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MatGPT was launched on March 22, 2023 and I am amazed at how many times it has been downloaded since then - close to 16,000 downloads in one year. When AI Chat Playground came out on MATLAB Central, I thought surely that people will stop using MatGPT. Boy I was wrong.
In early 2023 I was playing with the new shiny toy called ChatGPT like everyone else but instead of having it tell me jokes or haiku, I wanted to know how I can use it on MATLAB, and I started collecting the prompts that worked. Someone suggested I should turn that into an app, and MatGPT was born with help from other colleagues.
Here is the question - what should I do with it now? Some people suggested I could add other LLMs like Gemini or Claude, but I am more interested in learning how people actually use it.
If you are a MatGPT user, do you mind sharing how you use the app?
I feel like no one at UC San Diego knows this page, let alone this server, is still live. For the younger generation, this is what the whole internet used to look like :)
In short: support varying color in at least the plot, plot3, fplot, and fplot3 functions.
This has been a thing that's come up quite a few times, and includes questions/requests by users, workarounds by the community, and workarounds presented by MathWorks -- examples of each below. It's a feature that exists in Python's Matplotlib library and Sympy. Anyways, given that there are myriads of workarounds, it appears to be one of the most common requests for Matlab plots (Matlab's plotting is, IMO, one of the best features of the product), the request precedes the 21st century, and competitive tools provide the functionality, it would seem to me that this might be the next great feature for Matlab plotting.
I'm curious to get the rest of the community's thoughts... what's everyone else think about this?
---
User questions/requests
- https://www.mathworks.com/matlabcentral/answers/480389-colored-line-plot-according-to-a-third-variable
- https://www.mathworks.com/matlabcentral/answers/2092641-how-to-solve-a-problem-with-the-generation-of-multiple-colored-segments-on-one-line-in-matlab-plot
- https://www.mathworks.com/matlabcentral/answers/5042-how-do-i-vary-color-along-a-2d-line
- https://www.mathworks.com/matlabcentral/answers/1917650-how-to-plot-a-trajectory-with-varying-colour
- https://www.mathworks.com/matlabcentral/answers/1917650-how-to-plot-a-trajectory-with-varying-colour
- https://www.mathworks.com/matlabcentral/answers/511523-how-to-create-plot3-varying-color-figure
- https://www.mathworks.com/matlabcentral/answers/393810-multiple-colours-in-a-trajectory-plot
- https://www.mathworks.com/matlabcentral/answers/523135-creating-a-rainbow-colour-plot-trajectory
- https://www.mathworks.com/matlabcentral/answers/469929-how-to-vary-the-color-of-a-dynamic-line
- https://www.mathworks.com/matlabcentral/answers/585011-how-could-i-adjust-the-color-of-multiple-lines-within-a-graph-without-using-the-default-matlab-colo
- https://www.mathworks.com/matlabcentral/answers/517177-how-to-interpolate-color-along-a-curve-with-specific-colors
- https://www.mathworks.com/matlabcentral/answers/281645-variate-color-depending-on-the-y-value-in-plot
- https://www.mathworks.com/matlabcentral/answers/439176-how-do-i-vary-the-color-along-a-line-in-polar-coordinates
- https://www.mathworks.com/matlabcentral/answers/1849193-creating-rainbow-coloured-plots-in-3d
- https://groups.google.com/g/comp.soft-sys.matlab/c/cLgjSeEC15I?hl=en&pli=1 (a question asked in 1999!)
- ... the list goes on, and on, and on...
User-provided workarounds
- https://undocumentedmatlab.com/articles/plot-line-transparency-and-color-gradient
- https://www.mathworks.com/matlabcentral/fileexchange/19476-colored-line-or-scatter-plot
- https://www.mathworks.com/matlabcentral/fileexchange/23566-3d-colored-line-plot
- https://www.mathworks.com/matlabcentral/fileexchange/30423-conditionally-colored-line-plot
- https://www.mathworks.com/matlabcentral/fileexchange/14677-cline
- https://www.mathworks.com/matlabcentral/fileexchange/8597-plot-3d-color-line
- https://www.mathworks.com/matlabcentral/fileexchange/39972-colormapline-color-changing-2d-or-3d-line
- https://www.mathworks.com/matlabcentral/fileexchange/37725-conditionally-colored-plot-ccplot
- https://www.mathworks.com/matlabcentral/fileexchange/11611-linear-2d-plot-with-rainbow-color
- https://www.mathworks.com/matlabcentral/fileexchange/26692-color_line
- https://www.mathworks.com/matlabcentral/fileexchange/32911-plot3rgb
- And perhaps more?
MathWorks-provided workarounds
- https://www.mathworks.com/videos/coloring-a-line-based-on-height-gradient-or-some-other-value-in-matlab-97128.html
- https://www.mathworks.com/videos/making-a-multi-color-line-in-matlab-97127.html
- https://www.mathworks.com/matlabcentral/fileexchange/95663-color-trajectory-plot (contributed by a MathWorks staff member)
- And perhaps more?
I was in a meeting the other day and a coworker shared a smiley face they created using the AI Chat Playground. The image looked something like this:
And I suspect the prompt they used was something like this:
"Create a smiley face"
I imagine this output wasn't what my coworker had expected so he was left thinking that this was as good as it gets without manually editing the code, and that the AI Chat Playground couldn't do any better.
I thought I could get a better result using the Playground so I tried a more detailed prompt using a multi-step technique like this:
"Follow these instructions:
- Create code that plots a circle
- Create two smaller circles as eyes within the first circle
- Create an arc that looks like a smile in the lower part of the first circle"
The output of this prompt was better in my opinion.
These queries/prompts are examples of 'zero-shot' prompts, the expectation being a good result with just one query. As opposed to a back-and-forth chat session working towards a desired outcome.
I wonder how many attempts everyone tries before they decide they can't anything more from the AI/LLM. There are times I'll send dozens of chat queries if I feel like I'm getting close to my goal, while other times I'll try just one or two. One thing I always find useful is seeing how others interact with AI models, which is what inspired me to share this.
Does anyone have examples of techniques that work well? I find multi-step instructions often produces good results.
The line integral 
, where C is the boundary of the square 
oriented counterclockwise, can be evaluated in two ways:
, where C is the boundary of the square oriented counterclockwise, can be evaluated in two ways: Using the definition of the line integral:
% Initialize the integral sum
integral_sum = 0;
% Segment C1: x = -1, y goes from -1 to 1
y = linspace(-1, 1);
x = -1 * ones(size(y));
dy = diff(y);
integral_sum = integral_sum + sum(-x(1:end-1) .* dy);
% Segment C2: y = 1, x goes from -1 to 1
x = linspace(-1, 1);
y = ones(size(x));
dx = diff(x);
integral_sum = integral_sum + sum(y(1:end-1).^2 .* dx);
% Segment C3: x = 1, y goes from 1 to -1
y = linspace(1, -1);
x = ones(size(y));
dy = diff(y);
integral_sum = integral_sum + sum(-x(1:end-1) .* dy);
% Segment C4: y = -1, x goes from 1 to -1
x = linspace(1, -1);
y = -1 * ones(size(x));
dx = diff(x);
integral_sum = integral_sum + sum(y(1:end-1).^2 .* dx);
disp(['Direct Method Integral: ', num2str(integral_sum)]);
Plotting the square path
% Define the square's vertices
vertices = [-1 -1; -1 1; 1 1; 1 -1; -1 -1];
% Plot the square
figure;
plot(vertices(:,1), vertices(:,2), '-o');
title('Square Path for Line Integral');
xlabel('x');
ylabel('y');
grid on;
axis equal;
% Add arrows to indicate the path direction (counterclockwise)
hold on;
for i = 1:size(vertices,1)-1
% Calculate direction
dx = vertices(i+1,1) - vertices(i,1);
dy = vertices(i+1,2) - vertices(i,2);
% Reduce the length of the arrow for better visibility
scale = 0.2;
dx = scale * dx;
dy = scale * dy;
% Calculate the start point of the arrow
startx = vertices(i,1) + (1 - scale) * dx;
starty = vertices(i,2) + (1 - scale) * dy;
% Plot the arrow
quiver(startx, starty, dx, dy, 'MaxHeadSize', 0.5, 'Color', 'r', 'AutoScale', 'off');
end
hold off;
Apply Green's Theorem for the line integral
% Define the partial derivatives of P and Q
f = @(x, y) -1 - 2*y; % derivative of -x with respect to x is -1, and derivative of y^2 with respect to y is 2y
% Compute the double integral over the square [-1,1]x[-1,1]
integral_value = integral2(f, -1, 1, 1, -1);
disp(['Green''s Theorem Integral: ', num2str(integral_value)]);
Plotting the vector field related to Green’s theorem
% Define the grid for the vector field
[x, y] = meshgrid(linspace(-2, 2, 20), linspace(-2 ,2, 20));
% Define the vector field components
P = y.^2; % y^2 component
Q = -x; % -x component
% Plot the vector field
figure;
quiver(x, y, P, Q, 'b');
hold on; % Hold on to plot the square on the same figure
% Define the square's vertices
vertices = [-1 -1; -1 1; 1 1; 1 -1; -1 -1];
% Plot the square path
plot(vertices(:,1), vertices(:,2), '-o', 'Color', 'k'); % 'k' for black color
title('Vector Field (P = y^2, Q = -x) with Square Path');
xlabel('x');
ylabel('y');
axis equal;
% Add arrows to indicate the path direction (counterclockwise)
for i = 1:size(vertices,1)-1
% Calculate direction
dx = vertices(i+1,1) - vertices(i,1);
dy = vertices(i+1,2) - vertices(i,2);
% Reduce the length of the arrow for better visibility
scale = 0.2;
dx = scale * dx;
dy = scale * dy;
% Calculate the start point of the arrow
startx = vertices(i,1) + (1 - scale) * dx;
starty = vertices(i,2) + (1 - scale) * dy;
% Plot the arrow
quiver(startx, starty, dx, dy, 'MaxHeadSize', 0.5, 'Color', 'r', 'AutoScale', 'off');
end
hold off;
To solve a surface integral for example the
over the sphere 
easily in MATLAB, you can leverage the symbolic toolbox for a direct and clear solution. Here is a tip to simplify the process:
over the sphere easily in MATLAB, you can leverage the symbolic toolbox for a direct and clear solution. Here is a tip to simplify the process:- Use Symbolic Variables and Functions: Define your variables symbolically, including the parameters of your spherical coordinates θ and ϕ and the radius r . This allows MATLAB to handle the expressions symbolically, making it easier to manipulate and integrate them.
