Y(:, 2);
% Plot the results
figure;
subplot(2,1,1);
plot(t, R, 'r', 'LineWidth', 1.5);
hold on;
plot(t, W, 'b', 'LineWidth', 1.5);
xlabel('Time (months)');
ylabel('Population');
legend('R', 'W');
grid on;
subplot(2,1,2);
plot(R, W, 'k', 'LineWidth', 1.5);
xlabel('R');
ylabel('W');
grid on;
hold on;
plot(Req, Weq, 'ro', 'MarkerSize', 8, 'MarkerFaceColor', 'r');
legend('Phase Trajectory', 'Equilibrium Point');
Now, we need to handle a modified version of the Lotka-Volterra equations. These modified equations incorporate logistic growth fot the prey population.
These equations are:
% Define parameters
alpha = 1.0;
K = 100; % Carrying Capacity of the prey population
beta = 0.1;
gamma = 1.5;
delta = 0.075;
% Define the symbolic variables
syms R W
% Define the equations
eq1 = alpha*R*(1 - R/K) - beta*R*W == 0;
eq2 = delta*R*W - gamma*W == 0;
% Solve the equations
equilibriumPoints = solve([eq1, eq2], [R, W]);
% Extract the equilibrium point values
Req = double(equilibriumPoints.R);
Weq = double(equilibriumPoints.W);
% Display the equilibrium points
equilibriumPointsValues = [Req, Weq]
% Solve the differential equations using ode45
modified_lv = @(t,Y)[alpha*Y(1)*(1-Y(1)/K)-beta*Y(1)*Y(2);
delta*Y(1)*Y(2)-gamma*Y(2)];
% Initial conditions
R0 = 40;
W0 = 9;
Y0 = [R0, W0];
tspan = [0, 100];
% Solve the differential equations
[t, Y] = ode45(modified_lv, tspan, Y0);
% Extract the populations
R = Y(:, 1);
W = Y(:, 2);
% Plot the results
figure;
subplot(2,1,1);
plot(t, R, 'r', 'LineWidth', 1.5);
hold on;
plot(t, W, 'b', 'LineWidth', 1.5);
xlabel('Time (months)');
ylabel('Population');
legend('R', 'W');
grid on;
subplot(2,1,2);
plot(R, W, 'k', 'LineWidth', 1.5);
xlabel('R');
ylabel('W');
grid on;
hold on;
plot(Req, Weq, 'ro', 'MarkerSize', 8, 'MarkerFaceColor', 'r');
legend('Phase Trajectory', 'Equilibrium Point');
Does your company or organization require that all your Word Documents and Excel workbooks be labeled with a Microsoft Azure Information Protection label or else they can't be saved? These are the labels that are right below the tool ribbon that apply a category label such as "Public", "Business Use", or "Highly Restricted". If so, you can either
- Create and save a "template file" with the desired label and then call copyfile to make a copy of that file and then write your results to the new copy, or
- If using Windows you can create and/or open the file using ActiveX and then apply the desired label from your MATLAB program's code.
For #1 you can do
copyfile(templateFileName, newDataFileName);
writematrix(myData, newDataFileName);
If the template has the AIP label applied to it, then the copy will also inherit the same label.
For #2, here is a demo for how to apply the code using ActiveX.
% Test to set the Microsoft Azure Information Protection label on an Excel workbook.
% Reference support article:
% https://www.mathworks.com/matlabcentral/answers/1901140-why-does-azure-information-protection-popup-pause-the-matlab-script-when-i-use-actxserver?s_tid=ta_ans_results
clc; % Clear the command window.
close all; % Close all figures (except those of imtool.)
clear; % Erase all existing variables. Or clearvars if you want.
workspace; % Make sure the workspace panel is showing.
format compact;
% Define your workbook file name.
excelFullFileName = fullfile(pwd, '\testAIP.xlsx');
% Make sure it exists. Open Excel as an ActiveX server if it does.
if isfile(excelFullFileName)
% If the workbook exists, launch Excel as an ActiveX server.
Excel = actxserver('Excel.Application');
Excel.visible = true; % Make the server visible.
fprintf('Excel opened successfully.\n');
fprintf('Your workbook file exists:\n"%s".\nAbout to try to open it.\n', excelFullFileName);
% Open up the existing workbook named in the variable fullFileName.
Excel.Workbooks.Open(excelFullFileName);
fprintf('Excel opened file successfully.\n');
else
% File does not exist. Alert the user.
warningMessage = sprintf('File does not exist:\n\n"%s"\n', excelFullFileName);
fprintf('%s\n', warningMessage);
errordlg(warningMessage);
return;
end
% If we get here, the workbook file exists and has been opened by Excel.
% Ask Excel for the Microsoft Azure Information Protection (AIP) label of the workbook we just opened.
label = Excel.ActiveWorkbook.SensitivityLabel.GetLabel
% See if there is a label already. If not, these will be null:
existingLabelID = label.LabelId
existingLabelName = label.LabelName
% Create a label.
label = Excel.ActiveWorkbook.SensitivityLabel.CreateLabelInfo
label.LabelId = "a518e53f-798e-43aa-978d-c3fda1f3a682";
label.LabelName = "Business Use";
% Assign the label to the workbook.
fprintf('Setting Microsoft Azure Information Protection to "Business Use", GUID of a518e53f-798e-43aa-978d-c3fda1f3a682\n');
Excel.ActiveWorkbook.SensitivityLabel.SetLabel(label, label);
% Save this workbook with the new AIP setting we just created.
Excel.ActiveWorkbook.Save;
% Shut down Excel.
Excel.ActiveWorkbook.Close;
Excel.Quit;
% Excel is now closed down. Delete the variable from the MATLAB workspace.
clear Excel;
% Now check to see if the AIP label has been set
% by opening up the file in Excel and looking at the AIP banner.
winopen(excelFullFileName)
Note that there is a line in there that gets an AIP label from the existing workbook, if there is one at all. If there is not one, you can set one. But to determine what the proper LabelId (that crazy long hexadecimal number) should be, you will probably need to open an existing document that already has the label that you want set (applied to it) and then read that label with this line:
label = Excel.ActiveWorkbook.SensitivityLabel.GetLabel
This stems purely from some play on my part. Suppose I asked you to work with the sequence formed as 2*n*F_n + 1, where F_n is the n'th Fibonacci number? Part of me would not be surprised to find there is nothing simple we could do. But, then it costs nothing to try, to see where MATLAB can take me in an explorative sense.
n = sym(0:100).';
Fn = fibonacci(n);
Sn = 2*n.*Fn + 1;
Sn(1:10) % A few elements
For kicks, I tried asking ChatGPT. Giving it nothing more than the first 20 members of thse sequence as integers, it decided this is a Perrin sequence, and gave me a recurrence relation, but one that is in fact incorrect. Good effort from the Ai, but a fail in the end.
Is there anything I can do? Try null! (Look carefully at the array generated by Toeplitz. It is at least a pretty way to generate the matrix I needed.)
X = toeplitz(Sn,[1,zeros(1,4)]);
rank(X(5:end,:))
Hmm. So there is no linear combination of those columns that yields all zeros, since the resulting matrix was full rank.
X = toeplitz(Sn,[1,zeros(1,5)]);
rank(X(6:end,:))
But if I take it one step further, we see the above matrix is now rank deficient. What does that tell me? It says there is some simple linear combination of the columns of X(6:end,:) that always yields zero. The previous test tells me there is no shorter constant coefficient recurrence releation, using fewer terms.
null(X(6:end,:))
Let me explain what those coefficients tell me. In fact, they yield a very nice recurrence relation for the sequence S_n, not unlike the original Fibonacci sequence it was based upon.
