Hi,

I have a 1D heat diffusion code which I was using on a timescale of 10s of years and I am now trying to use the same code to work on a scale of millions of years. Obviously if I keep my timestep the same this will take ages to calculate but if I increase my timestep I encounter numerical stability issues.

My questions are:

How should I approach this problem?
What affects the maximum stable timestep? And how do I calculate this?

Many thanks,

Alex

close all
clear all

dx = 4; % discretization step in m
dt = 0.0000001; % timestep in Myrs
h=1000; % height of box in m
nx=h/dx+1;
model_lenth=1; %length of model in Myrs
nt=ceil(model_lenth/dt)+1; % number of tsteps to reach end of model
kappa = 1e-6; % thermal diffusivity
x=0:dx:0+h; % finite difference mesh
T=38+0.05.*x; % initial T=Tm everywhere …
time=zeros(1,nt);
t=0;
Tnew = zeros(1,nx);

%Lower sill
sill_1_thickness=18;
Sill_1_top_position=590;
Sill_1_top=ceil(Sill_1_top_position/dx);
Sill_1_bottom=ceil((Sill_1_top_position+sill_1_thickness)/dx);

%Upper sill
sill_2_thickness=10;
Sill_2_top_position=260;
Sill_2_top=ceil(Sill_2_top_position/dx);
Sill_2_bottom=ceil((Sill_2_top_position+sill_2_thickness)/dx);

%Temperature of dolerite intrusions
Tm=1300;

T(Sill_1_top:Sill_1_bottom)=Tm; %Apply temperature to intrusion 1

% unit conversion to SI:
secinmyr=24*3600*365*1000000; % dt in sec
dt=dt*secinmyr;

%Plot initial conditions
figure(1), clf
f1 = figure(1); %Make full screen
set(f1,’Units’, ‘Normalized’, ‘OuterPosition’, [0 0 1 1]);
plot (T,x,’LineWidth’,2)
xlabel(‘T [^oC]’)
ylabel(‘x[m]’)
axis([0 1310 0 1000])
title(‘ Initial Conditions’)
set(gca,’YDir’,’reverse’);

%Main calculation
for it=1:nt

%Apply temperature to upper intrusion
if it==10;
T(Sill_2_top:Sill_2_bottom)=Tm;
end

for i = 2:nx-1
Tnew(i) = T(i) + kappa*dt*(T(i+1) – 2*T(i) + T(i-1))/dx/dx;
end

Tnew(1) = T(1);
Tnew(nx) = T(nx);

time(it) = t;

T = Tnew; %Set old Temp to = new temp for next loop
tmyears=(t/secinmyr);

%Plot a figure which updates in the loop of temperature against depth
figure(2), clf
plot (T,x,’LineWidth’,2)
xlabel(‘T [^oC]’)
ylabel(‘x[m]’)
title([‘ Temperature against Depth after ‘,num2str(tmyears),’ Myrs’])
axis([0 1300 0 1000])
set(gca,’YDir’,’reverse’);%Reverse y axis

%Make full screen
f2 = figure(2);
set(f2,’Units’, ‘Normalized’, ‘OuterPosition’, [0 0 1 1]);

drawnow

t=t+dt;
endHi,

I have a 1D heat diffusion code which I was using on a timescale of 10s of years and I am now trying to use the same code to work on a scale of millions of years. Obviously if I keep my timestep the same this will take ages to calculate but if I increase my timestep I encounter numerical stability issues.

My questions are:

How should I approach this problem?
What affects the maximum stable timestep? And how do I calculate this?

Many thanks,

Alex

close all
clear all

dx = 4; % discretization step in m
dt = 0.0000001; % timestep in Myrs
h=1000; % height of box in m
nx=h/dx+1;
model_lenth=1; %length of model in Myrs
nt=ceil(model_lenth/dt)+1; % number of tsteps to reach end of model
kappa = 1e-6; % thermal diffusivity
x=0:dx:0+h; % finite difference mesh
T=38+0.05.*x; % initial T=Tm everywhere …
time=zeros(1,nt);
t=0;
Tnew = zeros(1,nx);

%Lower sill
sill_1_thickness=18;
Sill_1_top_position=590;
Sill_1_top=ceil(Sill_1_top_position/dx);
Sill_1_bottom=ceil((Sill_1_top_position+sill_1_thickness)/dx);

%Upper sill
sill_2_thickness=10;
Sill_2_top_position=260;
Sill_2_top=ceil(Sill_2_top_position/dx);
Sill_2_bottom=ceil((Sill_2_top_position+sill_2_thickness)/dx);

%Temperature of dolerite intrusions
Tm=1300;

T(Sill_1_top:Sill_1_bottom)=Tm; %Apply temperature to intrusion 1

% unit conversion to SI:
secinmyr=24*3600*365*1000000; % dt in sec
dt=dt*secinmyr;