- Express in Spherical Coordinates Directly: Since you already know the sphere's equation and the relationship in spherical coordinates, define x, y, and z in terms of r , θ and ϕ directly.
- Perform Symbolic Integration: Use MATLAB's `int` function to integrate symbolically. Since the sphere and the function
are symmetric, you can exploit these symmetries to simplify the calculation.
Here’s how you can apply this tip in MATLAB code:
% Include the symbolic math toolbox
syms theta phi
% Define the limits for theta and phi
theta_limits = [0, pi];
phi_limits = [0, 2*pi];
% Define the integrand function symbolically
integrand = 16 * sin(theta)^3 * cos(phi)^2;
% Perform the symbolic integral for the surface integral
surface_integral = int(int(integrand, theta, theta_limits(1), theta_limits(2)), phi, phi_limits(1), phi_limits(2));
% Display the result of the surface integral symbolically
disp(['The surface integral of x^2 over the sphere is ', char(surface_integral)]);
% Number of points for plotting
num_points = 100;
% Define theta and phi for the sphere's surface
[theta_mesh, phi_mesh] = meshgrid(linspace(double(theta_limits(1)), double(theta_limits(2)), num_points), ...
linspace(double(phi_limits(1)), double(phi_limits(2)), num_points));
% Spherical to Cartesian conversion for plotting
r = 2; % radius of the sphere
x = r * sin(theta_mesh) .* cos(phi_mesh);
y = r * sin(theta_mesh) .* sin(phi_mesh);
z = r * cos(theta_mesh);
% Plot the sphere
figure;
surf(x, y, z, 'FaceColor', 'interp', 'EdgeColor', 'none');
colormap('jet'); % Color scheme
shading interp; % Smooth shading
camlight headlight; % Add headlight-type lighting
lighting gouraud; % Use Gouraud shading for smooth color transitions
title('Sphere: x^2 + y^2 + z^2 = 4');
xlabel('x-axis');
ylabel('y-axis');
zlabel('z-axis');
colorbar; % Add color bar to indicate height values
axis square; % Maintain aspect ratio to be square
view([-30, 20]); % Set a nice viewing angle
I am often talking to new MATLAB users. I have put together one script. If you know how this script works, why, and what each line means, you will be well on your way on your MATLAB learning journey.
% Clear existing variables and close figures
clear;
close all;
% Print to the Command Window
disp('Hello, welcome to MATLAB!');
% Create a simple vector and matrix
vector = [1, 2, 3, 4, 5];
matrix = [1, 2, 3; 4, 5, 6; 7, 8, 9];
% Display the created vector and matrix
disp('Created vector:');
disp(vector);
disp('Created matrix:');
disp(matrix);
% Perform element-wise multiplication
result = vector .* 2;
% Display the result of the operation
disp('Result of element-wise multiplication of the vector by 2:');
disp(result);
% Create plot
x = 0:0.1:2*pi; % Generate values from 0 to 2*pi
y = sin(x); % Calculate the sine of these values
% Plotting
figure; % Create a new figure window
plot(x, y); % Plot x vs. y
title('Simple Plot of sin(x)'); % Give the plot a title
xlabel('x'); % Label the x-axis
ylabel('sin(x)'); % Label the y-axis
grid on; % Turn on the grid
disp('This is the end of the script. Explore MATLAB further to learn more!');
I would like to propose the creation of MATLAB EduHub, a dedicated channel within the MathWorks community where educators, students, and professionals can share and access a wealth of educational material that utilizes MATLAB. This platform would act as a central repository for articles, teaching notes, and interactive learning modules that integrate MATLAB into the teaching and learning of various scientific fields.
Key Features:
1. Resource Sharing: Users will be able to upload and share their own educational materials, such as articles, tutorials, code snippets, and datasets.
2. Categorization and Search: Materials can be categorized for easy searching by subject area, difficulty level, and MATLAB version..
3. Community Engagement: Features for comments, ratings, and discussions to encourage community interaction.
4. Support for Educators: Special sections for educators to share teaching materials and track engagement.
Benefits:
- Enhanced Educational Experience: The platform will enrich the learning experience through access to quality materials.
- Collaboration and Networking: It will promote collaboration and networking within the MATLAB community.
- Accessibility of Resources: It will make educational materials available to a wider audience.
By establishing MATLAB EduHub, I propose a space where knowledge and experience can be freely shared, enhancing the educational process and the MATLAB community as a whole.
In one line of MATLAB code, compute how far you can see at the seashore. In otherwords, how far away is the horizon from your eyes? You can assume you know your height and the diameter or radius of the earth.
Keep calm and study PDEs
Me at the beginning of every meeting
A bit late. Compliments to Chris for sharing.
can you relate?
Happy Pi Day!
3.14 π Day has arrived, and this post provides some very cool pi implementations and complete MATLAB code.
Firstly, in order to obtain the first n decimal places of pi, we need to write the following code (to prevent inaccuracies, we need to take a few more tails and perform another operation of taking the first n decimal places when needed):
function Pi=getPi(n)
if nargin<1,n=3;end
Pi=char(vpa(sym(pi),n+10));
Pi=abs(Pi)-48;
Pi=Pi(3:n+2);
end
With this function to obtain the decimal places of pi, our visualization journey has begun~Step by step, from simple to complex~(Please try to use newer versions of MATLAB to run, at least R17b)
1 Pie chart
Just calculate the proportion of each digit to the first 1500 decimal places:
% 获取pi前1500位小数
Pi=getPi(1500);
% 统计各个数字出现次数
numNum=find([diff(sort(Pi)),1]);
numNum=[numNum(1),diff(numNum)];
% 配色列表
CM=[20,164,199;43,187,170;53,165,81;189,190,28;248,167,22;
232,74,27;244,57,99;240,118,177;168,109,195;78,125,187]./255;
% 绘图并修饰
pieHdl=pie(numNum);
set(gcf,'Color',[1,1,1],'Position',[200,100,620,620]);
for i=1:2:20
pieHdl(i).EdgeColor=[1,1,1];
pieHdl(i).LineWidth=1;
pieHdl(i).FaceColor=CM((i+1)/2,:);
end
for i=2:2:20
pieHdl(i).Color=CM(i/2,:);
pieHdl(i).FontWeight='bold';
pieHdl(i).FontSize=14;
end
% 绘制图例并修饰
lgdHdl=legend(num2cell('0123456789'));
lgdHdl.FontWeight='bold';
lgdHdl.FontSize=11;
lgdHdl.TextColor=[.5,.5,.5];
lgdHdl.Location='southoutside';
lgdHdl.Box='off';
lgdHdl.NumColumns=10;
lgdHdl.ItemTokenSize=[20,15];
title("VISUALIZING \pi 'Pi' Chart | 1500 digits",'FontSize',18,...