S(n+1) = 3*S(n) - S_(n-1) - 3*S(n-2) + S(n-3) + S(n-4)
where the first 5 members of that sequence are given as [1 3 5 13 25]. So a 6 term linear constant coefficient recurrence relation. If it reminds you of the generating relation for the Fibonacci sequence, that is good, because it should. (Remember I started the sequence at n==0, IF you decide to test it out.) We can test it out, like this:
SfunM = memoize(@(N) Sfun(N));
SfunM(25)
2*25*fibonacci(sym(25)) + 1
And indeed, it works as expected.
function Sn = Sfun(n)
switch n
case 0
Sn = 1;
case 1
Sn = 3;
case 2
Sn = 5;
case 3
Sn = 13;
case 4
Sn = 25;
otherwise
Sn = Sfun(n-5) + Sfun(n-4) - 3*Sfun(n-3) - Sfun(n-2) +3*Sfun(n-1);
end
end
A beauty of this, is I started from nothing but a sequence of integers, derived from an expression where I had no rational expectation of finding a formula, and out drops something pretty. I might call this explorational mathematics.
The next step of course is to go in the other direction. That is, given the derived recurrence relation, if I substitute the formula for S_n in terms of the Fibonacci numbers, can I prove it is valid in general? (Yes.) After all, without some proof, it may fail for n larger than 100. (I'm not sure how much I can cram into a single discussion, so I'll stop at this point for now. If I see interest in the ideas here, I can proceed further. For example, what was I doing with that sequence in the first place? And of course, can I prove the relation is valid? Can I do so using MATLAB?)
(I'll be honest, starting from scratch, I'm not sure it would have been obvious to find that relation, so null was hugely useful here.)
We are modeling the introduction of a novel pathogen into a completely susceptible population. In the cells below, I have provided you with the Matlab code for a simple stochastic SIR model, implemented using the "GillespieSSA" function
Simulating the stochastic model 100 times for 
Since γ is 0.4 per day, 
per day
per day% Define the parameters
beta = 0.36;
gamma = 0.4;
n_sims = 100;
tf = 100; % Time frame changed to 100
% Calculate R0
R0 = beta / gamma
% Initial state values
initial_state_values = [1000000; 1; 0; 0]; % S, I, R, cum_inc
% Define the propensities and state change matrix
a = @(state) [beta * state(1) * state(2) / 1000000, gamma * state(2)];
nu = [-1, 0; 1, -1; 0, 1; 0, 0];
% Define the Gillespie algorithm function
function [t_values, state_values] = gillespie_ssa(initial_state, a, nu, tf)
t = 0;
state = initial_state(:); % Ensure state is a column vector
t_values = t;
state_values = state';
while t < tf
rates = a(state);
rate_sum = sum(rates);
if rate_sum == 0
break;
end
tau = -log(rand) / rate_sum;
t = t + tau;
r = rand * rate_sum;
cum_sum_rates = cumsum(rates);
reaction_index = find(cum_sum_rates >= r, 1);
state = state + nu(:, reaction_index);
% Update cumulative incidence if infection occurred
if reaction_index == 1
state(4) = state(4) + 1; % Increment cumulative incidence
end
t_values = [t_values; t];
state_values = [state_values; state'];
end
end
% Function to simulate the stochastic model multiple times and plot results
function simulate_stoch_model(beta, gamma, n_sims, tf, initial_state_values, R0, plot_type)
% Define the propensities and state change matrix
a = @(state) [beta * state(1) * state(2) / 1000000, gamma * state(2)];
nu = [-1, 0; 1, -1; 0, 1; 0, 0];
% Set random seed for reproducibility
rng(11);
% Initialize plot
figure;
hold on;
for i = 1:n_sims
[t, output] = gillespie_ssa(initial_state_values, a, nu, tf);
% Check if the simulation had only one step and re-run if necessary
while length(t) == 1
[t, output] = gillespie_ssa(initial_state_values, a, nu, tf);
end
if strcmp(plot_type, 'cumulative_incidence')
plot(t, output(:, 4), 'LineWidth', 2, 'Color', rand(1, 3));
elseif strcmp(plot_type, 'prevalence')
plot(t, output(:, 2), 'LineWidth', 2, 'Color', rand(1, 3));
end
end
xlabel('Time (days)');
if strcmp(plot_type, 'cumulative_incidence')
ylabel('Cumulative Incidence');
ylim([0 inf]);
elseif strcmp(plot_type, 'prevalence')
ylabel('Prevalence of Infection');
ylim([0 50]);
end
title(['Stochastic model output for R0 = ', num2str(R0)]);
subtitle([num2str(n_sims), ' simulations']);
xlim([0 tf]);
grid on;
hold off;
end
% Simulate the model 100 times and plot cumulative incidence
simulate_stoch_model(beta, gamma, n_sims, tf, initial_state_values, R0, 'cumulative_incidence');
% Simulate the model 100 times and plot prevalence
simulate_stoch_model(beta, gamma, n_sims, tf, initial_state_values, R0, 'prevalence');
Base case:
Suppose you need to do a computation many times. We are going to assume that this computation cannot be vectorized. The simplest case is to use a for loop:
number_of_elements = 1e6;
test_fcn = @(x) sqrt(x) / x;
tic
for i = 1:number_of_elements
x(i) = test_fcn(i);
end
t_forward = toc;
disp(t_forward + " seconds")
Preallocation:
This can easily be sped up by preallocating the variable that houses results:
tic
x = zeros(number_of_elements, 1);
for i = 1:number_of_elements
x(i) = test_fcn(i);
end
t_forward_prealloc = toc;
disp(t_forward_prealloc + " seconds")
In this example, preallocation speeds up the loop by a factor of about three to four (running in R2024a). Comment below if you get dramatically different results.
disp(sprintf("%.1f", t_forward / t_forward_prealloc))
Run it in reverse:
Is there a way to skip the explicit preallocation and still be fast? Indeed, there is.
clear x
tic
for i = number_of_elements:-1:1
x(i) = test_fcn(i);
end
t_backward = toc;
disp(t_backward + " seconds")
By running the loop backwards, the preallocation is implicitly performed during the first iteration and the loop runs in about the same time (within statistical noise):
disp(sprintf("%.2f", t_forward_prealloc / t_backward))
Do you get similar results when running this code? Let us know your thoughts in the comments below.
Beneficial side effect:
Have you ever had to use a for loop to delete elements from a vector? If so, keeping track of index offsets can be tricky, as deleting any element shifts all those that come after. By running the for loop in reverse, you don't need to worry about index offsets while deleting elements.
Many times when ploting, we not only need to set the color of the plot, but also its
transparency, Then how we set the alphaData of colorbar at the same time ?
It seems easy to do so :
data = rand(12,12);
% Transparency range 0-1, .3-1 for better appearance here
AData = rescale(- data, .3, 1);
% Draw an imagesc with numerical control over colormap and transparency
imagesc(data, 'AlphaData',AData);
colormap(jet);
ax = gca;
ax.DataAspectRatio = [1,1,1];
ax.TickDir = 'out';
ax.Box = 'off';
% get colorbar object
CBarHdl = colorbar;
pause(1e-16)
% Modify the transparency of the colorbar
CData = CBarHdl.Face.Texture.CData;
ALim = [min(min(AData)), max(max(AData))];
CData(4,:) = uint8(255.*rescale(1:size(CData, 2), ALim(1), ALim(2)));
CBarHdl.Face.Texture.ColorType = 'TrueColorAlpha';
CBarHdl.Face.Texture.CData = CData;
But !!!!!!!!!!!!!!! We cannot preserve the changes when saving them as images ：
It seems that when saving plots, the `Texture` will be refresh, but the `Face` will not :
however, object Face only have 4 colors to change(The four corners of a quadrilateral), how
can we set more colors ??