%Plot initial conditions
figure(1), clf
f1 = figure(1); %Make full screen
set(f1,’Units’, ‘Normalized’, ‘OuterPosition’, [0 0 1 1]);
plot (T,x,’LineWidth’,2)
xlabel(‘T [^oC]’)
ylabel(‘x[m]’)
axis([0 1310 0 1000])
title(‘ Initial Conditions’)
set(gca,’YDir’,’reverse’);

%Main calculation
for it=1:nt

%Apply temperature to upper intrusion
if it==10;
T(Sill_2_top:Sill_2_bottom)=Tm;
end

for i = 2:nx-1
Tnew(i) = T(i) + kappa*dt*(T(i+1) – 2*T(i) + T(i-1))/dx/dx;
end

Tnew(1) = T(1);
Tnew(nx) = T(nx);

time(it) = t;

T = Tnew; %Set old Temp to = new temp for next loop
tmyears=(t/secinmyr);

%Plot a figure which updates in the loop of temperature against depth
figure(2), clf
plot (T,x,’LineWidth’,2)
xlabel(‘T [^oC]’)
ylabel(‘x[m]’)
title([‘ Temperature against Depth after ‘,num2str(tmyears),’ Myrs’])
axis([0 1300 0 1000])
set(gca,’YDir’,’reverse’);%Reverse y axis

%Make full screen
f2 = figure(2);
set(f2,’Units’, ‘Normalized’, ‘OuterPosition’, [0 0 1 1]);

drawnow

t=t+dt;
end Hi,

I have a 1D heat diffusion code which I was using on a timescale of 10s of years and I am now trying to use the same code to work on a scale of millions of years. Obviously if I keep my timestep the same this will take ages to calculate but if I increase my timestep I encounter numerical stability issues.

My questions are:

How should I approach this problem?
What affects the maximum stable timestep? And how do I calculate this?

Many thanks,

Alex

close all
clear all

dx = 4; % discretization step in m
dt = 0.0000001; % timestep in Myrs
h=1000; % height of box in m
nx=h/dx+1;
model_lenth=1; %length of model in Myrs
nt=ceil(model_lenth/dt)+1; % number of tsteps to reach end of model
kappa = 1e-6; % thermal diffusivity
x=0:dx:0+h; % finite difference mesh
T=38+0.05.*x; % initial T=Tm everywhere …
time=zeros(1,nt);
t=0;
Tnew = zeros(1,nx);

%Lower sill
sill_1_thickness=18;
Sill_1_top_position=590;
Sill_1_top=ceil(Sill_1_top_position/dx);
Sill_1_bottom=ceil((Sill_1_top_position+sill_1_thickness)/dx);

%Upper sill
sill_2_thickness=10;
Sill_2_top_position=260;
Sill_2_top=ceil(Sill_2_top_position/dx);
Sill_2_bottom=ceil((Sill_2_top_position+sill_2_thickness)/dx);

%Temperature of dolerite intrusions
Tm=1300;

T(Sill_1_top:Sill_1_bottom)=Tm; %Apply temperature to intrusion 1

% unit conversion to SI:
secinmyr=24*3600*365*1000000; % dt in sec
dt=dt*secinmyr;

%Plot initial conditions
figure(1), clf
f1 = figure(1); %Make full screen
set(f1,’Units’, ‘Normalized’, ‘OuterPosition’, [0 0 1 1]);
plot (T,x,’LineWidth’,2)
xlabel(‘T [^oC]’)
ylabel(‘x[m]’)
axis([0 1310 0 1000])
title(‘ Initial Conditions’)
set(gca,’YDir’,’reverse’);

%Main calculation
for it=1:nt

%Apply temperature to upper intrusion
if it==10;
T(Sill_2_top:Sill_2_bottom)=Tm;
end

for i = 2:nx-1
Tnew(i) = T(i) + kappa*dt*(T(i+1) – 2*T(i) + T(i-1))/dx/dx;
end

Tnew(1) = T(1);
Tnew(nx) = T(nx);

time(it) = t;

T = Tnew; %Set old Temp to = new temp for next loop
tmyears=(t/secinmyr);

%Plot a figure which updates in the loop of temperature against depth
figure(2), clf
plot (T,x,’LineWidth’,2)
xlabel(‘T [^oC]’)
ylabel(‘x[m]’)
title([‘ Temperature against Depth after ‘,num2str(tmyears),’ Myrs’])
axis([0 1300 0 1000])
set(gca,’YDir’,’reverse’);%Reverse y axis

