Hi,
I have implemented a toolchain based on costum target file cxc_ert.tlc
When I use external mode tcpip -> MEX-file name: ext_comm
The code is generated without problems.
Now I am changing for external mode using XCP on TCP/IP. For that on the sl_costumization.m script I add the line:
cm.ExtModeTransports.add(‘cxc_ert.tlc’, ‘XCP on TCP/IP’,’ext_xcp’, ‘Level1’);
After choose XCP on TCP/IP and try to generate the code the following error appears (see image below).

How can I provide a custom Platform Abstraction Layer? What is the meaning of that? I am new on the Embedded Coder. My target HW is ubuntu Core.

Can you help me?
Many thanks in advance
Miguel PintoHi,
I have implemented a toolchain based on costum target file cxc_ert.tlc
When I use external mode tcpip -> MEX-file name: ext_comm
The code is generated without problems.
Now I am changing for external mode using XCP on TCP/IP. For that on the sl_costumization.m script I add the line:
cm.ExtModeTransports.add(‘cxc_ert.tlc’, ‘XCP on TCP/IP’,’ext_xcp’, ‘Level1’);
After choose XCP on TCP/IP and try to generate the code the following error appears (see image below).

How can I provide a custom Platform Abstraction Layer? What is the meaning of that? I am new on the Embedded Coder. My target HW is ubuntu Core.

Can you help me?
Many thanks in advance
Miguel Pinto Hi,
I have implemented a toolchain based on costum target file cxc_ert.tlc
When I use external mode tcpip -> MEX-file name: ext_comm
The code is generated without problems.
Now I am changing for external mode using XCP on TCP/IP. For that on the sl_costumization.m script I add the line:
cm.ExtModeTransports.add(‘cxc_ert.tlc’, ‘XCP on TCP/IP’,’ext_xcp’, ‘Level1’);
After choose XCP on TCP/IP and try to generate the code the following error appears (see image below).

How can I provide a custom Platform Abstraction Layer? What is the meaning of that? I am new on the Embedded Coder. My target HW is ubuntu Core.

Can you help me?
Many thanks in advance
Miguel Pinto external mode, xcp on tcp/ip MATLAB Answers — New Questions

​

Please, how can I read an image of size 2048*2048 pixels using Matlab 2013 by the "imread()" command ? thank you.Please, how can I read an image of size 2048*2048 pixels using Matlab 2013 by the "imread()" command ? thank you. Please, how can I read an image of size 2048*2048 pixels using Matlab 2013 by the "imread()" command ? thank you. image processing, image acquisition MATLAB Answers — New Questions

​

for i = 1:length(files)
file = fullfile(‘c:sar’, files(i).name);
disp(file);
% Load and preprocess the data
source = loadProduct(file); % Define or replace this function with actual data loading
it gives following errors
>> SAR_preprocessing
c:sarS1A_IW_GRDH_1SDV_20240823T005527_20240823T005552_055333_06BF4D_21CD.SAFE.zip
Error in SAR_preprocessing>main (line 44)
source = loadProduct(file); % Define or replace this function with actual data loading
Unrecognized function or variable ‘loadProduct’.

Please suggest me how to load zip file for further processing.
Kuldeepfor i = 1:length(files)
file = fullfile(‘c:sar’, files(i).name);
disp(file);
% Load and preprocess the data
source = loadProduct(file); % Define or replace this function with actual data loading
it gives following errors
>> SAR_preprocessing
c:sarS1A_IW_GRDH_1SDV_20240823T005527_20240823T005552_055333_06BF4D_21CD.SAFE.zip
Error in SAR_preprocessing>main (line 44)
source = loadProduct(file); % Define or replace this function with actual data loading
Unrecognized function or variable ‘loadProduct’.

Please suggest me how to load zip file for further processing.
Kuldeep for i = 1:length(files)
file = fullfile(‘c:sar’, files(i).name);
disp(file);
% Load and preprocess the data
source = loadProduct(file); % Define or replace this function with actual data loading
it gives following errors
>> SAR_preprocessing
c:sarS1A_IW_GRDH_1SDV_20240823T005527_20240823T005552_055333_06BF4D_21CD.SAFE.zip
Error in SAR_preprocessing>main (line 44)
source = loadProduct(file); % Define or replace this function with actual data loading
Unrecognized function or variable ‘loadProduct’.