`Face` is a quadrilateral object, and we can change the `VertexData` to draw more than one little quadrilaterals:
data = rand(12,12);
% Transparency range 0-1, .3-1 for better appearance here
AData = rescale(- data, .3, 1);
%Draw an imagesc with numerical control over colormap and transparency
imagesc(data, 'AlphaData',AData);
colormap(jet);
ax = gca;
ax.DataAspectRatio = [1,1,1];
ax.TickDir = 'out';
ax.Box = 'off';
% get colorbar object
CBarHdl = colorbar;
pause(1e-16)
% Modify the transparency of the colorbar
CData = CBarHdl.Face.Texture.CData;
ALim = [min(min(AData)), max(max(AData))];
CData(4,:) = uint8(255.*rescale(1:size(CData, 2), ALim(1), ALim(2)));
warning off
CBarHdl.Face.ColorType = 'TrueColorAlpha';
VertexData = CBarHdl.Face.VertexData;
tY = repmat((1:size(CData,2))./size(CData,2), [4,1]);
tY1 = tY(:).'; tY2 = tY - tY(1,1); tY2(3:4,:) = 0; tY2 = tY2(:).';
tM1 = [tY1.*0 + 1; tY1; tY1.*0 + 1];
tM2 = [tY1.*0; tY2; tY1.*0];
CBarHdl.Face.VertexData = repmat(VertexData, [1,size(CData,2)]).*tM1 + tM2;
CBarHdl.Face.ColorData = reshape(repmat(CData, [4,1]), 4, []);
The higher the value, the more transparent it becomes
data = rand(12,12);
AData = rescale(- data, .3, 1);
imagesc(data, 'AlphaData',AData);
colormap(jet);
ax = gca;
ax.DataAspectRatio = [1,1,1];
ax.TickDir = 'out';
ax.Box = 'off';
CBarHdl = colorbar;
pause(1e-16)
CData = CBarHdl.Face.Texture.CData;
ALim = [min(min(AData)), max(max(AData))];
CData(4,:) = uint8(255.*rescale(size(CData, 2):-1:1, ALim(1), ALim(2)));
warning off
CBarHdl.Face.ColorType = 'TrueColorAlpha';
VertexData = CBarHdl.Face.VertexData;
tY = repmat((1:size(CData,2))./size(CData,2), [4,1]);
tY1 = tY(:).'; tY2 = tY - tY(1,1); tY2(3:4,:) = 0; tY2 = tY2(:).';
tM1 = [tY1.*0 + 1; tY1; tY1.*0 + 1];
tM2 = [tY1.*0; tY2; tY1.*0];
CBarHdl.Face.VertexData = repmat(VertexData, [1,size(CData,2)]).*tM1 + tM2;
CBarHdl.Face.ColorData = reshape(repmat(CData, [4,1]), 4, []);
More transparent in the middle
data = rand(12,12) - .5;
AData = rescale(abs(data), .1, .9);
imagesc(data, 'AlphaData',AData);
colormap(jet);
ax = gca;
ax.DataAspectRatio = [1,1,1];
ax.TickDir = 'out';
ax.Box = 'off';
CBarHdl = colorbar;
pause(1e-16)
CData = CBarHdl.Face.Texture.CData;
ALim = [min(min(AData)), max(max(AData))];
CData(4,:) = uint8(255.*rescale(abs((1:size(CData, 2)) - (1 + size(CData, 2))/2), ALim(1), ALim(2)));
warning off
CBarHdl.Face.ColorType = 'TrueColorAlpha';
VertexData = CBarHdl.Face.VertexData;
tY = repmat((1:size(CData,2))./size(CData,2), [4,1]);
tY1 = tY(:).'; tY2 = tY - tY(1,1); tY2(3:4,:) = 0; tY2 = tY2(:).';
tM1 = [tY1.*0 + 1; tY1; tY1.*0 + 1];
tM2 = [tY1.*0; tY2; tY1.*0];
CBarHdl.Face.VertexData = repmat(VertexData, [1,size(CData,2)]).*tM1 + tM2;
CBarHdl.Face.ColorData = reshape(repmat(CData, [4,1]), 4, []);
The code will work if the plot have AlphaData property
data = peaks(30);
AData = rescale(data, .2, 1);
surface(data, 'FaceAlpha','flat','AlphaData',AData);
colormap(jet(100));
ax = gca;
ax.DataAspectRatio = [1,1,1];
ax.TickDir = 'out';
ax.Box = 'off';
view(3)
CBarHdl = colorbar;
pause(1e-16)
CData = CBarHdl.Face.Texture.CData;
ALim = [min(min(AData)), max(max(AData))];
CData(4,:) = uint8(255.*rescale(1:size(CData, 2), ALim(1), ALim(2)));
warning off
CBarHdl.Face.ColorType = 'TrueColorAlpha';
VertexData = CBarHdl.Face.VertexData;
tY = repmat((1:size(CData,2))./size(CData,2), [4,1]);
tY1 = tY(:).'; tY2 = tY - tY(1,1); tY2(3:4,:) = 0; tY2 = tY2(:).';
tM1 = [tY1.*0 + 1; tY1; tY1.*0 + 1];
tM2 = [tY1.*0; tY2; tY1.*0];
CBarHdl.Face.VertexData = repmat(VertexData, [1,size(CData,2)]).*tM1 + tM2;
CBarHdl.Face.ColorData = reshape(repmat(CData, [4,1]), 4, []);
While searching the internet for some books on ordinary differential equations, I came across a link that I believe is very useful for all math students and not only. If you are interested in ODEs, it's worth taking the time to study it.
A First Look at Ordinary Differential Equations by Timothy S. Judson is an excellent resource for anyone looking to understand ODEs better. Here's a brief overview of the main topics covered:
- Introduction to ODEs: Basic concepts, definitions, and initial differential equations.
- Methods of Solution:
- Separable equations
- First-order linear equations
- Exact equations
- Transcendental functions
- Applications of ODEs: Practical examples and applications in various scientific fields.
- Systems of ODEs: Analysis and solutions of systems of differential equations.
- Series and Numerical Methods: Use of series and numerical methods for solving ODEs.
This book provides a clear and comprehensive introduction to ODEs, making it suitable for students and new researchers in mathematics. If you're interested, you can explore the book in more detail here: A First Look at Ordinary Differential Equations.
How to leave feedback on a doc page
Leaving feedback is a two-step process. At the bottom of most pages in the MATLAB documentation is a star rating.
Start by selecting a star that best answers the question. After selecting a star rating, an edit box appears where you can offer specific feedback.
When you press "Submit" you'll see the confirmation dialog below. You cannot go back and edit your content, although you can refresh the page to go through that process again.
Tips on leaving feedback
- Be productive. The reader should clearly understand what action you'd like to see, what was unclear, what you think needs work, or what areas were really helpful.
- Positive feedback is also helpful. By nature, feedback often focuses on suggestions for changes but it also helps to know what was clear and what worked well.
- Point to specific areas of the page. This helps the reader to narrow the focus of the page to the area described by your feedback.
What happens to that feedback?
Before working at MathWorks I often left feedback on documentation pages but I never knew what happens after that. One day in 2021 I shared my speculation on the process:
> That feedback is received by MathWorks Gnomes which are never seen nor heard but visit the MathWorks documentation team at night while they are sleeping and whisper selected suggestions into their ears to manipulate their dreams. Occassionally this causes them to wake up with a Eureka moment that leads to changes in the documentation.
I'd like to let you in on the secret which is much less fanciful. Feedback left in the star rating and edit box are collected and periodically reviewed by the doc writers who look for trends on highly trafficked pages and finer grain feedback on less visited pages. Your feedback is important and often results in improvements.
📚 New Book Announcement: "Image Processing Recipes in MATLAB" 📚
I am delighted to share the release of my latest book, "Image Processing Recipes in MATLAB," co-authored by my dear friend and colleague Gustavo Benvenutti Borba.
This 'cookbook' contains 30 practical recipes for image processing, ranging from foundational techniques to recently published algorithms. It serves as a concise and readable reference for quickly and efficiently deploying image processing pipelines in MATLAB.
Gustavo and I are immensely grateful to the MathWorks Book Program for their support. We also want to thank Randi Slack and her fantastic team at CRC Press for their patience, expertise, and professionalism throughout the process.
___________
A colleague said that you can search the Help Center using the phrase 'Introduced in' followed by a release version. Such as, 'Introduced in R2022a'. Doing this yeilds search results specific for that release.
Seems pretty handy so I thought I'd share.