%Make full screen
f2 = figure(2);
set(f2,’Units’, ‘Normalized’, ‘OuterPosition’, [0 0 1 1]);

drawnow

t=t+dt;
end time step; heat diffusion MATLAB Answers — New Questions

​

Hi, I want to do Friedman test on a test data where data from an individual has been collected for 18 days. I have 4 groups with 5, 7, 8, and 6 individuals in each group. If I am understanding correctly, then the data should be in the following format:
[D11 D11
D21 D21
D31 D31
D12 D12];
So basically , each column would have 18*n group individuals and each column would represent different group. I want to make sure I am understanding the reqs correctly. Also, how would I represent the reps in this case? The code gives me bad rep number.
Thank you.Hi, I want to do Friedman test on a test data where data from an individual has been collected for 18 days. I have 4 groups with 5, 7, 8, and 6 individuals in each group. If I am understanding correctly, then the data should be in the following format:
[D11 D11
D21 D21
D31 D31
D12 D12];
So basically , each column would have 18*n group individuals and each column would represent different group. I want to make sure I am understanding the reqs correctly. Also, how would I represent the reps in this case? The code gives me bad rep number.
Thank you. Hi, I want to do Friedman test on a test data where data from an individual has been collected for 18 days. I have 4 groups with 5, 7, 8, and 6 individuals in each group. If I am understanding correctly, then the data should be in the following format:
[D11 D11
D21 D21
D31 D31
D12 D12];
So basically , each column would have 18*n group individuals and each column would represent different group. I want to make sure I am understanding the reqs correctly. Also, how would I represent the reps in this case? The code gives me bad rep number.
Thank you. friedman test, non-parametric, statistics MATLAB Answers — New Questions

​

Hello
i have two column one is number of hiarachy the second is the name of that hiarachy like this :

can you help make a code that will generate what is manually wrote in the third column , note that the file is large and contain long WBS path like this (2.2.3.5.4.5.7)
RegardsHello
i have two column one is number of hiarachy the second is the name of that hiarachy like this :

can you help make a code that will generate what is manually wrote in the third column , note that the file is large and contain long WBS path like this (2.2.3.5.4.5.7)
Regards Hello
i have two column one is number of hiarachy the second is the name of that hiarachy like this :

can you help make a code that will generate what is manually wrote in the third column , note that the file is large and contain long WBS path like this (2.2.3.5.4.5.7)
Regards data, excel, matlab, matrix, importing excel data, code MATLAB Answers — New Questions

​

For a while loop how I can calculate the values at each hour before do a break? As we know while loop only performs the condition statment and breaks the loop, then gives you the value at that specific hour only. I provide te example below :
for i=1:Time
while water_tank_soc(i-1) < Water_tank_capacity_max
m_net_o(i) =………
m_net_i(i) = ……
% Update SOC
water_tank_soc(i) = water_tank_soc(i-1) + m_net_i(i) – m_net_o(i);
if water_tank_soc(i) >= Water_tank_capacity_max
break
end
% Update for next time step
water_tank_soc(i-1) = water_tank_soc(i); % Update previous state for next iteration
end
endFor a while loop how I can calculate the values at each hour before do a break? As we know while loop only performs the condition statment and breaks the loop, then gives you the value at that specific hour only. I provide te example below :
for i=1:Time
while water_tank_soc(i-1) < Water_tank_capacity_max
m_net_o(i) =………
m_net_i(i) = ……
% Update SOC
water_tank_soc(i) = water_tank_soc(i-1) + m_net_i(i) – m_net_o(i);
if water_tank_soc(i) >= Water_tank_capacity_max
break
end
% Update for next time step
water_tank_soc(i-1) = water_tank_soc(i); % Update previous state for next iteration
end
end For a while loop how I can calculate the values at each hour before do a break? As we know while loop only performs the condition statment and breaks the loop, then gives you the value at that specific hour only. I provide te example below :
for i=1:Time
while water_tank_soc(i-1) < Water_tank_capacity_max
m_net_o(i) =………
m_net_i(i) = ……
% Update SOC
water_tank_soc(i) = water_tank_soc(i-1) + m_net_i(i) – m_net_o(i);
if water_tank_soc(i) >= Water_tank_capacity_max
break
end
% Update for next time step
water_tank_soc(i-1) = water_tank_soc(i); % Update previous state for next iteration
end
end save a values, while loop MATLAB Answers — New Questions