'FontName','Times New Roman','Color',[.5,.5,.5])
2 line chart
Calculate the change in the proportion of each number:
% 获取pi前1500位小数
Pi=getPi(1500);
% 计算比例变化
Ratio=cumsum(Pi==(0:9)',2);
Ratio=Ratio./sum(Ratio);
D=1:length(Ratio);
% 配色列表
CM=[20,164,199;43,187,170;53,165,81;189,190,28;248,167,22;
232,74,27;244,57,99;240,118,177;168,109,195;78,125,187]./255;
hold on
% 循环绘图
for i=1:10
plot(D(20:end),Ratio(i,20:end),'Color',[CM(i,:),.6],'LineWidth',1.8)
end
% 坐标区域修饰
ax=gca;box on;grid on
ax.YLim=[0,.2];
ax.YTick=0:.05:.2;
ax.XTick=0:200:1400;
ax.YTickLabel={'0%','5%','10%','15%','20%'};
ax.XMinorTick='on';
ax.YMinorTick='on';
ax.LineWidth=.8;
ax.GridLineStyle='-.';
ax.FontName='Cambria';
ax.FontSize=11;
ax.XLabel.String='Decimals';
ax.YLabel.String='Proportion';
ax.XLabel.FontSize=13;
ax.YLabel.FontSize=13;
% 绘制图例并修饰
lgdHdl=legend(num2cell('0123456789'));
lgdHdl.NumColumns=5;
lgdHdl.FontWeight='bold';
lgdHdl.FontSize=11;
lgdHdl.TextColor=[.5,.5,.5];
3 stacked area diagram
% 获取pi前500位小数
Pi=getPi(500);
% 计算比例变化
Ratio=cumsum(Pi==(0:9)',2);
Ratio=Ratio./sum(Ratio);
% 配色列表
CM=[231,98,84;239,138,71;247,170,88;255,208,111;255,230,183;
170,220,224;114,188,213;82,143,173;55,103,149;30,70,110]./255;
% 绘制堆叠面积图
hold on
areaHdl=area(Ratio');
for i=1:10
areaHdl(i).FaceColor=CM(i,:);
areaHdl(i).FaceAlpha=.9;
end
% 图窗和坐标区域修饰
set(gcf,'Position',[200,100,720,420]);
ax=gca;
ax.YLim=[0,1];
ax.XMinorTick='on';
ax.YMinorTick='on';
ax.LineWidth=.8;
ax.FontName='Cambria';
ax.FontSize=11;
ax.TickDir='out';
ax.XLabel.String='Decimals';
ax.YLabel.String='Proportion';
ax.XLabel.FontSize=13;
ax.YLabel.FontSize=13;
ax.Title.String='Area Chart of Proportion — 500 digits';
ax.Title.FontSize=14;
% 绘制图例并修饰
lgdHdl=legend(num2cell('0123456789'));
lgdHdl.NumColumns=5;
lgdHdl.FontSize=11;
lgdHdl.Location='southeast';
4 connected stacked bar chart
% 获取pi前100位小数
Pi=getPi(100);
% 计算比例变化
Ratio=cumsum(Pi==(0:9)',2);
Ratio=Ratio./sum(Ratio);
X=Ratio(:,10:10:80)';
barHdl=bar(X,'stacked','BarWidth',.2);
CM=[231,98,84;239,138,71;247,170,88;255,208,111;255,230,183;
170,220,224;114,188,213;82,143,173;55,103,149;30,70,110]./255;
for i=1:10
barHdl(i).FaceColor=CM(i,:);
end
% 以下是生成连接的部分
hold on;axis tight
yEndPoints=reshape([barHdl.YEndPoints]',length(barHdl(1).YData),[])';
zeros(1,length(barHdl(1).YData));
yEndPoints=[zeros(1,length(barHdl(1).YData));yEndPoints];
barWidth=barHdl(1).BarWidth;
for i=1:length(barHdl)
for j=1:length(barHdl(1).YData)-1
y1=min(yEndPoints(i,j),yEndPoints(i+1,j));
y2=max(yEndPoints(i,j),yEndPoints(i+1,j));
if y1*y2<0
ty=yEndPoints(find(yEndPoints(i+1,j)*yEndPoints(1:i,j)>=0,1,'last'),j);
y1=min(ty,yEndPoints(i+1,j));
y2=max(ty,yEndPoints(i+1,j));
end
y3=min(yEndPoints(i,j+1),yEndPoints(i+1,j+1));
y4=max(yEndPoints(i,j+1),yEndPoints(i+1,j+1));
if y3*y4<0
ty=yEndPoints(find(yEndPoints(i+1,j+1)*yEndPoints(1:i,j+1)>=0,1,'last'),j+1);
y3=min(ty,yEndPoints(i+1,j+1));
y4=max(ty,yEndPoints(i+1,j+1));
end
fill([j+.5.*barWidth,j+1-.5.*barWidth,j+1-.5.*barWidth,j+.5.*barWidth],...
[y1,y3,y4,y2],barHdl(i).FaceColor,'FaceAlpha',.4,'EdgeColor','none');
end
end
% 图窗和坐标区域修饰
set(gcf,'Position',[200,100,720,420]);
ax=gca;box off
ax.YLim=[0,1];
ax.XMinorTick='on';
ax.YMinorTick='on';
ax.LineWidth=.8;
ax.FontName='Cambria';
ax.FontSize=11;
ax.TickDir='out';
ax.XTickLabel={'10','20','30','40','50','60','70','80'};
ax.XLabel.String='Decimals';
ax.YLabel.String='Proportion';
ax.XLabel.FontSize=13;
ax.YLabel.FontSize=13;
ax.Title.String='Area Chart of Proportion — 10-80 digits';
ax.Title.FontSize=14;
% 绘制图例并修饰
lgdHdl=legend(barHdl,num2cell('0123456789'));
lgdHdl.NumColumns=5;
lgdHdl.FontSize=11;
lgdHdl.Location='southeast';
5 bichord chart
Need to use this tool:
% 构建连接矩阵
dataMat=zeros(10,10);
Pi=getPi(1001);
for i=1:1000
dataMat(Pi(i)+1,Pi(i+1)+1)=dataMat(Pi(i)+1,Pi(i+1)+1)+1;
end
BCC=biChordChart(dataMat,'Arrow','on','Label',num2cell('0123456789'));
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)
set(gcf,'Position',[200,100,820,820]);
6 Gravity simulation diagram
Imagine each decimal as a small ball with a mass of
For example, if
, the weight of ball 0 is 1, ball 9 is 1.2589, the initial velocity of the ball is 0, and it is attracted by other balls. Gravity follows the inverse square law, and if the balls are close enough, they will collide and their value will become 
, the weight of ball 0 is 1, ball 9 is 1.2589, the initial velocity of the ball is 0, and it is attracted by other balls. Gravity follows the inverse square law, and if the balls are close enough, they will collide and their value will become After adding, take the mod, add the velocity direction proportionally, and recalculate the weight.