Bringing the beauty of MathWorks Natick's tulips to life through code!
Remix challenge: create and share with us your new breeds of MATLAB tulips!
From Alpha Vantage's website: API Documentation | Alpha Vantage
Try using the built-in Matlab function webread(URL)... for example:
% copy a URL from the examples on the site
URL = 'https://www.alphavantage.co/query?function=TIME_SERIES_DAILY&symbol=IBM&apikey=demo'
% or use the pattern to create one
tickers = [{'IBM'} {'SPY'} {'DJI'} {'QQQ'}]; i = 1;
URL = ...
['https://www.alphavantage.co/query?function=TIME_SERIES_DAILY_ADJUSTED&outputsize=full&symbol=', ...
+ tickers{i}, ...
+ '&apikey=***Put Your API Key here***'];
X = webread(URL);
You can access any of the data available on the site as per the Alpha Vantage documentation using these two lines of code but with different designations for the requested data as per the documentation.
It's fun!
This cheat sheet is here:
reference:
- https://github.com/peijin94/matlabPlotCheatsheet
- https://github.com/mathworks/visualization-cheat-sheet
- https://www.mathworks.com/products/matlab/plot-gallery.html
- https://www.mathworks.com/help/matlab/release-notes.html
MATLAB used to have official visualization-cheat-sheet, but there have been quite a few new updates in MATLAB versions recently. Therefore, I made my own cheat sheet and marked the versions of each new thing that were released :
Dear MATLAB contest enthusiasts,
I believe many of you have been captivated by the innovative entries from Zhaoxu Liu / slanderer, in the 2023 MATLAB Flipbook Mini Hack contest.
Ever wondered about the person behind these creative entries? What drives a MATLAB user to such levels of skill? And what inspired his participation in the contest? We were just as curious as you are!
We were delighted to catch up with him and learn more about his use of MATLAB. The interview has recently been published in MathWorks Blogs. For an in-depth look into his insights and experiences, be sure to read our latest blog post: Community Q&A – Zhaoxu Liu.
But the conversation doesn't end here! Who would you like to see featured in our next interview? Drop their name in the comments section below and let us know who we should reach out to next!
Temporary print statements are often helpful during debugging but it's easy to forget to remove the statements or sometimes you may not have writing privileges for the file. This tip uses conditional breakpoints to add print statements without ever editing the file!
What are conditional breakpoints?
Conditional breakpoints allow you to write a conditional statement that is executed when the selected line is hit and if the condition returns true, MATLAB pauses at that line. Otherwise, it continues.
The Hack: use ~fprintf() as the condition
fprintf prints information to the command window and returns the size of the message in bytes. The message size will always be greater than 0 which will always evaluate as true when converted to logical. Therefore, by negating an fprintf statement within a conditional breakpoint, the fprintf command will execute, print to the command window, and evalute as false which means the execution will continue uninterupted!
How to set a conditional break point
1. Right click the line number where you want the condition to be evaluated and select "Set Conditional Breakpoint"
2. Enter a valid MATLAB expression that returns a logical scalar value in the editor dialog.
Handy one-liners
Check if a line is reached: Don't forget the negation (~) and the line break (\n)!
~fprintf('Entered callback function\n')
Display the call stack from the break point line: one of my favorites!
~fprintf('%s\n',formattedDisplayText(struct2table(dbstack)))
Inspect variable values: For scalar values,
~fprintf('v = %.5f\n', v)
~fprintf('%s\n', formattedDisplayText(v)).
Make sense of frequent hits: In some situations such as responses to listeners or interactive callbacks, a line can be executed 100s of times per second. Incorporate a timestamp to differentiate messages during rapid execution.
~fprintf('WindowButtonDownFcn - %s\n', datetime('now'))
Closing
This tip not only keeps your code clean but also offers a dynamic way to monitor code execution and variable states without permanent modifications. Interested in digging deeper? @Steve Eddins takes this tip to the next level with his Code Trace for MATLAB tool available on the File Exchange (read more).
Summary animation
To reproduce the events in this animation:
% buttonDownFcnDemo.m
fig = figure();
tcl = tiledlayout(4,4,'TileSpacing','compact');
for i = 1:16
ax = nexttile(tcl);
title(ax,"#"+string(i))
ax.ButtonDownFcn = @axesButtonDownFcn;
xlim(ax,[-1 1])
ylim(ax,[-1,1])
hold(ax,'on')
end
function axesButtonDownFcn(obj,event)
colors = lines(16);
plot(obj,event.IntersectionPoint(1),event.IntersectionPoint(2),...
'ko','MarkerFaceColor',colors(obj.Layout.Tile,:))
end
The beautiful and elegant chord diagrams were all created using MATLAB?
Indeed, they were all generated using the chord diagram plotting toolkit that I developed myself:
- - Chord chart: [chord chart](https://www.mathworks.com/matlabcentral/fileexchange/116550-chord-chart)
- - Directed graph chord chart: [digraph chord chart]:(https://www.mathworks.com/matlabcentral/fileexchange/121043-digraph-chord-chart)
You can download these toolkits from the provided links.
The reason for writing this article is that many people have started using the chord diagram plotting toolkit that I developed. However, some users are unsure about customizing certain styles. As the developer, I have a good understanding of the implementation principles of the toolkit and can apply it flexibly. This has sparked the idea of challenging myself to create various styles of chord diagrams. Currently, the existing code is quite lengthy. In the future, I may integrate some of this code into the toolkit, enabling users to achieve the effects of many lines of code with just a few lines.
Without further ado, let's see the extent to which this MATLAB toolkit can currently perform.
demo 1
rng(2)
dataMat = randi([0,5], [11,5]);
dataMat(1:6,1) = 0;
dataMat([11,7],1) = [45,25];
dataMat([1,4,5,7],2) = [20,20,30,30];
dataMat(:,3) = 0;
dataMat(6,3) = 45;
dataMat(1:5,4) = 0;
dataMat([6,7],4) = [25,25];
dataMat([5,6,9],5) = [25,25,25];
colName = {'Fly', 'Beetle', 'Leaf', 'Soil', 'Waxberry'};
rowName = {'Bartomella', 'Bradyrhizobium', 'Dysgomonas', 'Enterococcus',...
'Lactococcus', 'norank', 'others', 'Pseudomonas', 'uncultured',...
'Vibrionimonas', 'Wolbachia'};
figure('Units','normalized', 'Position',[.02,.05,.6,.85])
CC = chordChart(dataMat, 'rowName',rowName, 'colName',colName, 'Sep',1/80);
CC = CC.draw();
% 修改上方方块颜色(Modify the color of the blocks above)
CListT = [0.7765 0.8118 0.5216; 0.4431 0.4706 0.3843; 0.5804 0.2275 0.4549;
0.4471 0.4039 0.6745; 0.0157 0 0 ];
for i = 1:size(dataMat, 2)
CC.setSquareT_N(i, 'FaceColor',CListT(i,:))
end
% 修改下方方块颜色(Modify the color of the blocks below)
CListF = [0.5843 0.6863 0.7843; 0.1098 0.1647 0.3255; 0.0902 0.1608 0.5373;
0.6314 0.7961 0.2118; 0.0392 0.2078 0.1059; 0.0157 0 0 ;
0.8549 0.9294 0.8745; 0.3882 0.3255 0.4078; 0.5020 0.7216 0.3843;
0.0902 0.1843 0.1804; 0.8196 0.2314 0.0706];
for i = 1:size(dataMat, 1)
CC.setSquareF_N(i, 'FaceColor',CListF(i,:))
end
% 修改弦颜色(Modify chord color)
for i = 1:size(dataMat, 1)
for j = 1:size(dataMat, 2)
CC.setChordMN(i,j, 'FaceColor',CListT(j,:), 'FaceAlpha',.5)
end
end
CC.tickState('on')
CC.labelRotate('on')
CC.setFont('FontSize',17, 'FontName','Cambria')
% CC.labelRotate('off')
% textHdl = findobj(gca,'Tag','ChordLabel');
% for i = 1:length(textHdl)
% if textHdl(i).Position(2) < 0
% if abs(textHdl(i).Position(1)) > .7
% textHdl(i).Rotation = textHdl(i).Rotation + 45;
% textHdl(i).HorizontalAlignment = 'right';
% if textHdl(i).Rotation > 90
% textHdl(i).Rotation = textHdl(i).Rotation + 180;
% textHdl(i).HorizontalAlignment = 'left';
% end
% else
% textHdl(i).Rotation = textHdl(i).Rotation + 10;
% textHdl(i).HorizontalAlignment = 'right';
% end
% end
% end
demo 2
rng(3)
dataMat = randi([1,15], [7,22]);
dataMat(dataMat < 11) = 0;
dataMat(1, sum(dataMat, 1) == 0) = 15;
colName = {'A2M', 'FGA', 'FGB', 'FGG', 'F11', 'KLKB1', 'SERPINE1', 'VWF',...