Please suggest me how to load zip file for further processing.
Kuldeep how do i load zip file in matlab code? MATLAB Answers — New Questions

​

%%%% Jeffry-Hamel equation stability analysis %%%%
%%% ode:(U”’ +2sUU’ = 0) such that U’ = g, g’ = h, h’ = -2*s*U*g %%%
%%% where s = S/a and a = slope of the wall %%%
%%% BC: U =1,g =0 at y =0 && U = 0, g= 10 (Kn = 0.1) at y = 1 %%%
%%% Find h(0) such that g(1) = 10 %%%
%% Range of Variable %%
y_cl = 0;
y_chw = 1;
Re = 0.68e5;
S = 6;
%% Boundary conditions %%
U0 = 1;
U1 = 0;
g0 = 0;
g1 = 10;
h0 = [0.1 0.3];
%% Improved step size and error tolerance %%
d_y = 0.001; % Reduce step size for better accuracy
N = (y_chw -y_cl)/d_y;
err = 1;
max_iter = 6e5; % Limit the maximum number of iterations
iter_count = 0;
%% initializing solution %%
y = y_cl : d_y : y_chw; % y coordinate for plot function
while err > 1e-15 && iter_count < max_iter
iter_count = iter_count + 1;
for j = 1:2
F = zeros(N+1, 3); % Initialize solution matrix
F(1, 🙂 = [U0 g0 h0(j)];

% Runge-Kutta 4th order method
for i = 1:N
k1 = d_y * [F(i, 2) F(i, 3) -(F(i, 1) * F(i, 2)) * 2 * S];
k2 = d_y * [F(i, 2) + k1(2)/2 F(i, 3) + k1(3)/2 -( (F(i, 1) + k1(1)/2) * (F(i, 2) + k1(2)/2) ) * 2 * S];
k3 = d_y * [F(i, 2) + k2(2)/2 F(i, 3) + k2(3)/2 -( (F(i, 1) + k2(1)/2) * (F(i, 2) + k2(2)/2) ) * 2 * S];
k4 = d_y * [F(i, 2) + k3(2) F(i, 3) + k3(3) -( (F(i, 1) + k3(1)) * (F(i, 2) + k3(2)) ) * 2 * S];
F(i+1, 🙂 = F(i, 🙂 + (1/6) * (k1 + 2*k2 + 2*k3 + k4);
end

g_1(j) = F(end, 2); % Store g(1) for each trial
end

% Compute error and update h0
[err, index] = max(abs(g1 – g_1));
P = diff(g_1);
if P ~= 0
h0_new = h0(1) + (diff(h0) / diff(g_1)) * (g1 – g_1(1));
h0(index) = h0_new;
end
end
% Final check on iteration count
if iter_count >= max_iter
warning(‘Max iterations reached. Convergence may not be achieved.’);
end
%% plotting
f1=figure;
f1.Units = ‘normalized’;
f1.Position = [0.1 0.1 0.8 0.6];
plot(F,y’,’LineWidth’,1);
xlim([0 1]);
ylim([0 1]);
axis square
yticks(0:0.2:1);
yticklabels({‘0′,’0.2′,’0.4′,’0.6′,’0.8′,’1’});
xticks(0:0.2:1);
xticklabels({‘0′,’0.2′,’0.4′,’0.6′,’0.8′,’1’});
grid on
title(‘Jeffry Hamel solution’);
xlabel(‘y’);
legend(‘U’,’U"’,’U"”’);

I’m also inserting a plot the I want to see:

please help me in getting this resolved.
Thanks in advance.%%%% Jeffry-Hamel equation stability analysis %%%%
%%% ode:(U”’ +2sUU’ = 0) such that U’ = g, g’ = h, h’ = -2*s*U*g %%%
%%% where s = S/a and a = slope of the wall %%%
%%% BC: U =1,g =0 at y =0 && U = 0, g= 10 (Kn = 0.1) at y = 1 %%%
%%% Find h(0) such that g(1) = 10 %%%
%% Range of Variable %%
y_cl = 0;
y_chw = 1;
Re = 0.68e5;
S = 6;
%% Boundary conditions %%
U0 = 1;
U1 = 0;
g0 = 0;
g1 = 10;
h0 = [0.1 0.3];
%% Improved step size and error tolerance %%
d_y = 0.001; % Reduce step size for better accuracy
N = (y_chw -y_cl)/d_y;
err = 1;
max_iter = 6e5; % Limit the maximum number of iterations
iter_count = 0;
%% initializing solution %%
y = y_cl : d_y : y_chw; % y coordinate for plot function
while err > 1e-15 && iter_count < max_iter
iter_count = iter_count + 1;
for j = 1:2
F = zeros(N+1, 3); % Initialize solution matrix
F(1, 🙂 = [U0 g0 h0(j)];