Pi=[3,getPi(71)];K=.18;
% 基础配置
CM=[239,32,120;239,60,52;247,98,32;255,182,60;247,235,44;
142,199,57;55,180,70;0,170,239;40,56,146;147,37,139]./255;
T=linspace(0,2*pi,length(Pi)+1)';
T=T(1:end-1);
ct=linspace(0,2*pi,100);
cx=cos(ct).*.027;
cy=sin(ct).*.027;
% 初始数据
Pi=Pi(:);
N=Pi;
X=cos(T);Y=sin(T);
VX=T.*0;VY=T.*0;
PX=X;PY=Y;
% 未碰撞时初始质量
getM=@(x)(x+1).^K;
M=getM(N);
% 绘制初始圆圈
hold on
for i=1:length(N)
fill(cx+X(i),cy+Y(i),CM(N(i)+1,:),'EdgeColor','w','LineWidth',1)
end
for k=1:800
% 计算加速度
Rn2=1./squareform(pdist([X,Y])).^2;
Rn2(eye(length(X))==1)=0;
MRn2=Rn2.*(M');
AX=X'-X;AY=Y'-Y;
normXY=sqrt(AX.^2+AY.^2);
AX=AX./normXY;AX(eye(length(X))==1)=0;
AY=AY./normXY;AY(eye(length(X))==1)=0;
AX=sum(AX.*MRn2,2)./150000;
AY=sum(AY.*MRn2,2)./150000;
% 计算速度及新位置
VX=VX+AX;X=X+VX;PX=[PX,X];
VY=VY+AY;Y=Y+VY;PY=[PY,Y];
% 检测是否有碰撞
R=squareform(pdist([X,Y]));
R(triu(ones(length(X)))==1)=inf;
[row,col]=find(R<=0.04);
if length(X)==1
break;
end
if ~isempty(row)
% 碰撞的点合为一体
XC=(X(row)+X(col))./2;YC=(Y(row)+Y(col))./2;
VXC=(VX(row).*M(row)+VX(col).*M(col))./(M(row)+M(col));
VYC=(VY(row).*M(row)+VY(col).*M(col))./(M(row)+M(col));
PC=nan(length(row),size(PX,2));
NC=mod(N(row)+N(col),10);
% 删除碰撞点并绘图
uniNum=unique([row;col]);
X(uniNum)=[];VX(uniNum)=[];
Y(uniNum)=[];VY(uniNum)=[];
for i=1:length(uniNum)
plot(PX(uniNum(i),:),PY(uniNum(i),:),'LineWidth',2,'Color',CM(N(uniNum(i))+1,:))
end
PX(uniNum,:)=[];PY(uniNum,:)=[];N(uniNum,:)=[];
% 绘制圆形
for i=1:length(XC)
fill(cx+XC(i),cy+YC(i),CM(NC(i)+1,:),'EdgeColor','w','LineWidth',1)
end
% 补充合体点
X=[X;XC];Y=[Y;YC];VX=[VX;VXC];VY=[VY;VYC];
PX=[PX;PC];PY=[PY;PC];N=[N;NC];M=getM(N);
end
end
for i=1:size(PX,1)
plot(PX(i,:),PY(i,:),'LineWidth',2,'Color',CM(N(i)+1,:))
end
text(-1,1,{['Num=',num2str(length(Pi))];['K=',num2str(K)]},'FontSize',13,'FontName','Cambria')
% 图窗及坐标区域修饰
set(gcf,'Position',[200,100,820,820]);
ax=gca;
ax.Position=[0,0,1,1];
ax.DataAspectRatio=[1,1,1];
ax.XLim=[-1.1,1.1];
ax.YLim=[-1.1,1.1];
ax.XTick=[];
ax.YTick=[];
ax.XColor='none';
ax.YColor='none';
7 forest chart
The method comes from
The digits of π are shown as a forest. Each tree in the forest represents the digits of π up to the next 9. The first 10 trees are "grown" from the digit sets 314159, 2653589, 79, 3238462643383279, 50288419, 7169, 39, 9, 3751058209, and 749.
BRANCHES
The first digit of a tree controls how many branches grow from the trunk of the tree. For example, the first tree's first digit is 3, so you see 3 branches growing from the trunk.
The next digit's branches grow from the end of a branch of the previous digit in left-to-right order. This process continues until all the tree's digits have been used up.
Each tree grows from a set of consecutive digits sampled from the digits of π up to the next 9. The first tree, shown here, grows from 314159. Each of the digits determine how many branches grow at each fork in the tree — the branches here are colored by their corresponding digit to illustrate this. Leaves encode the digits in a left-to-right order. The digit 9 spawns a flower on one of the branches of the previous digit. The branching exception is 0, which terminates the current branch — 0 branches grow!
LEAVES AND FLOWERS
The tree's digits themselves are drawn as circular leaves, color-coded by the digit.
The leaf exception is 9, which causes one of the branches of the previous digit to sprout a flower! The petals of the flower are colored by the digit before the 9 and the center is colored by the digit after the 9, which is on the next tree. This is how the forest propagates.
The colors of a flower are determined by the first digit of the next tree and the penultimate digit of the current tree. If the current tree only has one digit, then that digit is used. Leaves are placed at the tips of branches in a left-to-right order — you can "easily" read them off. Additionally, the leaves are distributed within the tree (without disturbing their left-to-right order) to spread them out as much as possible and avoid overlap. This order is deterministic.
The leaf placement exception are the branch set that sprouted the flower. These are not used to grow leaves — the flower needs space!
function PiTree(X,pos,D)
lw=2;
theta=pi/2+(rand(1)-.5).*pi./12;
% 树叶及花朵颜色
CM=[237,32,121;237,62,54;247,99,33;255,183,59;245,236,43;
141,196,63;57,178,74;0,171,238;40,56,145;146,39,139]./255;
hold on
if all(X(1:end-2)==0)
endSet=[pos,pos,theta];
else
kplot(pos(1)+[0,cos(theta)],pos(2)+[0,sin(theta)],lw./.6)
endSet=[pos,pos+[cos(theta),sin(theta)],theta];
% 计算层级
Layer=0;
for i=1:length(X)
Layer=[Layer,ones(1,X(i)).*i];
end
% 计算树枝
if D
for i=1:length(X)-2
if X(i)==0 % 若数值为0则不长树枝
newSet=endSet(1,:);
elseif X(i)==1 % 若数值为1则一长一短两个树枝
tTheta=endSet(1,5);
tTheta=linspace(tTheta+pi/8,tTheta-pi/8,2)'+(rand([2,1])-.5).*pi./8;
newSet=repmat(endSet(1,3:4),[X(i),1]);
newSet=[newSet.*[1;1],newSet+[cos(tTheta),sin(tTheta)].*.7^Layer(i).*[1;.1],tTheta];
else % 其他情况数值为几长几个树枝
tTheta=endSet(1,5);
tTheta=linspace(tTheta+pi/5,tTheta-pi/5,X(i))'+(rand([X(i),1])-.5).*pi./8;
newSet=repmat(endSet(1,3:4),[X(i),1]);
newSet=[newSet,newSet+[cos(tTheta),sin(tTheta)].*.7^Layer(i),tTheta];
end
% 绘制树枝
for j=1:size(newSet,1)
kplot(newSet(j,[1,3]),newSet(j,[2,4]),lw.*.6^Layer(i))
end
endSet=[endSet;newSet];
endSet(1,:)=[];
end
end
end
% 计算叶子和花朵位置
FLSet=endSet(:,3:4);
[~,FLInd]=sort(FLSet(:,1));
FLSet=FLSet(FLInd,:);
[~,tempInd]=sort(rand([1,size(FLSet,1)]));
tempInd=sort(tempInd(1:length(X)-2));
flowerInd=tempInd(randi([1,length(X)-2],[1,1]));
leafInd=tempInd(tempInd~=flowerInd);
% 绘制树叶
for i=1:length(leafInd)
scatter(FLSet(leafInd(i),1),FLSet(leafInd(i),2),70,'filled','CData',CM(X(i)+1,:))
end
% 绘制花朵
for i=1:5
% if ~D
% tC=CM(X(end)+1,:);
% else
% tC=CM(X(end-2)+1,:);
% end
scatter(FLSet(flowerInd,1)+cos(pi*2*i/5).*.18,FLSet(flowerInd,2)+sin(pi*2*i/5).*.18,60,...
'filled','CData',CM(X(end-2)+1,:),'MarkerEdgeColor',[1,1,1])
end
scatter(FLSet(flowerInd,1),FLSet(flowerInd,2),60,'filled','CData',CM(X(end)+1,:),'MarkerEdgeColor',[1,1,1])
drawnow;%axis tight
% =========================================================================
function kplot(XX,YY,LW,varargin)
LW=linspace(LW,LW*.6,10);%+rand(1,20).*LW./10;
XX=linspace(XX(1),XX(2),11)';
XX=[XX(1:end-1),XX(2:end)];
YY=linspace(YY(1),YY(2),11)';
YY=[YY(1:end-1),YY(2:end)];
for ii=1:10
plot(XX(ii,:),YY(ii,:),'LineWidth',LW(ii),'Color',[.1,.1,.1])
end
end
end
main part:
Pi=[3,getPi(800)];
pos9=[0,find(Pi==9)];
set(gcf,'Position',[200,50,900,900],'Color',[1,1,1]);
ax=gca;hold on
ax.Position=[0,0,1,1];
ax.DataAspectRatio=[1,1,1];
ax.XLim=[.5,36];
ax.XTick=[];
ax.YTick=[];
ax.XColor='none';
ax.YColor='none';
for j=1:8
for i=1:11
n=i+(j-1)*11;
if n<=85
tPi=Pi((pos9(n)+1):pos9(n+1)+1);
if length(tPi)>2
PiTree(tPi,[0+i*3,0-j*4],true);
else
PiTree([Pi(pos9(n)),tPi],[0+i*3,0-j*4],false);
end
end
end
end
8 random walk
n=1200;
% 获取pi前n位小数
Pi=getPi(n);
CM=[239,65,75;230,115,48;229,158,57;232,136,85;239,199,97;
144,180,116;78,166,136;81,140,136;90,118,142;43,121,159]./255;
hold on
endPoint=[0,0];
t=linspace(0,2*pi,100);
T=linspace(0,2*pi,11)+pi/2;
fill(endPoint(1)+cos(t).*.5,endPoint(2)+sin(t).*.5,CM(Pi(1)+1,:),'EdgeColor','none')
for i=1:n
theta=T(Pi(i)+1);
plot(endPoint(1)+[0,cos(theta)],endPoint(2)+[0,sin(theta)],'Color',[CM(Pi(i)+1,:),.8],'LineWidth',1.2);
endPoint=endPoint+[cos(theta),sin(theta)];
end
fill(endPoint(1)+cos(t).*.5,endPoint(2)+sin(t).*.5,CM(Pi(n)+1,:),'EdgeColor','none')
% 图窗和坐标区域修饰
set(gcf,'Position',[200,100,820,820]);
ax=gca;
ax.XTick=[];
ax.YTick=[];
ax.Color=[0,0,0];
ax.DataAspectRatio=[1,1,1];
ax.XLim=[-30,5];
ax.YLim=[-5,40];
% 绘制图例
endPoint=[1,35];
for i=1:10
theta=T(i);
plot(endPoint(1)+[0,cos(theta).*2],endPoint(2)+[0,sin(theta).*2],'Color',[CM(i,:),.8],'LineWidth',3);
text(endPoint(1)+cos(theta).*2.7,endPoint(2)+sin(theta).*2.7,num2str(i-1),'Color',[1,1,1].*.7,...