'THBD', 'TFPI', 'PLAT', 'SERPINA5', 'SERPIND1', 'F2', 'PLG', 'F12',...
'SERPINC1', 'SERPINA1', 'PROS1', 'SERPINF2', 'F13A1', 'PROC'};
rowName = {'Lung', 'Spleen', 'Liver', 'Heart',...
'Renal cortex', 'Renal medulla', 'Thyroid'};
figure('Units','normalized', 'Position',[.02,.05,.6,.85])
CC = chordChart(dataMat, 'rowName',rowName, 'colName',colName, 'Sep',1/80, 'LRadius',1.21);
CC = CC.draw();
CC.labelRotate('on')
% 单独设置每一个弦末端方块(Set individual end blocks for each chord)
% Use obj.setEachSquareF_Prop
% or obj.setEachSquareT_Prop
% F means from (blocks below)
% T means to (blocks above)
CListT = [173,70,65; 79,135,136]./255;
% Upregulated:1 | Downregulated:2
Regulated = rand([7, 22]);
Regulated = (Regulated < .8) + 1;
for i = 1:size(Regulated, 1)
for j = 1:size(Regulated, 2)
CC.setEachSquareT_Prop(i, j, 'FaceColor', CListT(Regulated(i,j),:))
end
end
% 绘制图例(Draw legend)
H1 = fill([0,1,0] + 100, [1,0,1] + 100, CListT(1,:), 'EdgeColor','none');
H2 = fill([0,1,0] + 100, [1,0,1] + 100, CListT(2,:), 'EdgeColor','none');
lgdHdl = legend([H1,H2], {'Upregulated','Downregulated'}, 'AutoUpdate','off', 'Location','best');
lgdHdl.ItemTokenSize = [12,12];
lgdHdl.Box = 'off';
lgdHdl.FontSize = 13;
% 修改下方方块颜色(Modify the color of the blocks below)
CListF = [128,108,171; 222,208,161; 180,196,229; 209,150,146; 175,201,166;
134,156,118; 175,175,173]./255;
for i = 1:size(dataMat, 1)
CC.setSquareF_N(i, 'FaceColor',CListF(i,:))
end
% 修改弦颜色(Modify chord color)
for i = 1:size(dataMat, 1)
for j = 1:size(dataMat, 2)
CC.setChordMN(i,j, 'FaceColor',CListF(i,:), 'FaceAlpha',.45)
end
end
demo 3
dataMat = rand([15,15]);
dataMat(dataMat > .15) = 0;
CList = [ 75,146,241; 252,180, 65; 224, 64, 10; 5,100,146; 191,191,191;
26, 59,105; 255,227,130; 18,156,221; 202,107, 75; 0, 92,219;
243,210,136; 80, 99,129; 241,185,168; 224,131, 10; 120,147,190]./255;
figure('Units','normalized', 'Position',[.02,.05,.6,.85])
BCC = biChordChart(dataMat, 'Arrow','on', 'CData',CList);
BCC = BCC.draw();
% 添加刻度
BCC.tickState('on')
% 修改字体，字号及颜色
BCC.setFont('FontName','Cambria', 'FontSize',17, 'Color',[0,0,.8])
demo 4
rng(5)
dataMat = randi([1,20], [5,5]);
dataMat(1,1) = 110;
dataMat(2,2) = 40;
dataMat(3,3) = 50;
dataMat(5,5) = 50;
CList1 = [164,190,158; 216,213,153; 177,192,208; 238,238,227; 249,217,153]./255;
CList2 = [247,204,138; 128,187,185; 245,135,124; 140,199,197; 252,223,164]./255;
CList = CList2;
NameList={'CHORD','CHART','MADE','BY','SLANDARER'};
figure('Units','normalized', 'Position',[.02,.05,.6,.85])
BCC = biChordChart(dataMat, 'Arrow','on', 'CData',CList, 'Sep',1/30, 'Label',NameList, 'LRadius',1.33);
BCC = BCC.draw();
% 添加刻度
BCC.tickState('on')
% 修改弦颜色(Modify chord color)
for i = 1:size(dataMat, 1)
for j = 1:size(dataMat, 2)
if dataMat(i,j) > 0
BCC.setChordMN(i,j, 'FaceAlpha',.7, 'EdgeColor',CList(i,:)./1.1)
end
end
end
% 修改方块颜色(Modify the color of the blocks)
for i = 1:size(dataMat, 1)
BCC.setSquareN(i, 'EdgeColor',CList(i,:)./1.7)
end
% 修改字体，字号及颜色
BCC.setFont('FontName','Cambria', 'FontSize',17)
BCC.tickLabelState('on')
BCC.setTickFont('FontName','Cambria', 'FontSize',9)
demo 5
dataMat=randi([1,20], [14,3]);
dataMat(11:14,1) = 0;
dataMat(6:10,2) = 0;
dataMat(1:5,3) = 0;
colName = compose('C%d', 1:3);
rowName = [compose('A%d', 1:7), compose('B%d', 7:-1:1)];
figure('Units','normalized', 'Position',[.02,.05,.6,.85])
CC = chordChart(dataMat, 'rowName',rowName, 'colName',colName, 'Sep',1/80);
CC = CC.draw();
% 修改上方方块颜色(Modify the color of the blocks above)
for i = 1:size(dataMat, 2)
CC.setSquareT_N(i, 'FaceColor',[190,190,190]./255)
end
% 修改下方方块颜色(Modify the color of the blocks below)
CListF=[255,244,138; 253,220,117; 254,179, 78; 253,190, 61;
252, 78, 41; 228, 26, 26; 178, 0, 36; 4, 84,119;
1,113,137; 21,150,155; 67,176,173; 68,173,158;
123,204,163; 184,229,162]./255;
for i = 1:size(dataMat, 1)
CC.setSquareF_N(i, 'FaceColor',CListF(i,:))
end
% 修改弦颜色(Modify chord color)
for i = 1:size(dataMat, 1)
for j = 1:size(dataMat, 2)
CC.setChordMN(i,j, 'FaceColor',CListF(i,:), 'FaceAlpha',.5)
end
end
CC.tickState('on')
CC.tickLabelState('on')
demo 6
rng(2)
dataMat = randi([0,40], [20,4]);
dataMat(rand([20,4]) < .2) = 0;
dataMat(1,3) = 500;
dataMat(20,1:4) = [140; 150; 80; 90];
colName = compose('T%d', 1:4);
rowName = compose('SL%d', 1:20);
figure('Units','normalized', 'Position',[.02,.05,.6,.85])
CC = chordChart(dataMat, 'rowName',rowName, 'colName',colName, 'Sep',1/80, 'LRadius',1.23);
CC = CC.draw();
% 修改上方方块颜色(Modify the color of the blocks above)
CListT = [0.62,0.49,0.27; 0.28,0.57,0.76
0.25,0.53,0.30; 0.86,0.48,0.34];
for i = 1:size(dataMat, 2)
CC.setSquareT_N(i, 'FaceColor',CListT(i,:))
end
% 修改下方方块颜色(Modify the color of the blocks below)
CListF = [0.94,0.84,0.60; 0.16,0.50,0.67; 0.92,0.62,0.49;
0.48,0.44,0.60; 0.48,0.44,0.60; 0.71,0.79,0.73;
0.96,0.98,0.98; 0.51,0.82,0.95; 0.98,0.70,0.82;
0.97,0.85,0.84; 0.55,0.64,0.62; 0.94,0.93,0.60;
0.98,0.90,0.85; 0.72,0.84,0.81; 0.85,0.45,0.49;
0.76,0.76,0.84; 0.59,0.64,0.62; 0.62,0.14,0.15;
0.75,0.75,0.75; 1.00,1.00,1.00];
for i = 1:size(dataMat, 1)
CC.setSquareF_N(i, 'FaceColor',CListF(i,:))
end
CC.setSquareF_N(size(dataMat, 1), 'EdgeColor','k', 'LineWidth',1)
% 修改弦颜色(Modify chord color)
for i = 1:size(dataMat, 1)
for j = 1:size(dataMat, 2)
CC.setChordMN(i,j, 'FaceColor',CListT(j,:), 'FaceAlpha',.46)
end
end
CC.tickState('on')
CC.labelRotate('on')
CC.setFont('FontSize',17, 'FontName','Cambria')
demo 7
dataMat = randi([10,10000], [10,10]);
dataMat(6:10,:) = 0;
dataMat(:,1:5) = 0;
NameList = {'BOC', 'ICBC', 'ABC', 'BOCM', 'CCB', ...