​

Hi everyone,
I’m having some trouble understanding Gaussian Process Regression (GPR) options in the Regression Learner App. There are three main choices for GPR models:
Predefined Kernel: I can directly choose a kernel (Rational Quadratic, Squared Exponential, Matern 5/2, or Exponential) if I know which one suits my data best.
All GPR Models (non-optimizable): If I’m unsure which kernel to use, I can select this option to try all non-optimizable GPR models.
Optimizable GPR: This option allows the hyperparameters to be optimized and has even more Kernels available.
Regardless of whether I choose the optimizable or non-optimizable version, each kernel has hyperparameters that can be tuned.
Here are my questions:
Automatic Hyperparameter Selection: For the standard model without optimization, the kernel parameters (hyperparameters) are automatically selected and are an initial best guess rather than the optimal ones to my understanding. If so, how are they estimated/selected, and why aren’t they optimal?
Optimization Process: When the optimization option is selected, the hyperparameters are found by maximizing the Log Marginal Likelihood function. So, they are not guessed, but these are the hyperparameters that best describe the data. So finding the best value is basically the optimization process?
Choosing Non-Optimal Hyperparameters: Why would someone use the version without the best hyperparameters? One reason I’ve encountered is that optimization can take too long with the same data. Are there other reasons to choose non-optimizable models?
Best Practices: Are there standard practices or guidelines for when to use each version of the GPR models? Or one usually starts with non-optimizable version and then go to the optimizable one if the results are not sufficient enough?
Output Differences: Both versions output forecasted data and the covariance matrix. Does the optimized version provide any additional information or benefits?
Plotting Covariance: How can I plot the covariance that quantifies the uncertainty of my predictions? This is one of the main advantages of using GPR, and I want to visualize it. I have found some sample code here https://ch.mathworks.com/help/stats/gaussian-process-regression-models.html But I am a bit confused, because my input is a table actually, so I do not have only one predictor but a table. So far I have always used “predictedData = trainedModel.predictFcn(Table); but this gives me only the forecasted data without the 95% prediction intervals
I’m generally confused about the differences between both versions and the results they produce. Any insights or explanations would be greatly appreciated.
Thanks in advance!
Best regards,
GeorgiHi everyone,
I’m having some trouble understanding Gaussian Process Regression (GPR) options in the Regression Learner App. There are three main choices for GPR models:
Predefined Kernel: I can directly choose a kernel (Rational Quadratic, Squared Exponential, Matern 5/2, or Exponential) if I know which one suits my data best.
All GPR Models (non-optimizable): If I’m unsure which kernel to use, I can select this option to try all non-optimizable GPR models.
Optimizable GPR: This option allows the hyperparameters to be optimized and has even more Kernels available.
Regardless of whether I choose the optimizable or non-optimizable version, each kernel has hyperparameters that can be tuned.
Here are my questions:
Automatic Hyperparameter Selection: For the standard model without optimization, the kernel parameters (hyperparameters) are automatically selected and are an initial best guess rather than the optimal ones to my understanding. If so, how are they estimated/selected, and why aren’t they optimal?
Optimization Process: When the optimization option is selected, the hyperparameters are found by maximizing the Log Marginal Likelihood function. So, they are not guessed, but these are the hyperparameters that best describe the data. So finding the best value is basically the optimization process?
Choosing Non-Optimal Hyperparameters: Why would someone use the version without the best hyperparameters? One reason I’ve encountered is that optimization can take too long with the same data. Are there other reasons to choose non-optimizable models?
Best Practices: Are there standard practices or guidelines for when to use each version of the GPR models? Or one usually starts with non-optimizable version and then go to the optimizable one if the results are not sufficient enough?
Output Differences: Both versions output forecasted data and the covariance matrix. Does the optimized version provide any additional information or benefits?
Plotting Covariance: How can I plot the covariance that quantifies the uncertainty of my predictions? This is one of the main advantages of using GPR, and I want to visualize it. I have found some sample code here https://ch.mathworks.com/help/stats/gaussian-process-regression-models.html But I am a bit confused, because my input is a table actually, so I do not have only one predictor but a table. So far I have always used “predictedData = trainedModel.predictFcn(Table); but this gives me only the forecasted data without the 95% prediction intervals
I’m generally confused about the differences between both versions and the results they produce. Any insights or explanations would be greatly appreciated.
Thanks in advance!
Best regards,
Georgi Hi everyone,
I’m having some trouble understanding Gaussian Process Regression (GPR) options in the Regression Learner App. There are three main choices for GPR models:
Predefined Kernel: I can directly choose a kernel (Rational Quadratic, Squared Exponential, Matern 5/2, or Exponential) if I know which one suits my data best.
All GPR Models (non-optimizable): If I’m unsure which kernel to use, I can select this option to try all non-optimizable GPR models.
Optimizable GPR: This option allows the hyperparameters to be optimized and has even more Kernels available.
Regardless of whether I choose the optimizable or non-optimizable version, each kernel has hyperparameters that can be tuned.
Here are my questions:
Automatic Hyperparameter Selection: For the standard model without optimization, the kernel parameters (hyperparameters) are automatically selected and are an initial best guess rather than the optimal ones to my understanding. If so, how are they estimated/selected, and why aren’t they optimal?
Optimization Process: When the optimization option is selected, the hyperparameters are found by maximizing the Log Marginal Likelihood function. So, they are not guessed, but these are the hyperparameters that best describe the data. So finding the best value is basically the optimization process?
Choosing Non-Optimal Hyperparameters: Why would someone use the version without the best hyperparameters? One reason I’ve encountered is that optimization can take too long with the same data. Are there other reasons to choose non-optimizable models?
Best Practices: Are there standard practices or guidelines for when to use each version of the GPR models? Or one usually starts with non-optimizable version and then go to the optimizable one if the results are not sufficient enough?
Output Differences: Both versions output forecasted data and the covariance matrix. Does the optimized version provide any additional information or benefits?
Plotting Covariance: How can I plot the covariance that quantifies the uncertainty of my predictions? This is one of the main advantages of using GPR, and I want to visualize it. I have found some sample code here https://ch.mathworks.com/help/stats/gaussian-process-regression-models.html But I am a bit confused, because my input is a table actually, so I do not have only one predictor but a table. So far I have always used “predictedData = trainedModel.predictFcn(Table); but this gives me only the forecasted data without the 95% prediction intervals
I’m generally confused about the differences between both versions and the results they produce. Any insights or explanations would be greatly appreciated.
Thanks in advance!
Best regards,
Georgi regression, optimization MATLAB Answers — New Questions