'FontSize',12,'FontWeight','bold','FontName','Cambria','HorizontalAlignment','center')
end
text(-15,35,'Random walk of \pi digits','Color',[1,1,1],'FontName','Cambria',...
'HorizontalAlignment','center','FontSize',25,'FontAngle','italic')
9 grid chart
Pi=[3,getPi(399)];
% 配色数据
CM=[248,65,69;246,152,36;249,198,81;67,170,139;87,118,146]./255;
% 绘制圆圈
hold on
t=linspace(0,2*pi,100);
x=cos(t).*.8.*.5;
y=sin(t).*.8.*.5;
for i=1:400
[col,row]=ind2sub([20,20],i);
if mod(Pi(i),2)==0
fill(x+col,y+row,CM(round((Pi(i)+1)/2),:),'LineWidth',1,'EdgeAlpha',.8)
else
fill(x+col,y+row,[0,0,0],'EdgeColor',CM(round((Pi(i)+1)/2),:),'LineWidth',1,'EdgeAlpha',.7)
end
end
text(10.5,-.4,'\pi on a grid — 400 digits','Color',[1,1,1],'FontName','Cambria',...
'HorizontalAlignment','center','FontSize',25,'FontAngle','italic')
% 图窗和坐标区域修饰
set(gcf,'Position',[200,100,820,820]);
ax=gca;
ax.YDir='reverse';
ax.XLim=[.5,20.5];
ax.YLim=[-1,20.5];
ax.XTick=[];
ax.YTick=[];
ax.Color=[0,0,0];
ax.DataAspectRatio=[1,1,1];
10 scale grid diagram
Let's still put the numbers in the form of circles, but the difference is that six numbers are grouped together, and the pure purple circle at the end is the six 9s that we are familiar with decimal places 762-767
Pi=[3,getPi(767)];
% 762-767
% 配色数据
CM=[239,32,120;239,60,52;247,98,32;255,182,60;247,235,44;
142,199,57;55,180,70;0,170,239;40,56,146;147,37,139]./255;
% 绘制圆圈
hold on
t=linspace(0,2*pi,100);
x=cos(t).*.9.*.5;
y=sin(t).*.9.*.5;
for i=1:6:length(Pi)
n=round((i-1)/6+1);
[col,row]=ind2sub([10,13],n);
tNum=Pi(i:i+5);
numNum=find([diff(sort(tNum)),1]);
numNum=[numNum(1),diff(numNum)];
cumNum=cumsum(numNum);
uniNum=unique(tNum);
for j=length(cumNum):-1:1
fill(x./6.*cumNum(j)+col,y./6.*cumNum(j)+row,CM(uniNum(j)+1,:),'EdgeColor','none')
end
end
% 绘制图例
for i=1:10
fill(x./4+5.5+(i-5.5)*.32,y./4+14.5,CM(i,:),'EdgeColor','none')
text(5.5+(i-5.5)*.32,14.9,num2str(i-1),'Color',[1,1,1],'FontSize',...
9,'FontName','Cambria','HorizontalAlignment','center')
end
text(5.5,-.2,'FEYNMAN POINT of \pi','Color',[1,1,1],'FontName','Cambria',...
'HorizontalAlignment','center','FontSize',25,'FontAngle','italic')
% 图窗和坐标区域修饰
set(gcf,'Position',[200,100,600,820]);
ax=gca;
ax.YDir='reverse';
ax.Position=[0,0,1,1];
ax.XLim=[.3,10.7];
ax.YLim=[-1,15.5];
ax.XTick=[];
ax.YTick=[];
ax.Color=[0,0,0];
ax.DataAspectRatio=[1,1,1];
11 text chart
First, write a code to generate an image of each letter:
function getLogo
if ~exist('image','dir')
mkdir('image\')
end
logoSet=['.',char(65:90)];
for i=1:27
figure();
ax=gca;
ax.XLim=[-1,1];
ax.YLim=[-1,1];
ax.XColor='none';
ax.YColor='none';
ax.DataAspectRatio=[1,1,1];
logo=logoSet(i);
hold on
text(0,0,logo,'HorizontalAlignment','center','FontSize',320,'FontName','Segoe UI Black')
exportgraphics(ax,['image\',logo,'.png'])
close
end
dotPic=imread('image\..png');
newDotPic=uint8(ones([400,size(dotPic,2),3]).*255);
newDotPic(end-size(dotPic,1)+1:end,:,1)=dotPic(:,:,1);
newDotPic(end-size(dotPic,1)+1:end,:,2)=dotPic(:,:,2);
newDotPic(end-size(dotPic,1)+1:end,:,3)=dotPic(:,:,3);
imwrite(newDotPic,'image\..png')
S=20;
for i=1:27
logo=logoSet(i);
tPic=imread(['image\',logo,'.png']);
sz=size(tPic,[1,2]);
sz=round(sz./sz(1).*400);
tPic=imresize(tPic,sz);
tBox=uint8(255.*ones(size(tPic,[1,2])+S));
tBox(S+1:S+size(tPic,1),S+1:S+size(tPic,2))=tPic(:,:,1);
imwrite(cat(3,tBox,tBox,tBox),['image\',logo,'.png'])
end
end
Pi=[3,-1,getPi(150)];
CM=[109,110,113;224,25,33;244,126,26;253,207,2;154,203,57;111,150,124;
121,192,235;6,109,183;190,168,209;151,118,181;233,93,163]./255;
ST={'.','ZERO','ONE','TWO','THREE','FOUR','FIVE','SIX','SEVEN','EIGHT','NINE'};
n=1;
hold on
% 循环绘制字母
for i=1:20%:10
STList='';
NMList=[];
PicListR=uint8(zeros(400,0));
PicListG=uint8(zeros(400,0));
PicListB=uint8(zeros(400,0));
% PicListA=uint8(zeros(400,0));
for j=1:6
STList=[STList,ST{Pi(n)+2}];
NMList=[NMList,ones(size(ST{Pi(n)+2})).*(Pi(n)+2)];
n=n+1;
if length(STList)>15&&length(STList)+length(ST{Pi(n)+2})>20
break;
end
end
for k=1:length(STList)
tPic=imread(['image\',STList(k),'.png']);
% PicListA=[PicListA,tPic(:,:,1)];
PicListR=[PicListR,(255-tPic(:,:,1)).*CM(NMList(k),1)];
PicListG=[PicListG,(255-tPic(:,:,2)).*CM(NMList(k),2)];
PicListB=[PicListB,(255-tPic(:,:,3)).*CM(NMList(k),3)];
end
PicList=cat(3,PicListR,PicListG,PicListB);
image([-1200,1200],[0,150]-(i-1)*150,flipud(PicList))
end
% 图窗及坐标区域修饰
set(gcf,'Position',[200,100,600,820]);
ax=gca;
ax.DataAspectRatio=[1,1,1];
ax.XLim=[-1300,1300];
ax.Position=[0,0,1,1];
ax.XTick=[];
ax.YTick=[];
ax.Color=[0,0,0];
ax.YLim=[-19*150-80,230];
12 spiral chart
Pi=getPi(600);
% 配色列表
CM=[78,121,167;242,142,43;225,87,89;118,183,178;89,161,79;
237,201,72;176,122,161;255,157,167;156,117,95;186,176,172]./255;
% 绘制圆圈
hold on
t=linspace(0,2*pi,100);
x=cos(t).*.8;
y=sin(t).*.8;
for i=1:600
X=i.*cos(i./10)./10;
Y=i.*sin(i./10)./10;
fill(X+x,Y+y,CM(Pi(i)+1,:),'EdgeColor','none','FaceAlpha',.9)
end
text(0,65,'The Circle of \pi','Color',[1,1,1],'FontName','Cambria',...