% Runge-Kutta 4th order method
for i = 1:N
k1 = d_y * [F(i, 2) F(i, 3) -(F(i, 1) * F(i, 2)) * 2 * S];
k2 = d_y * [F(i, 2) + k1(2)/2 F(i, 3) + k1(3)/2 -( (F(i, 1) + k1(1)/2) * (F(i, 2) + k1(2)/2) ) * 2 * S];
k3 = d_y * [F(i, 2) + k2(2)/2 F(i, 3) + k2(3)/2 -( (F(i, 1) + k2(1)/2) * (F(i, 2) + k2(2)/2) ) * 2 * S];
k4 = d_y * [F(i, 2) + k3(2) F(i, 3) + k3(3) -( (F(i, 1) + k3(1)) * (F(i, 2) + k3(2)) ) * 2 * S];
F(i+1, 🙂 = F(i, 🙂 + (1/6) * (k1 + 2*k2 + 2*k3 + k4);
end

g_1(j) = F(end, 2); % Store g(1) for each trial
end

% Compute error and update h0
[err, index] = max(abs(g1 – g_1));
P = diff(g_1);
if P ~= 0
h0_new = h0(1) + (diff(h0) / diff(g_1)) * (g1 – g_1(1));
h0(index) = h0_new;
end
end
% Final check on iteration count
if iter_count >= max_iter
warning(‘Max iterations reached. Convergence may not be achieved.’);
end
%% plotting
f1=figure;
f1.Units = ‘normalized’;
f1.Position = [0.1 0.1 0.8 0.6];
plot(F,y’,’LineWidth’,1);
xlim([0 1]);
ylim([0 1]);
axis square
yticks(0:0.2:1);
yticklabels({‘0′,’0.2′,’0.4′,’0.6′,’0.8′,’1’});
xticks(0:0.2:1);
xticklabels({‘0′,’0.2′,’0.4′,’0.6′,’0.8′,’1’});
grid on
title(‘Jeffry Hamel solution’);
xlabel(‘y’);
legend(‘U’,’U"’,’U"”’);

I’m also inserting a plot the I want to see:

please help me in getting this resolved.
Thanks in advance. %%%% Jeffry-Hamel equation stability analysis %%%%
%%% ode:(U”’ +2sUU’ = 0) such that U’ = g, g’ = h, h’ = -2*s*U*g %%%
%%% where s = S/a and a = slope of the wall %%%
%%% BC: U =1,g =0 at y =0 && U = 0, g= 10 (Kn = 0.1) at y = 1 %%%
%%% Find h(0) such that g(1) = 10 %%%
%% Range of Variable %%
y_cl = 0;
y_chw = 1;
Re = 0.68e5;
S = 6;
%% Boundary conditions %%
U0 = 1;
U1 = 0;
g0 = 0;
g1 = 10;
h0 = [0.1 0.3];
%% Improved step size and error tolerance %%
d_y = 0.001; % Reduce step size for better accuracy
N = (y_chw -y_cl)/d_y;
err = 1;
max_iter = 6e5; % Limit the maximum number of iterations
iter_count = 0;
%% initializing solution %%
y = y_cl : d_y : y_chw; % y coordinate for plot function
while err > 1e-15 && iter_count < max_iter
iter_count = iter_count + 1;
for j = 1:2
F = zeros(N+1, 3); % Initialize solution matrix
F(1, 🙂 = [U0 g0 h0(j)];

% Runge-Kutta 4th order method
for i = 1:N
k1 = d_y * [F(i, 2) F(i, 3) -(F(i, 1) * F(i, 2)) * 2 * S];
k2 = d_y * [F(i, 2) + k1(2)/2 F(i, 3) + k1(3)/2 -( (F(i, 1) + k1(1)/2) * (F(i, 2) + k1(2)/2) ) * 2 * S];
k3 = d_y * [F(i, 2) + k2(2)/2 F(i, 3) + k2(3)/2 -( (F(i, 1) + k2(1)/2) * (F(i, 2) + k2(2)/2) ) * 2 * S];
k4 = d_y * [F(i, 2) + k3(2) F(i, 3) + k3(3) -( (F(i, 1) + k3(1)) * (F(i, 2) + k3(2)) ) * 2 * S];
F(i+1, 🙂 = F(i, 🙂 + (1/6) * (k1 + 2*k2 + 2*k3 + k4);
end