​

Hello,
I am trying to define some predefined figure styles by changing groot properties (I was inspired by the Default Property Values documentation). I am using the following syntax:
set(groot,"DefaultObjectTypePropertyName")
This works well for many cases, but I am unable to set correctly the Default Axes Title Font Size when I use LaTeX fonts.
I am using the "mwa_cmr10" font found in <MATLAB root folder>R2023asysfontsttf.
I actually can set the font size to the value I want, but after I divide it by 10.
Here is a MWE.
%% Dummy data
x1 = linspace(0,1,100);
y1 = sin(2*pi*x1);
x2 = linspace(0,1,100);
y2 = sin(4*pi*x2);

set( groot, "DefaultAxesTitleFontSize", 16/10 ); % Dividing the font size I want (16) by 10
set( groot, "DefaultTextInterpreter", "latex" );
set( groot, "DefaultAxesFontName", "mwa_cmr10" );

figure;
hold all
plot(x1, y1);
plot(x2, y2);
xlabel("X")
ylabel("Y")
title("Title")
legend;
gca().Title.FontSize % The font size is 16, but should be 1.6 as it is 16/10
I do not know if this is a bug or not, maybe it has something to do with the 10 in the "mwa_cmr10" font?
I wanted to ask this here first before signaling it as a bug. I think some more obscure graphics properties can be tricky to handle.
Thank you for your attention.
Regards,
Pedro
Edit: stylingHello,
I am trying to define some predefined figure styles by changing groot properties (I was inspired by the Default Property Values documentation). I am using the following syntax:
set(groot,"DefaultObjectTypePropertyName")
This works well for many cases, but I am unable to set correctly the Default Axes Title Font Size when I use LaTeX fonts.
I am using the "mwa_cmr10" font found in <MATLAB root folder>R2023asysfontsttf.
I actually can set the font size to the value I want, but after I divide it by 10.
Here is a MWE.
%% Dummy data
x1 = linspace(0,1,100);
y1 = sin(2*pi*x1);
x2 = linspace(0,1,100);
y2 = sin(4*pi*x2);

set( groot, "DefaultAxesTitleFontSize", 16/10 ); % Dividing the font size I want (16) by 10
set( groot, "DefaultTextInterpreter", "latex" );
set( groot, "DefaultAxesFontName", "mwa_cmr10" );

figure;
hold all
plot(x1, y1);
plot(x2, y2);
xlabel("X")
ylabel("Y")
title("Title")
legend;
gca().Title.FontSize % The font size is 16, but should be 1.6 as it is 16/10
I do not know if this is a bug or not, maybe it has something to do with the 10 in the "mwa_cmr10" font?
I wanted to ask this here first before signaling it as a bug. I think some more obscure graphics properties can be tricky to handle.
Thank you for your attention.
Regards,
Pedro
Edit: styling Hello,
I am trying to define some predefined figure styles by changing groot properties (I was inspired by the Default Property Values documentation). I am using the following syntax:
set(groot,"DefaultObjectTypePropertyName")
This works well for many cases, but I am unable to set correctly the Default Axes Title Font Size when I use LaTeX fonts.
I am using the "mwa_cmr10" font found in <MATLAB root folder>R2023asysfontsttf.
I actually can set the font size to the value I want, but after I divide it by 10.
Here is a MWE.
%% Dummy data
x1 = linspace(0,1,100);
y1 = sin(2*pi*x1);
x2 = linspace(0,1,100);
y2 = sin(4*pi*x2);

set( groot, "DefaultAxesTitleFontSize", 16/10 ); % Dividing the font size I want (16) by 10
set( groot, "DefaultTextInterpreter", "latex" );
set( groot, "DefaultAxesFontName", "mwa_cmr10" );

figure;
hold all
plot(x1, y1);
plot(x2, y2);
xlabel("X")
ylabel("Y")
title("Title")
legend;
gca().Title.FontSize % The font size is 16, but should be 1.6 as it is 16/10
I do not know if this is a bug or not, maybe it has something to do with the 10 in the "mwa_cmr10" font?
I wanted to ask this here first before signaling it as a bug. I think some more obscure graphics properties can be tricky to handle.
Thank you for your attention.
Regards,
Pedro
Edit: styling graphics, latex MATLAB Answers — New Questions