'HorizontalAlignment','center','FontSize',25,'FontAngle','italic')
% 图窗和坐标区域修饰
set(gcf,'Position',[200,100,820,820]);
ax=gca;
ax.XLim=[-60,60];
ax.YLim=[-60,70];
ax.XTick=[];
ax.YTick=[];
ax.Color=[0,0,0];
ax.DataAspectRatio=[1,1,1];
13 Archimedean spiral diagram
a=1;b=.227;
Pi=getPi(500);
% 配色列表
CM=[78,121,167;242,142,43;225,87,89;118,183,178;89,161,79;
237,201,72;176,122,161;255,157,167;156,117,95;186,176,172]./255;
% 绘制圆圈
hold on
T=0;R=1;
t=linspace(0,2*pi,100);
x=cos(t).*.7;
y=sin(t).*.7;
for i=1:500
X=R.*cos(T);Y=R.*sin(T);
fill(X+x,Y+y,CM(Pi(i)+1,:),'EdgeColor','none','FaceAlpha',.9)
T=T+1./R.*1.4;
R=a+b*T;
end
text(17.25,22,{'The Archimedes spiral of \pi';'—— 500 digits'},...
'Color',[1,1,1],'FontName','Cambria',...
'HorizontalAlignment','right','FontSize',25,'FontAngle','italic')
% 图窗和坐标区域修饰
set(gcf,'Position',[200,100,820,820]);
ax=gca;
ax.XLim=[-19,18.5];
ax.YLim=[-20,25];
ax.XTick=[];
ax.YTick=[];
ax.Color=[0,0,0];
ax.DataAspectRatio=[1,1,1];
14 proportional Archimedean spiral diagram
Pi=[3,getPi(1199)];
% 配色数据
CM=[239,32,120;239,60,52;247,98,32;255,182,60;247,235,44;
142,199,57;55,180,70;0,170,239;40,56,146;147,37,139]./255;
% CM=slanCM(184,10);
% 绘制圆圈
hold on
T=0;R=1;
t=linspace(0,2*pi,100);
x=cos(t).*.7;
y=sin(t).*.7;
for i=1:4:length(Pi)
X=R.*cos(T);Y=R.*sin(T);
tNum=Pi(i:i+3);
numNum=find([diff(sort(tNum)),1]);
numNum=[numNum(1),diff(numNum)];
cumNum=cumsum(numNum);
uniNum=unique(tNum);
for j=length(cumNum):-1:1
fill(x./4.*cumNum(j)+X,y./4.*cumNum(j)+Y,CM(uniNum(j)+1,:),'EdgeColor','none')
end
T=T+1./R.*1.4;
R=a+b*T;
end
text(14,16.5,{'The ratio of four numbers from \pi';'—— 1200 digits'},...
'Color',[1,1,1],'FontName','Cambria',...
'HorizontalAlignment','right','FontSize',23,'FontAngle','italic')
% 图窗和坐标区域修饰
set(gcf,'Position',[200,100,820,820]);
ax=gca;
ax.XLim=[-15,15.5];
ax.YLim=[-15,19];
ax.XTick=[];
ax.YTick=[];
ax.Color=[0,0,0];
ax.DataAspectRatio=[1,1,1];
15 graph
% 构建连接矩阵
corrMat=zeros(10,10);
Pi=getPi(401);
for i=1:400
corrMat(Pi(i)+1,Pi(i+1)+1)=corrMat(Pi(i)+1,Pi(i+1)+1)+1;
end
% 配色列表
colorList=[0.3725 0.2745 0.5647
0.1137 0.4118 0.5882
0.2196 0.6510 0.6471
0.0588 0.5216 0.3294
0.4510 0.6863 0.2824
0.9294 0.6784 0.0314
0.8824 0.4863 0.0196
0.8000 0.3137 0.2431
0.5804 0.2039 0.4314
0.4353 0.2510 0.4392];
t=linspace(0,2*pi,11);t=t(1:10)';
posXY=[cos(t),sin(t)];
maxWidth=max(corrMat(corrMat>0));
minWidth=min(corrMat(corrMat>0));
ttList=linspace(0,1,3)';
% 循环绘图
hold on
for i=1:size(corrMat,1)
for j=i+1:size(corrMat,2)
if corrMat(i,j)>0
tW=(corrMat(i,j)-minWidth)./(maxWidth-minWidth);
colorData=(1-ttList).*colorList(i,:)+ttList.*colorList(j,:);
CData(:,:,1)=colorData(:,1);
CData(:,:,2)=colorData(:,2);
CData(:,:,3)=colorData(:,3);
% 绘制连线
fill(linspace(posXY(i,1),posXY(j,1),3),...
linspace(posXY(i,2),posXY(j,2),3),[0,0,0],'LineWidth',tW.*12+1,...
'CData',CData,'EdgeColor','interp','EdgeAlpha',.7,'FaceAlpha',.7)
end
end
% 绘制圆点
scatter(posXY(i,1),posXY(i,2),200,'filled','LineWidth',1.2,...
'MarkerFaceColor',colorList(i,:),'MarkerEdgeColor',[.7,.7,.7]);
text(posXY(i,1).*1.13,posXY(i,2).*1.13,num2str(i-1),'Color',[1,1,1].*.7,...
'FontSize',30,'FontWeight','bold','FontName','Cambria','HorizontalAlignment','center')
end
text(0,1.3,'Numerical adjacency of \pi — 400 digits','Color',[1,1,1],'FontName','Cambria',...
'HorizontalAlignment','center','FontSize',25,'FontAngle','italic')
% 图窗和坐标区域修饰
set(gcf,'Position',[200,100,820,820]);
ax=gca;
ax.XLim=[-1.2,1.2];
ax.YLim=[-1.21,1.5];
ax.XTick=[];
ax.YTick=[];
ax.Color=[0,0,0];
ax.DataAspectRatio=[1,1,1];
16 circos chart
Need to use this tool:
Class=getPi(1001)+1;
Data=diag(ones(1,1000),-1);
className={'0','1','2','3','4','5','6','7','8','9'};
colorOrder=[239,65,75;230,115,48;229,158,57;232,136,85;239,199,97;
144,180,116;78,166,136;81,140,136;90,118,142;43,121,159]./255;
CC=circosChart(Data,Class,'ClassName',className,'ColorOrder',colorOrder);
CC=CC.draw();
ax=gca;
ax.Color=[0,0,0];
CC.setClassLabel('Color',[1,1,1],'FontSize',25,'FontName','Cambria')
CC.setLine('LineWidth',.7)
YOU CAN GET ALL CODE HERE:
Can you solve it?
Chord diagrams are very common in Python and R, but there are no related functions in MATLAB before. It is not easy to draw chord diagrams of the same quality as R language, But I created a MATLAB tool that could almost do it.
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Here is the help document:
1 Data Format
The data requirement is a numerical matrix with all values greater than or equal to 0, or a table array, or a numerical matrix and cell array for names. First, give an example of a numerical matrix:
1.1 Numerical Matrix
dataMat=randi([0,5],[5,4]);
% 绘图(draw)
CC=chordChart(dataMat);
CC=CC.draw();
Since each object is not named, it will be automatically named Rn and Cn
1.2 Numerical Matrix and Cell Array for Names
dataMat=[2 0 1 2 5 1 2;
3 5 1 4 2 0 1;
4 0 5 5 2 4 3];
colName={'G1','G2','G3','G4','G5','G6','G7'};
rowName={'S1','S2','S3'};
CC=chordChart(dataMat,'rowName',rowName,'colName',colName);
CC=CC.draw();
RowName should be the same size as the rows of the matrix
ColName should be the same size as the columns of the matrix
For this example, if the value in the second row and third column is 1, it indicates that there is an energy flow from S2 to G3, and a chord with a width of 1 is needed between these two.
1.3 Table Array
A table array in the following format is required:
2 Decorate Chord
2.1 Batch modification of chords
Batch modification of chords can be done using the setChordProp function, and all properties of the Patch object can be modified. For example, modifying the color of the string, edge color, edge line sstyle, etc.:
CC.setChordProp('EdgeColor',[.3,.3,.3],'LineStyle','--',...