'yama', 'nikoto', 'saki', 'koto', 'kawa'};
CList = [0.63,0.75,0.88
0.67,0.84,0.75
0.85,0.78,0.88
1.00,0.92,0.93
0.92,0.63,0.64
0.57,0.67,0.75
1.00,0.65,0.44
0.72,0.73,0.40
0.65,0.57,0.58
0.92,0.94,0.96];
figure('Units','normalized', 'Position',[.02,.05,.6,.85])
BCC = biChordChart(dataMat, 'Arrow','on', 'CData',CList, 'Label',NameList);
BCC = BCC.draw();
% 修改弦颜色(Modify chord color)
for i = 1:size(dataMat, 1)
for j = 1:size(dataMat, 2)
if dataMat(i,j) > 0
BCC.setChordMN(i,j, 'FaceAlpha',.85, 'EdgeColor',CList(i,:)./1.5, 'LineWidth',.8)
end
end
end
for i = 1:size(dataMat, 1)
BCC.setSquareN(i, 'EdgeColor',CList(i,:)./1.5, 'LineWidth',1)
end
% 添加刻度、修改字体
BCC.tickState('on')
BCC.setFont('FontName','Cambria', 'FontSize',17)
demo 8
dataMat = rand([11,4]);
dataMat = round(10.*dataMat.*((11:-1:1).'+1))./10;
colName = {'A','B','C','D'};
rowName = {'Acidobacteriota', 'Actinobacteriota', 'Proteobacteria', ...
'Chloroflexi', 'Bacteroidota', 'Firmicutes', 'Gemmatimonadota', ...
'Verrucomicrobiota', 'Patescibacteria', 'Planctomyetota', 'Others'};
figure('Units','normalized', 'Position',[.02,.05,.8,.85])
CC = chordChart(dataMat, 'colName',colName, 'Sep',1/80, 'SSqRatio',30/100);% -30/100
CC = CC.draw();
% 修改上方方块颜色(Modify the color of the blocks above)
CListT = [0.93,0.60,0.62
0.55,0.80,0.99
0.95,0.82,0.18
1.00,0.81,0.91];
for i = 1:size(dataMat, 2)
CC.setSquareT_N(i, 'FaceColor',CListT(i,:))
end
% 修改下方方块颜色(Modify the color of the blocks below)
CListF = [0.75,0.73,0.86
0.56,0.83,0.78
0.00,0.60,0.20
1.00,0.49,0.02
0.78,0.77,0.95
0.59,0.24,0.36
0.98,0.51,0.45
0.96,0.55,0.75
0.47,0.71,0.84
0.65,0.35,0.16
0.40,0.00,0.64];
for i = 1:size(dataMat, 1)
CC.setSquareF_N(i, 'FaceColor',CListF(i,:))
end
% 修改弦颜色(Modify chord color)
CListC = [0.55,0.83,0.76
0.75,0.73,0.86
0.00,0.60,0.19
1.00,0.51,0.04];
for i = 1:size(dataMat, 1)
for j = 1:size(dataMat, 2)
CC.setChordMN(i,j, 'FaceColor',CListC(j,:), 'FaceAlpha',.4)
end
end
% 单独设置每一个弦末端方块(Set individual end blocks for each chord)
% Use obj.setEachSquareF_Prop
% or obj.setEachSquareT_Prop
% F means from (blocks below)
% T means to (blocks above)
for i = 1:size(dataMat, 1)
for j = 1:size(dataMat, 2)
CC.setEachSquareT_Prop(i,j, 'FaceColor', CListF(i,:))
end
end
% 添加刻度
CC.tickState('on')
% 修改字体，字号及颜色
CC.setFont('FontName','Cambria', 'FontSize',17)
% 隐藏下方标签
textHdl = findobj(gca, 'Tag','ChordLabel');
for i = 1:length(textHdl)
if textHdl(i).Position(2) < 0
set(textHdl(i), 'Visible','off')
end
end
% 绘制图例(Draw legend)
scatterHdl = scatter(10.*ones(size(dataMat,1)),10.*ones(size(dataMat,1)), ...
55, 'filled');
for i = 1:length(scatterHdl)
scatterHdl(i).CData = CListF(i,:);
end
lgdHdl = legend(scatterHdl, rowName, 'Location','best', 'FontSize',16, 'FontName','Cambria', 'Box','off');
set(lgdHdl, 'Position',[.7482,.3577,.1658,.3254])
demo 9
dataMat = randi([0,10], [5,5]);
CList1 = [0.70,0.59,0.67
0.62,0.70,0.62
0.81,0.75,0.62
0.80,0.62,0.56
0.62,0.65,0.65];
CList2 = [0.02,0.02,0.02
0.59,0.26,0.33
0.38,0.49,0.38
0.03,0.05,0.03
0.29,0.28,0.32];
CList = CList2;
NameList={'CHORD','CHART','MADE','BY','SLANDARER'};
figure('Units','normalized', 'Position',[.02,.05,.6,.85])
BCC = biChordChart(dataMat, 'Arrow','on', 'CData',CList, 'Sep',1/30, 'Label',NameList, 'LRadius',1.33);
BCC = BCC.draw();
% 修改弦颜色(Modify chord color)
for i = 1:size(dataMat, 1)
for j = 1:size(dataMat, 2)
BCC.setChordMN(i,j, 'FaceAlpha',.5)
end
end
% 修改方块颜色(Modify the color of the blocks)
for i = 1:size(dataMat, 1)
BCC.setSquareN(i, 'EdgeColor',[0,0,0], 'LineWidth',5)
end
% 添加刻度
BCC.tickState('on')
% 修改字体，字号及颜色
BCC.setFont('FontSize',17, 'FontWeight','bold')
BCC.tickLabelState('on')
BCC.setTickFont('FontSize',9)
demo 10
rng(2)
dataMat = rand([14,5]) > .3;
colName = {'phosphorylation', 'vasculature development', 'blood vessel development', ...
'cell adhesion', 'plasma membrane'};
rowName = {'THY1', 'FGF2', 'MAP2K1', 'CDH2', 'HBEGF', 'CXCR4', 'ECSCR',...