​

A=[2 3 4 5];
How do I choose any one variable from the vector A randomly each time.A=[2 3 4 5];
How do I choose any one variable from the vector A randomly each time. A=[2 3 4 5];
How do I choose any one variable from the vector A randomly each time. rand, randi MATLAB Answers — New Questions

​

As per title, i’m unable to recreate the following example (k-means clustering – MATLAB kmeans – MathWorks Italia), the k means clustering of Fisher’s iris dataset.

This is the code in question
load fisheriris
X = meas(:,3:4);
figure;
plot(X(:,1),X(:,2),’k*’,’MarkerSize’,5);
title ‘Fisher”s Iris Data’;
xlabel ‘Petal Lengths (cm)’;
ylabel ‘Petal Widths (cm)’
rng(1); % For reproducibility
[idx,C] = kmeans(X,3);
x1 = min(X(:,1)):0.01:max(X(:,1));
x2 = min(X(:,2)):0.01:max(X(:,2));
[x1G,x2G] = meshgrid(x1,x2);
XGrid = [x1G(:),x2G(:)]; % Defines a fine grid on the plot
idx2Region = kmeans(XGrid,3,’MaxIter’,1,’Start’,C);
figure;
gscatter(XGrid(:,1),XGrid(:,2),idx2Region,…
[0,0.75,0.75;0.75,0,0.75;0.75,0.75,0],’..’);
hold on;
plot(X(:,1),X(:,2),’k*’,’MarkerSize’,5);
title ‘Fisher”s Iris Data’;
xlabel ‘Petal Lengths (cm)’;
ylabel ‘Petal Widths (cm)’;
legend(‘Region 1′,’Region 2′,’Region 3′,’Data’,’Location’,’SouthEast’);
hold off;
An error occurs at line 16 involving the kmeans function
Error in Kmeans_dimostrativo (line 16)
[idx,C] = kmeans(X,3);
So i started digging in the kmeans function
function [label, mu, energy] = kmeans(X, m)
% Perform kmeans clustering.
% Input:
% X: d x n data matrix
% m: initialization parameter
% Output:
% label: 1 x n sample labels
% mu: d x k center of clusters
% energy: optimization target value
% Written by Mo Chen (sth4nth@gmail.com).
label = init(X, m);
n = numel(label);
idx = 1:n;
last = zeros(1,n);
while any(label ~= last)
[~,~,last(:)] = unique(label); % remove empty clusters
mu = X*normalize(sparse(idx,last,1),1); % compute cluster centers
[val,label] = min(dot(mu,mu,1)’/2-mu’*X,[],1); % assign sample labels
end
energy = dot(X(:),X(:),1)+2*sum(val);
function label = init(X, m)
[d,n] = size(X);
if numel(m) == 1 % random initialization
mu = X(:,randperm(n,m));
[~,label] = min(dot(mu,mu,1)’/2-mu’*X,[],1);
elseif all(size(m) == [1,n]) % init with labels
label = m;
elseif size(m,1) == d % init with seeds (centers)
[~,label] = min(dot(m,m,1)’/2-m’*X,[],1);
end
The following error messages are the ones relative to the kmeans function
Error using randperm
K must be less than or equal to N.

Error in kmeans>init (line 25)
mu = X(:,randperm(n,m));

Error in kmeans (line 11)
label = init(X, m);
I honestly don’t know the reason for the errors, i took the script directly from the website.As per title, i’m unable to recreate the following example (k-means clustering – MATLAB kmeans – MathWorks Italia), the k means clustering of Fisher’s iris dataset.