'LineWidth',.1,'FaceColor',[.3,.3,.3])
2.2 Individual Modification of Chord
The individual modification of chord can be done using the setChordMN function, where the values of m and n correspond exactly to the rows and columns of the original numerical matrix. For example, changing the color of the strings flowing from S2 to G4 to red:
CC.setChordMN(2,4,'FaceColor',[1,0,0])
2.3 Color Mapping of Chords
Just use function colormap to do so:
% version 1.7.0更新
% 可使用colormap函数直接修改颜色
% Colors can be adjusted directly using the function colormap(demo4)
colormap(flipud(pink))
3 Arc Shaped Block Decoration
3.1 Batch Decoration of Arc-Shaped Blocks
use:
- setSquareT_Prop
- setSquareF_Prop
to modify the upper and lower blocks separately, and all attributes of the Patch object can be modified. For example, batch modify the upper blocks (change to black):
CC.setSquareT_Prop('FaceColor',[0,0,0])
3.2 Arc-Shaped Blocks Individually Decoration
use:
- setSquareT_N
- setSquareF_N
to modify the upper and lower blocks separately. For example, modify the second block above separately (changed to red):
CC.setSquareT_N(2,'FaceColor',[.8,0,0])
4 Font Adjustment
Use the setFont function to adjust the font, and all properties of the text object can be modified. For example, changing the font size, font, and color of the text:
CC.setFont('FontSize',25,'FontName','Cambria','Color',[0,0,.8])
5 Show and Hide Ticks
Usage:
CC.tickState('on')
% CC.tickState('off')
6 Attribute 'Sep' with Adjustable Square Spacing
If the matrix size is large, the drawing will be out of scale:
dataMat=randi([0,1],[20,10]);
CC=chordChart(dataMat);
CC=CC.draw();
% CC.tickState('on')
We can modify its Sep attribute:
dataMat=randi([0,1],[20,10]);
% use Sep to decrease space (separation)
% 使用 sep 减小空隙
CC=chordChart(dataMat,'Sep',1/120);
CC=CC.draw();
7 Modify Text Direction
dataMat=randi([0,1],[20,10]);
% use Sep to decrease space (separation)
% 使用 sep 减小空隙
CC=chordChart(dataMat,'Sep',1/120);
CC=CC.draw();
CC.tickState('on')
% version 1.7.0更新
% 函数labelRatato用来旋转标签
% The function labelRatato is used to rotate the label
CC.labelRotate('on')
8 Add Tick Labels
dataMat=[2 0 1 2 5 1 2;
3 5 1 4 2 0 1;
4 0 5 5 2 4 3];
colName={'G1','G2','G3','G4','G5','G6','G7'};
rowName={'S1','S2','S3'};
CC=chordChart(dataMat,'rowName',rowName,'colName',colName);
CC=CC.draw();
CC.setFont('FontSize',17,'FontName','Cambria')
% 显示刻度和数值
% Displays scales and numeric values
CC.tickState('on')
CC.tickLabelState('on')
% 调节标签半径
% Adjustable Label radius
CC.setLabelRadius(1.3);
% figure()
% dataMat=[2 0 1 2 5 1 2;
% 3 5 1 4 2 0 1;
% 4 0 5 5 2 4 3];
% dataMat=dataMat+rand(3,7);
% dataMat(dataMat<1)=0;
%
% CC=chordChart(dataMat,'rowName',rowName,'colName',colName);
% CC=CC.draw();
% CC.setFont('FontSize',17,'FontName','Cambria')
%
% % 显示刻度和数值
% % Displays scales and numeric values
% CC.tickState('on')
% CC.tickLabelState('on')
%
% % 调节标签半径
% % Adjustable Label radius
% CC.setLabelRadius(1.4);
9 Custom Tick Label Format
A function handle is required to input numeric output strings. The format can be set through the setTickLabelFormat function, such as Scientific notation:
dataMat=[2 0 1 2 5 1 2;
3 5 1 4 2 0 1;
4 0 5 5 2 4 3];
dataMat=dataMat+rand(3,7);
dataMat(dataMat<1)=0;
dataMat=dataMat.*1000;
CC=chordChart(dataMat);
CC=CC.draw();
CC.setFont('FontSize',17,'FontName','Cambria')
% 显示刻度和数值
% Displays scales and numeric values
CC.tickState('on')
CC.tickLabelState('on')
% 调节标签半径
% Adjustable Label radius
CC.setLabelRadius(1.4);
% 调整数值字符串格式
% Adjust numeric string format
CC.setTickLabelFormat(@(x)sprintf('%0.1e',x))
10 A Demo
rng(2)
dataMat=randi([1,7],[11,5]);
colName={'Fly','Beetle','Leaf','Soil','Waxberry'};
rowName={'Bartomella','Bradyrhizobium','Dysgomonas','Enterococcus',...
'Lactococcus','norank','others','Pseudomonas','uncultured',...
'Vibrionimonas','Wolbachia'};
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=1:5
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=1:11
CC.setSquareF_N(i,'FaceColor',CListF(i,:))
end
% 修改弦颜色(Modify chord color)
for i=1:5
for j=1:11
CC.setChordMN(j,i,'FaceColor',CListT(i,:),'FaceAlpha',.5)
end
end
CC.tickState('on')
CC.labelRotate('on')
CC.setFont('FontSize',17,'FontName','Cambria')
Hope to have your Reviews and Stars!!!
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Most of my projects are the tension between what exists and what I must create to make the project work. The more I leverage what exists, the faster I can complete my project in a shorter amount of time. I often find out later in the process that a MATLAB function already exists and has existed for 10 releases. The faster that I discover built-in functions, the faster I can get to the new stuff that I have to invent.
I asked the AI Chat Playground why to use MATLAB's builtin functions and I got some interesting ideas. What are your thoughts?
AI Chat Playground:
- Efficiency and Performance: MATLAB's built-in functions are highly optimized for performance. They are written in optimized and compiled languages like C or C++, which means they run faster than equivalent functions written directly in MATLAB language. When you use these functions, your code can execute more efficiently, which is especially beneficial for handling large datasets or complex mathematical computations.
- Reliability and Stability: Built-in functions have been rigorously tested and debugged by the developers at MathWorks (the company behind MATLAB). They are less likely to contain errors compared to custom-written code, which means you can rely on them to produce accurate and consistent results.
- Ease of Use: Built-in functions are designed to be user-friendly, with clear documentation and support materials available. This can make it easier to perform complex operations without needing a deep understanding of the underlying algorithms. It also reduces the amount of code you need to write and debug, saving time and reducing the potential for errors.
- Compatibility: Built-in functions are updated alongside MATLAB itself, ensuring compatibility with new features and versions of the software. Using built-in functions can help ensure that your code continues to work correctly as MATLAB evolves.
- Advanced Features: MATLAB’s built-in functions often include advanced features and options that are not readily available or would be difficult to implement on your own. For example, MATLAB’s mathematical functions can handle a wide range of special cases and data types, providing more flexibility than custom implementations.
- Community and Support: Using built-in functions means you're working with standard tools recognized by the MATLAB community. This can make it easier to get help from other users or from MathWorks support, as more people will be familiar with the functionality and potential issues of these functions.
Given a vector v whose order we would like to randomly permute, many would perform the permutation by explicitly querying the length/size of v, e.g.,
I=randperm(numel(v));
v=v(I);
However, one can instead do as follows, avoiding the size query.
v=v(randperm(end))
Analogous things can be done with matrices, e.g.,
A=A(randperm(end), randperm(end));
The creativity comes from the copper sulfate crystal heart made in junior high school. Copper sulfate is a triclinic crystal, and the same structure was not used here for convenience in drawing.