'ACVRL1', 'RECK', 'PNPLA6', 'CDH5', 'AMOT', 'EFNB2', 'CAV1'};
figure('Units','normalized', 'Position',[.02,.05,.9,.85])
CC = chordChart(dataMat, 'colName',colName, 'rowName',rowName, 'Sep',1/80, 'LRadius',1.2);
CC = CC.draw();
% 修改上方方块颜色(Modify the color of the blocks above)
CListT1 = [0.5686 0.1961 0.2275
0.2275 0.2863 0.3765
0.8431 0.7882 0.4118
0.4275 0.4510 0.2706
0.3333 0.2706 0.2510];
CListT2 = [0.4941 0.5490 0.4118
0.9059 0.6510 0.3333
0.8980 0.6157 0.4980
0.8902 0.5137 0.4667
0.4275 0.2824 0.2784];
CListT3 = [0.4745 0.5843 0.7569
0.4824 0.5490 0.5843
0.6549 0.7216 0.6510
0.9412 0.9216 0.9059
0.9804 0.7608 0.6863];
CListT = CListT3;
for i = 1:size(dataMat, 2)
CC.setSquareT_N(i, 'FaceColor',CListT(i,:), 'EdgeColor',[0,0,0])
end
% 修改弦颜色(Modify chord color)
for i = 1:size(dataMat, 1)
for j = 1:size(dataMat, 2)
CC.setChordMN(i,j, 'FaceColor',CListT(j,:), 'FaceAlpha',.9, 'EdgeColor',[0,0,0])
end
end
% 修改下方方块颜色(Modify the color of the blocks below)
logFC = sort(rand(1,14))*6 - 3;
for i = 1:size(dataMat, 1)
CC.setSquareF_N(i, 'CData',logFC(i), 'FaceColor','flat', 'EdgeColor',[0,0,0])
end
CMap = [ 0 0 1.0000; 0.0645 0.0645 1.0000; 0.1290 0.1290 1.0000; 0.1935 0.1935 1.0000
0.2581 0.2581 1.0000; 0.3226 0.3226 1.0000; 0.3871 0.3871 1.0000; 0.4516 0.4516 1.0000
0.5161 0.5161 1.0000; 0.5806 0.5806 1.0000; 0.6452 0.6452 1.0000; 0.7097 0.7097 1.0000
0.7742 0.7742 1.0000; 0.8387 0.8387 1.0000; 0.9032 0.9032 1.0000; 0.9677 0.9677 1.0000
1.0000 0.9677 0.9677; 1.0000 0.9032 0.9032; 1.0000 0.8387 0.8387; 1.0000 0.7742 0.7742
1.0000 0.7097 0.7097; 1.0000 0.6452 0.6452; 1.0000 0.5806 0.5806; 1.0000 0.5161 0.5161
1.0000 0.4516 0.4516; 1.0000 0.3871 0.3871; 1.0000 0.3226 0.3226; 1.0000 0.2581 0.2581
1.0000 0.1935 0.1935; 1.0000 0.1290 0.1290; 1.0000 0.0645 0.0645; 1.0000 0 0];
colormap(CMap);
try clim([-3,3]),catch,end
try caxis([-3,3]),catch,end
CBHdl = colorbar();
CBHdl.Position = [0.74,0.25,0.02,0.2];
% =========================================================================
% 交换XY轴(Swap XY axis)
patchHdl = findobj(gca, 'Type','patch');
for i = 1:length(patchHdl)
tX = patchHdl(i).XData;
tY = patchHdl(i).YData;
patchHdl(i).XData = tY;
patchHdl(i).YData = - tX;
end
txtHdl = findobj(gca, 'Type','text');
for i = 1:length(txtHdl)
txtHdl(i).Position([1,2]) = [1,-1].*txtHdl(i).Position([2,1]);
if txtHdl(i).Position(1) < 0
txtHdl(i).HorizontalAlignment = 'right';
else
txtHdl(i).HorizontalAlignment = 'left';
end
end
lineHdl = findobj(gca, 'Type','line');
for i = 1:length(lineHdl)
tX = lineHdl(i).XData;
tY = lineHdl(i).YData;
lineHdl(i).XData = tY;
lineHdl(i).YData = - tX;
end
% =========================================================================
txtHdl = findobj(gca, 'Type','text');
for i = 1:length(txtHdl)
if txtHdl(i).Position(1) > 0
txtHdl(i).Visible = 'off';
end
end
text(1.25,-.15, 'LogFC', 'FontSize',16)
text(1.25,1, 'Terms', 'FontSize',16)
patchHdl = [];
for i = 1:size(dataMat, 2)
patchHdl(i) = fill([10,11,12],[10,13,13], CListT(i,:), 'EdgeColor',[0,0,0]);
end
lgdHdl = legend(patchHdl, colName, 'Location','best', 'FontSize',14, 'FontName','Cambria', 'Box','off');
lgdHdl.Position = [.735,.53,.167,.27];
lgdHdl.ItemTokenSize = [18,8];
demo 11
rng(2)
dataMat = rand([12,12]);
dataMat(dataMat < .85) = 0;
dataMat(7,:) = 1.*(rand(1,12)+.1);
dataMat(11,:) = .6.*(rand(1,12)+.1);
dataMat(12,:) = [2.*(rand(1,10)+.1), 0, 0];
CList = [repmat([49,49,49],[10,1]); 235,28,34; 19,146,241]./255;
figure('Units','normalized', 'Position',[.02,.05,.6,.85])
BCC = biChordChart(dataMat, 'Arrow','off', 'CData',CList);
BCC = BCC.draw();
% 添加刻度
BCC.tickState('on')
% 修改字体，字号及颜色
BCC.setFont('FontName','Cambria', 'FontSize',17)
% 修改弦颜色(Modify chord color)
for i = 1:size(dataMat, 1)
for j = 1:size(dataMat, 2)
if dataMat(i,j) > 0
BCC.setChordMN(i,j, 'FaceAlpha',.78, 'EdgeColor',[0,0,0])
end
end
end
% 修改方块颜色(Modify the color of the blocks)
for i = 1:size(dataMat, 1)
BCC.setSquareN(i, 'EdgeColor',[0,0,0], 'LineWidth',2)
end
demo 12
dataMat = rand([9,9]);
dataMat(dataMat > .7) = 0;
dataMat(eye(9) == 1) = (rand([1,9])+.2).*3;
CList = [0.85,0.23,0.24
0.96,0.39,0.18
0.98,0.63,0.22
0.99,0.80,0.26
0.70,0.76,0.21
0.24,0.74,0.71
0.27,0.65,0.84
0.09,0.37,0.80
0.64,0.40,0.84];
figure('Units','normalized', 'Position',[.02,.05,.6,.85])
BCC = biChordChart(dataMat, 'Arrow','on', 'CData',CList);
BCC = BCC.draw();
% 添加刻度、刻度标签
BCC.tickState('on')
% 修改字体，字号及颜色
BCC.setFont('FontName','Cambria', 'FontSize',17)
% 修改弦颜色(Modify chord color)
for i = 1:size(dataMat, 1)
for j = 1:size(dataMat, 2)
if dataMat(i,j) > 0
BCC.setChordMN(i,j, 'FaceAlpha',.7)
end
end
end
demo 13
rng(2)
dataMat = randi([1,40], [7,4]);
dataMat(rand([7,4]) < .1) = 0;
colName = compose('MATLAB%d', 1:4);
rowName = compose('SL%d', 1:7);
figure('Units','normalized', 'Position',[.02,.05,.7,.85])
CC = chordChart(dataMat, 'rowName',rowName, 'colName',colName, 'Sep',1/80, 'LRadius',1.32);
CC = CC.draw();
% 修改上方方块颜色(Modify the color of the blocks above)
CListT = [0.49,0.64,0.53
0.75,0.39,0.35
0.80,0.74,0.42
0.40,0.55,0.66];
for i = 1:size(dataMat, 2)
CC.setSquareT_N(i, 'FaceColor',CListT(i,:))
end
% 修改下方方块颜色(Modify the color of the blocks below)
CListF = [0.91,0.91,0.97
0.62,0.95,0.66
0.91,0.61,0.20
0.54,0.45,0.82
0.99,0.76,0.81
0.91,0.85,0.83
0.53,0.42,0.43];
for i = 1:size(dataMat, 1)
CC.setSquareF_N(i, 'FaceColor',CListF(i,:))
end
% 修改弦颜色(Modify chord color)
for i = 1:size(dataMat, 1)
for j = 1:size(dataMat, 2)
CC.setChordMN(i,j, 'FaceColor',CListT(j,:), 'FaceAlpha',.46)
end
end
CC.tickState('on')
CC.tickLabelState('on')
CC.setFont('FontSize',17, 'FontName','Cambria')
CC.setTickFont('FontSize',8, 'FontName','Cambria')
% 绘制图例(Draw legend)
scatterHdl = scatter(10.*ones(size(dataMat,1)),10.*ones(size(dataMat,1)), ...