This is the code in question
load fisheriris
X = meas(:,3:4);
figure;
plot(X(:,1),X(:,2),’k*’,’MarkerSize’,5);
title ‘Fisher”s Iris Data’;
xlabel ‘Petal Lengths (cm)’;
ylabel ‘Petal Widths (cm)’
rng(1); % For reproducibility
[idx,C] = kmeans(X,3);
x1 = min(X(:,1)):0.01:max(X(:,1));
x2 = min(X(:,2)):0.01:max(X(:,2));
[x1G,x2G] = meshgrid(x1,x2);
XGrid = [x1G(:),x2G(:)]; % Defines a fine grid on the plot
idx2Region = kmeans(XGrid,3,’MaxIter’,1,’Start’,C);
figure;
gscatter(XGrid(:,1),XGrid(:,2),idx2Region,…
[0,0.75,0.75;0.75,0,0.75;0.75,0.75,0],’..’);
hold on;
plot(X(:,1),X(:,2),’k*’,’MarkerSize’,5);
title ‘Fisher”s Iris Data’;
xlabel ‘Petal Lengths (cm)’;
ylabel ‘Petal Widths (cm)’;
legend(‘Region 1′,’Region 2′,’Region 3′,’Data’,’Location’,’SouthEast’);
hold off;
An error occurs at line 16 involving the kmeans function
Error in Kmeans_dimostrativo (line 16)
[idx,C] = kmeans(X,3);
So i started digging in the kmeans function
function [label, mu, energy] = kmeans(X, m)
% Perform kmeans clustering.
% Input:
% X: d x n data matrix
% m: initialization parameter
% Output:
% label: 1 x n sample labels
% mu: d x k center of clusters
% energy: optimization target value
% Written by Mo Chen (sth4nth@gmail.com).
label = init(X, m);
n = numel(label);
idx = 1:n;
last = zeros(1,n);
while any(label ~= last)
[~,~,last(:)] = unique(label); % remove empty clusters
mu = X*normalize(sparse(idx,last,1),1); % compute cluster centers
[val,label] = min(dot(mu,mu,1)’/2-mu’*X,[],1); % assign sample labels
end
energy = dot(X(:),X(:),1)+2*sum(val);
function label = init(X, m)
[d,n] = size(X);
if numel(m) == 1 % random initialization
mu = X(:,randperm(n,m));
[~,label] = min(dot(mu,mu,1)’/2-mu’*X,[],1);
elseif all(size(m) == [1,n]) % init with labels
label = m;
elseif size(m,1) == d % init with seeds (centers)
[~,label] = min(dot(m,m,1)’/2-m’*X,[],1);
end
The following error messages are the ones relative to the kmeans function
Error using randperm
K must be less than or equal to N.

Error in kmeans>init (line 25)
mu = X(:,randperm(n,m));

Error in kmeans (line 11)
label = init(X, m);
I honestly don’t know the reason for the errors, i took the script directly from the website. As per title, i’m unable to recreate the following example (k-means clustering – MATLAB kmeans – MathWorks Italia), the k means clustering of Fisher’s iris dataset.

This is the code in question
load fisheriris
X = meas(:,3:4);
figure;
plot(X(:,1),X(:,2),’k*’,’MarkerSize’,5);
title ‘Fisher”s Iris Data’;
xlabel ‘Petal Lengths (cm)’;
ylabel ‘Petal Widths (cm)’
rng(1); % For reproducibility
[idx,C] = kmeans(X,3);
x1 = min(X(:,1)):0.01:max(X(:,1));
x2 = min(X(:,2)):0.01:max(X(:,2));
[x1G,x2G] = meshgrid(x1,x2);
XGrid = [x1G(:),x2G(:)]; % Defines a fine grid on the plot
idx2Region = kmeans(XGrid,3,’MaxIter’,1,’Start’,C);
figure;
gscatter(XGrid(:,1),XGrid(:,2),idx2Region,…
[0,0.75,0.75;0.75,0,0.75;0.75,0.75,0],’..’);
hold on;
plot(X(:,1),X(:,2),’k*’,’MarkerSize’,5);
title ‘Fisher”s Iris Data’;
xlabel ‘Petal Lengths (cm)’;
ylabel ‘Petal Widths (cm)’;
legend(‘Region 1′,’Region 2′,’Region 3′,’Data’,’Location’,’SouthEast’);
hold off;
An error occurs at line 16 involving the kmeans function
Error in Kmeans_dimostrativo (line 16)
[idx,C] = kmeans(X,3);
So i started digging in the kmeans function
function [label, mu, energy] = kmeans(X, m)
% Perform kmeans clustering.
% Input:
% X: d x n data matrix
% m: initialization parameter
% Output:
% label: 1 x n sample labels
% mu: d x k center of clusters
% energy: optimization target value
% Written by Mo Chen (sth4nth@gmail.com).
label = init(X, m);
n = numel(label);
idx = 1:n;
last = zeros(1,n);
while any(label ~= last)
[~,~,last(:)] = unique(label); % remove empty clusters
mu = X*normalize(sparse(idx,last,1),1); % compute cluster centers
[val,label] = min(dot(mu,mu,1)’/2-mu’*X,[],1); % assign sample labels
end
energy = dot(X(:),X(:),1)+2*sum(val);
function label = init(X, m)
[d,n] = size(X);
if numel(m) == 1 % random initialization
mu = X(:,randperm(n,m));
[~,label] = min(dot(mu,mu,1)’/2-mu’*X,[],1);
elseif all(size(m) == [1,n]) % init with labels
label = m;
elseif size(m,1) == d % init with seeds (centers)
[~,label] = min(dot(m,m,1)’/2-m’*X,[],1);
end
The following error messages are the ones relative to the kmeans function
Error using randperm
K must be less than or equal to N.