Part 1. Coordinate transformation
To draw a crystal heart, one must first be able to draw crystal clusters. To draw a crystal cluster, one must first be able to draw a crystal. To draw a crystal, we need this kind of structure:
We first need a point with a certain distance from the straight line and a perpendicular point of cutPnt, which is very easy to find, for example, cutPnt=[x0, y0, z0]; The direction of the central axis is V=[x1, y1, z1]; If the distance to the straight line is L, the following points clearly meet the conditions:
v2=[z1,z1,-x1-y1];
v2=v2./norm(v2).*L;
pnt=cutPnt+v2;
But finding only one point is not enough. We need to find four points, and each point is obtained by rotating the previous point around a straight line by 
degrees. Therefore, we need to obtain our point rotation transformation matrix around a straight line
degrees. Therefore, we need to obtain our point rotation transformation matrix around a straight line
quite complex,right?
rotateMat=[u^2+(v^2+w^2)*cos(theta) , u*v*(1-cos(theta))-w*sin(theta), u*w*(1-cos(theta))+v*sin(theta), (a*(v^2+w^2)-u*(b*v+c*w))*(1-cos(theta))+(b*w-c*v)*sin(theta);
u*v*(1-cos(theta))+w*sin(theta), v^2+(u^2+w^2)*cos(theta) , v*w*(1-cos(theta))-u*sin(theta), (b*(u^2+w^2)-v*(a*u+c*w))*(1-cos(theta))+(c*u-a*w)*sin(theta);
u*w*(1-cos(theta))-v*sin(theta), v*w*(1-cos(theta))+u*sin(theta), w^2+(u^2+v^2)*cos(theta) , (c*(u^2+v^2)-w*(a*u+b*v))*(1-cos(theta))+(a*v-b*u)*sin(theta);
0 , 0 , 0 , 1];
Where [u, v, w] is the directional unit vector, and [a, b, c] is the initial coordinate of the axis:
Part 2. Crystal Cluster Drawing
function crystall
hold on
for i=1:50
len=rand(1)*8+5;
tempV=rand(1,3)-0.5;
tempV(3)=abs(tempV(3));
tempV=tempV./norm(tempV).*len;
tempEpnt=tempV;
drawCrystal([0 0 0],tempEpnt,pi/6,0.8,0.1,rand(1).*0.2+0.2)
disp(i)
end
ax=gca;
ax.XLim=[-15,15];
ax.YLim=[-15,15];
ax.ZLim=[-2,15];
grid on
ax.GridLineStyle='--';
ax.LineWidth=1.2;
ax.XColor=[1,1,1].*0.4;
ax.YColor=[1,1,1].*0.4;
ax.ZColor=[1,1,1].*0.4;
ax.DataAspectRatio=[1,1,1];
ax.DataAspectRatioMode='manual';
ax.CameraPosition=[-67.6287 -204.5276 82.7879];
function drawCrystal(Spnt,Epnt,theta,cl,w,alpha)
%plot3([Spnt(1),Epnt(1)],[Spnt(2),Epnt(2)],[Spnt(3),Epnt(3)])
mainV=Epnt-Spnt;
cutPnt=cl.*(mainV)+Spnt;
cutV=[mainV(3),mainV(3),-mainV(1)-mainV(2)];
cutV=cutV./norm(cutV).*w.*norm(mainV);
cornerPnt=cutPnt+cutV;
cornerPnt=rotateAxis(Spnt,Epnt,cornerPnt,theta);
cornerPntSet(1,:)=cornerPnt';
for ii=1:3
cornerPnt=rotateAxis(Spnt,Epnt,cornerPnt,pi/2);
cornerPntSet(ii+1,:)=cornerPnt';
end
F = [1,3,4;1,4,5;1,5,6;1,6,3;...
2,3,4;2,4,5;2,5,6;2,6,3];
V = [Spnt;Epnt;cornerPntSet];
patch('Faces',F,'Vertices',V,'FaceColor',[0 71 177]./255,...
'FaceAlpha',alpha,'EdgeColor',[0 71 177]./255.*0.8,...
'EdgeAlpha',0.6,'LineWidth',0.5,'EdgeLighting',...
'gouraud','SpecularStrength',0.3)
end
function newPnt=rotateAxis(Spnt,Epnt,cornerPnt,theta)
V=Epnt-Spnt;V=V./norm(V);
u=V(1);v=V(2);w=V(3);
a=Spnt(1);b=Spnt(2);c=Spnt(3);
cornerPnt=[cornerPnt(:);1];
rotateMat=[u^2+(v^2+w^2)*cos(theta) , u*v*(1-cos(theta))-w*sin(theta), u*w*(1-cos(theta))+v*sin(theta), (a*(v^2+w^2)-u*(b*v+c*w))*(1-cos(theta))+(b*w-c*v)*sin(theta);
u*v*(1-cos(theta))+w*sin(theta), v^2+(u^2+w^2)*cos(theta) , v*w*(1-cos(theta))-u*sin(theta), (b*(u^2+w^2)-v*(a*u+c*w))*(1-cos(theta))+(c*u-a*w)*sin(theta);
u*w*(1-cos(theta))-v*sin(theta), v*w*(1-cos(theta))+u*sin(theta), w^2+(u^2+v^2)*cos(theta) , (c*(u^2+v^2)-w*(a*u+b*v))*(1-cos(theta))+(a*v-b*u)*sin(theta);
0 , 0 , 0 , 1];
newPnt=rotateMat*cornerPnt;
newPnt(4)=[];
end
end
Part 3. Drawing of Crystal Heart
function crystalHeart
clc;clear;close all
hold on
% drawCrystal([1,1,1],[3,3,3],pi/6,0.8,0.14)
sep=pi/8;
t=[0:0.2:sep,sep:0.02:pi-sep,pi-sep:0.2:pi+sep,pi+sep:0.02:2*pi-sep,2*pi-sep:0.2:2*pi];
x=16*sin(t).^3;
y=13*cos(t)-5*cos(2*t)-2*cos(3*t)-cos(4*t);
z=zeros(size(t));
plot3(x,y,z,'Color',[186,110,64]./255,'LineWidth',1)
for i=1:length(t)
for j=1:6
len=rand(1)*2.5+1.5;
tempV=rand(1,3)-0.5;
tempV=tempV./norm(tempV).*len;
tempSpnt=[x(i),y(i),z(i)];
tempEpnt=tempV+tempSpnt;
drawCrystal(tempSpnt,tempEpnt,pi/6,0.8,0.14)
disp([i,j])
end
end
ax=gca;
ax.XLim=[-22,22];
ax.YLim=[-20,20];
ax.ZLim=[-10,10];
grid on
ax.GridLineStyle='--';
ax.LineWidth=1.2;
ax.XColor=[1,1,1].*0.4;
ax.YColor=[1,1,1].*0.4;
ax.ZColor=[1,1,1].*0.4;
ax.DataAspectRatio=[1,1,1];
ax.DataAspectRatioMode='manual';
function drawCrystal(Spnt,Epnt,theta,cl,w)
%plot3([Spnt(1),Epnt(1)],[Spnt(2),Epnt(2)],[Spnt(3),Epnt(3)])
mainV=Epnt-Spnt;
cutPnt=cl.*(mainV)+Spnt;
cutV=[mainV(3),mainV(3),-mainV(1)-mainV(2)];
cutV=cutV./norm(cutV).*w.*norm(mainV);
cornerPnt=cutPnt+cutV;
cornerPnt=rotateAxis(Spnt,Epnt,cornerPnt,theta);
cornerPntSet(1,:)=cornerPnt';
for ii=1:3
cornerPnt=rotateAxis(Spnt,Epnt,cornerPnt,pi/2);
cornerPntSet(ii+1,:)=cornerPnt';
end
F = [1,3,4;1,4,5;1,5,6;1,6,3;...
2,3,4;2,4,5;2,5,6;2,6,3];
V = [Spnt;Epnt;cornerPntSet];
patch('Faces',F,'Vertices',V,'FaceColor',[0 71 177]./255,...
'FaceAlpha',0.2,'EdgeColor',[0 71 177]./255.*0.9,...
'EdgeAlpha',0.25,'LineWidth',0.01,'EdgeLighting',...
'gouraud','SpecularStrength',0.3)
end
function newPnt=rotateAxis(Spnt,Epnt,cornerPnt,theta)
V=Epnt-Spnt;V=V./norm(V);
u=V(1);v=V(2);w=V(3);
a=Spnt(1);b=Spnt(2);c=Spnt(3);
cornerPnt=[cornerPnt(:);1];
rotateMat=[u^2+(v^2+w^2)*cos(theta) , u*v*(1-cos(theta))-w*sin(theta), u*w*(1-cos(theta))+v*sin(theta), (a*(v^2+w^2)-u*(b*v+c*w))*(1-cos(theta))+(b*w-c*v)*sin(theta);
u*v*(1-cos(theta))+w*sin(theta), v^2+(u^2+w^2)*cos(theta) , v*w*(1-cos(theta))-u*sin(theta), (b*(u^2+w^2)-v*(a*u+c*w))*(1-cos(theta))+(c*u-a*w)*sin(theta);
u*w*(1-cos(theta))-v*sin(theta), v*w*(1-cos(theta))+u*sin(theta), w^2+(u^2+v^2)*cos(theta) , (c*(u^2+v^2)-w*(a*u+b*v))*(1-cos(theta))+(a*v-b*u)*sin(theta);
0 , 0 , 0 , 1];
newPnt=rotateMat*cornerPnt;
newPnt(4)=[];
end
end
また、以下のリストから Web サイトを選択することもできます。
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