55, 'filled');
for i = 1:length(scatterHdl)
scatterHdl(i).CData = CListF(i,:);
end
lgdHdl = legend(scatterHdl, rowName, 'Location','best', 'FontSize',16, 'FontName','Cambria', 'Box','off');
set(lgdHdl, 'Position',[.77,.38,.1658,.27])
demo 14
rng(6)
dataMat = randi([1,20], [8,8]);
dataMat(dataMat > 5) = 0;
dataMat(1,:) = randi([1,15], [1,8]);
dataMat(1,8) = 40;
dataMat(8,8) = 60;
dataMat = dataMat./sum(sum(dataMat));
CList = [0.33,0.53,0.86
0.94,0.50,0.42
0.92,0.58,0.30
0.59,0.47,0.45
0.37,0.76,0.82
0.82,0.68,0.29
0.75,0.62,0.87
0.43,0.69,0.57];
NameList={'CHORD', 'CHART', 'AND', 'BICHORD',...
'CHART', 'MADE', 'BY', 'SLANDARER'};
figure('Units','normalized', 'Position',[.02,.05,.6,.85])
BCC = biChordChart(dataMat, 'Arrow','on', 'CData',CList, 'Sep',1/12, 'Label',NameList, 'LRadius',1.33);
BCC = BCC.draw();
% 添加刻度
BCC.tickState('on')
% 修改弦颜色(Modify chord color)
for i = 1:size(dataMat, 1)
for j = 1:size(dataMat, 2)
if dataMat(i,j) > 0
BCC.setChordMN(i,j, 'FaceAlpha',.7, 'EdgeColor',CList(i,:)./1.1)
end
end
end
% 修改方块颜色(Modify the color of the blocks)
for i = 1:size(dataMat, 1)
BCC.setSquareN(i, 'EdgeColor',CList(i,:)./1.7)
end
% 修改字体，字号及颜色
BCC.setFont('FontName','Cambria', 'FontSize',17)
BCC.tickLabelState('on')
BCC.setTickFont('FontName','Cambria', 'FontSize',9)
% 调整数值字符串格式
% Adjust numeric string format
BCC.setTickLabelFormat(@(x)[num2str(round(x*100)),'%'])
demo 15
CList = [0.81,0.72,0.83
0.69,0.82,0.89
0.17,0.44,0.64
0.70,0.85,0.55
0.03,0.57,0.13
0.97,0.67,0.64
0.84,0.09,0.12
1.00,0.80,0.46
0.98,0.52,0.01
];
figure('Units','normalized', 'Position',[.02,.05,.53,.85], 'Color',[1,1,1])
% =========================================================================
ax1 = axes('Parent',gcf, 'Position',[0,1/2,1/2,1/2]);
dataMat = rand([9,9]);
dataMat(dataMat > .4) = 0;
BCC = biChordChart(dataMat, 'Arrow','on', 'CData',CList);
BCC = BCC.draw();
BCC.tickState('on')
BCC.setFont('Visible','off')
% 修改弦颜色(Modify chord color)
for i = 1:size(dataMat, 1)
for j = 1:size(dataMat, 2)
if dataMat(i,j) > 0
BCC.setChordMN(i,j, 'FaceAlpha',.6)
end
end
end
text(-1.2,1.2, 'a', 'FontName','Times New Roman', 'FontSize',35)
% =========================================================================
ax2 = axes('Parent',gcf, 'Position',[1/2,1/2,1/2,1/2]);
dataMat = rand([9,9]);
dataMat(dataMat > .4) = 0;
dataMat = dataMat.*(1:9);
BCC = biChordChart(dataMat, 'Arrow','on', 'CData',CList);
BCC = BCC.draw();
BCC.tickState('on')
BCC.setFont('Visible','off')
% 修改弦颜色(Modify chord color)
for i = 1:size(dataMat, 1)
for j = 1:size(dataMat, 2)
if dataMat(i,j) > 0
BCC.setChordMN(i,j, 'FaceAlpha',.6)
end
end
end
text(-1.2,1.2, 'b', 'FontName','Times New Roman', 'FontSize',35)
% =========================================================================
ax3 = axes('Parent',gcf, 'Position',[0,0,1/2,1/2]);
dataMat = rand([9,9]);
dataMat(dataMat > .4) = 0;
dataMat = dataMat.*(1:9).';
BCC = biChordChart(dataMat, 'Arrow','on', 'CData',CList);
BCC = BCC.draw();
BCC.tickState('on')
BCC.setFont('Visible','off')
% 修改弦颜色(Modify chord color)
for i = 1:size(dataMat, 1)
for j = 1:size(dataMat, 2)
if dataMat(i,j) > 0
BCC.setChordMN(i,j, 'FaceAlpha',.6)
end
end
end
text(-1.2,1.2, 'c', 'FontName','Times New Roman', 'FontSize',35)
% =========================================================================
ax4 = axes('Parent',gcf, 'Position',[1/2,0,1/2,1/2]);
ax4.XColor = 'none'; ax4.YColor = 'none';
ax4.XLim = [-1,1]; ax4.YLim = [-1,1];
hold on
NameList = {'Food supply', 'Biodiversity', 'Water quality regulation', ...
'Air quality regulation', 'Erosion control', 'Carbon storage', ...
'Water retention', 'Recreation', 'Soil quality regulation'};
patchHdl = [];
for i = 1:size(dataMat, 2)
patchHdl(i) = fill([10,11,12],[10,13,13], CList(i,:), 'EdgeColor',[0,0,0]);
end
lgdHdl = legend(patchHdl, NameList, 'Location','best', 'FontSize',14, 'FontName','Cambria', 'Box','off');
lgdHdl.Position = [.625,.11,.255,.27];
lgdHdl.ItemTokenSize = [18,8];
demo 16
dataMat = rand([15,15]);
dataMat(dataMat > .2) = 0;
CList = [ 75,146,241; 252,180, 65; 224, 64, 10; 5,100,146; 191,191,191;
26, 59,105; 255,227,130; 18,156,221; 202,107, 75; 0, 92,219;
243,210,136; 80, 99,129; 241,185,168; 224,131, 10; 120,147,190]./255;
CListC = [54,69,92]./255;
CList = CList.*.6 + CListC.*.4;
figure('Units','normalized', 'Position',[.02,.05,.6,.85])
BCC = biChordChart(dataMat, 'Arrow','on', 'CData',CList);
BCC = BCC.draw();
% 添加刻度
BCC.tickState('on')
% 修改字体，字号及颜色
BCC.setFont('FontName','Cambria', 'FontSize',17, 'Color',[0,0,0])
% 修改弦颜色(Modify chord color)
for i = 1:size(dataMat, 1)
for j = 1:size(dataMat, 2)
if dataMat(i,j) > 0
BCC.setChordMN(i,j, 'FaceColor',CListC ,'FaceAlpha',.07)
end
end
end
[~, N] = max(sum(dataMat > 0, 2));
for j = 1:size(dataMat, 2)
BCC.setChordMN(N,j, 'FaceColor',CList(N,:) ,'FaceAlpha',.6)
end
You need to download following tools:
- - Chord chart: [chord chart](https://www.mathworks.com/matlabcentral/fileexchange/116550-chord-chart)
- - Directed graph chord chart: [digraph chord chart]:(https://www.mathworks.com/matlabcentral/fileexchange/121043-digraph-chord-chart)
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