Error in kmeans>init (line 25)
mu = X(:,randperm(n,m));

Error in kmeans (line 11)
label = init(X, m);
I honestly don’t know the reason for the errors, i took the script directly from the website. k-means clustering, statistics MATLAB Answers — New Questions

​

Dear all
I have created this function to do implement a fitting to some data:
function [fitresult,gof]=Fit(s,mag,in_mag,in_cond)
if round(in_mag(1))==1 && round(in_mag(end))==-1
fitting_function=’cos(2*atan(exp((x-x0)/Delta)))’;
elseif round(in_mag(1))==-1 && round(in_mag(end))==1
fitting_function=’cos(2*atan(exp(-(x-x0)/Delta)))’;
end
ft=fittype(fitting_function,’independent’,’x’,’dependent’,’y’);
opts=fitoptions(‘Method’,’NonlinearLeastSquares’);
opts.Display=’Off’;
opts.StartPoint=in_cond;
[fitresult,gof]=fit(space,mag,ft,opts);
When trying to fit some data:
initial_conditions=[1 1];
[fitresult,gof]=Fit(some_column_vector,other_column_vector,another_column_vector,initial_conditions’);
where "some_column_vector", "other_column_vector", and "another_column_vector" are 1000×1 arrays.
I received the following error:
Error using fit>iFit
Too many start points. You need 2 start points for this model.
Any ideas on what could be happening here?Dear all
I have created this function to do implement a fitting to some data:
function [fitresult,gof]=Fit(s,mag,in_mag,in_cond)
if round(in_mag(1))==1 && round(in_mag(end))==-1
fitting_function=’cos(2*atan(exp((x-x0)/Delta)))’;
elseif round(in_mag(1))==-1 && round(in_mag(end))==1
fitting_function=’cos(2*atan(exp(-(x-x0)/Delta)))’;
end
ft=fittype(fitting_function,’independent’,’x’,’dependent’,’y’);
opts=fitoptions(‘Method’,’NonlinearLeastSquares’);
opts.Display=’Off’;
opts.StartPoint=in_cond;
[fitresult,gof]=fit(space,mag,ft,opts);
When trying to fit some data:
initial_conditions=[1 1];
[fitresult,gof]=Fit(some_column_vector,other_column_vector,another_column_vector,initial_conditions’);
where "some_column_vector", "other_column_vector", and "another_column_vector" are 1000×1 arrays.
I received the following error:
Error using fit>iFit
Too many start points. You need 2 start points for this model.
Any ideas on what could be happening here? Dear all
I have created this function to do implement a fitting to some data:
function [fitresult,gof]=Fit(s,mag,in_mag,in_cond)
if round(in_mag(1))==1 && round(in_mag(end))==-1
fitting_function=’cos(2*atan(exp((x-x0)/Delta)))’;
elseif round(in_mag(1))==-1 && round(in_mag(end))==1
fitting_function=’cos(2*atan(exp(-(x-x0)/Delta)))’;
end
ft=fittype(fitting_function,’independent’,’x’,’dependent’,’y’);
opts=fitoptions(‘Method’,’NonlinearLeastSquares’);
opts.Display=’Off’;
opts.StartPoint=in_cond;
[fitresult,gof]=fit(space,mag,ft,opts);
When trying to fit some data:
initial_conditions=[1 1];
[fitresult,gof]=Fit(some_column_vector,other_column_vector,another_column_vector,initial_conditions’);
where "some_column_vector", "other_column_vector", and "another_column_vector" are 1000×1 arrays.
I received the following error:
Error using fit>iFit
Too many start points. You need 2 start points for this model.
Any ideas on what could be happening here? fitting MATLAB Answers — New Questions

​

I have multiple if conditions that is being checked to provide pulse and gate signals to switches I want that these conditions be checked every 1/2000th of a second. I tried for loop and tic toc methods but the condition is being checked at a frequency of 200kHz. I am not able to solve this issue kindly helpI have multiple if conditions that is being checked to provide pulse and gate signals to switches I want that these conditions be checked every 1/2000th of a second. I tried for loop and tic toc methods but the condition is being checked at a frequency of 200kHz. I am not able to solve this issue kindly help I have multiple if conditions that is being checked to provide pulse and gate signals to switches I want that these conditions be checked every 1/2000th of a second. I tried for loop and tic toc methods but the condition is being checked at a frequency of 200kHz. I am not able to solve this issue kindly help loop condition, if statement, for loop MATLAB Answers — New Questions

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