g_1(j) = F(end, 2); % Store g(1) for each trial
end

% Compute error and update h0
[err, index] = max(abs(g1 – g_1));
P = diff(g_1);
if P ~= 0
h0_new = h0(1) + (diff(h0) / diff(g_1)) * (g1 – g_1(1));
h0(index) = h0_new;
end
end
% Final check on iteration count
if iter_count >= max_iter
warning(‘Max iterations reached. Convergence may not be achieved.’);
end
%% plotting
f1=figure;
f1.Units = ‘normalized’;
f1.Position = [0.1 0.1 0.8 0.6];
plot(F,y’,’LineWidth’,1);
xlim([0 1]);
ylim([0 1]);
axis square
yticks(0:0.2:1);
yticklabels({‘0′,’0.2′,’0.4′,’0.6′,’0.8′,’1’});
xticks(0:0.2:1);
xticklabels({‘0′,’0.2′,’0.4′,’0.6′,’0.8′,’1’});
grid on
title(‘Jeffry Hamel solution’);
xlabel(‘y’);
legend(‘U’,’U"’,’U"”’);

I’m also inserting a plot the I want to see:

please help me in getting this resolved.
Thanks in advance. rk4, matlab, channelflow MATLAB Answers — New Questions

​

A1 = 1; wp1 = 10;
A2 = -2; wp2 = 20;
s = tf(‘s’);
H1 = A1 / (1 + s/wp1); H2 = A2 / (1 + s/wp2);
H = H1 + H2;

figure;
bp = bodeplot(H, H1, H2);
hLines = findobj(gcf, Type=’line’);
set(hLines, LineWidth=2); % not working in R2024b?
set(hLines(6), LineWidth=6, DisplayName=’H2′); % not working in R2024b?
set(hLines(3), LineWidth=6, DisplayName=’H2′); % not working in R2024b?
set(findall(gcf, ‘Type’, ‘axes’), LineWidth=1.5, Box="off", fontsize=12); % not working in R2024b?
% Any correct/equivalent/recommended practice in R2024b?

figure;
pzmap(H, H1, H2);
set(gca, YTick=[], Box="off", LineWidth=1.5, XAxisLocation=’origin’, YAxisLocation=’origin’);
% get an R2024b Error using controllib.chart.PZPlot/set Unrecognized property YTick for class PZPlot.
% Any correct/equivalent/recommended practice in R2024b?

R2024a:

R2024b:

The "Version History" of "BodePlot" and "PZPlot" does point out changes in "gca", but for the "Chart Object" mentioned in it, I Googled it and couldn’t find the relevant documentation, so I don’t know how to get axes from Chart Object. Since the R2024b version is too new, I don’t know where to find the information.A1 = 1; wp1 = 10;
A2 = -2; wp2 = 20;
s = tf(‘s’);
H1 = A1 / (1 + s/wp1); H2 = A2 / (1 + s/wp2);
H = H1 + H2;

figure;
bp = bodeplot(H, H1, H2);
hLines = findobj(gcf, Type=’line’);
set(hLines, LineWidth=2); % not working in R2024b?
set(hLines(6), LineWidth=6, DisplayName=’H2′); % not working in R2024b?
set(hLines(3), LineWidth=6, DisplayName=’H2′); % not working in R2024b?
set(findall(gcf, ‘Type’, ‘axes’), LineWidth=1.5, Box="off", fontsize=12); % not working in R2024b?
% Any correct/equivalent/recommended practice in R2024b?

figure;
pzmap(H, H1, H2);
set(gca, YTick=[], Box="off", LineWidth=1.5, XAxisLocation=’origin’, YAxisLocation=’origin’);
% get an R2024b Error using controllib.chart.PZPlot/set Unrecognized property YTick for class PZPlot.
% Any correct/equivalent/recommended practice in R2024b?

R2024a:

R2024b:

The "Version History" of "BodePlot" and "PZPlot" does point out changes in "gca", but for the "Chart Object" mentioned in it, I Googled it and couldn’t find the relevant documentation, so I don’t know how to get axes from Chart Object. Since the R2024b version is too new, I don’t know where to find the information. A1 = 1; wp1 = 10;
A2 = -2; wp2 = 20;
s = tf(‘s’);
H1 = A1 / (1 + s/wp1); H2 = A2 / (1 + s/wp2);
H = H1 + H2;

figure;
bp = bodeplot(H, H1, H2);
hLines = findobj(gcf, Type=’line’);
set(hLines, LineWidth=2); % not working in R2024b?
set(hLines(6), LineWidth=6, DisplayName=’H2′); % not working in R2024b?
set(hLines(3), LineWidth=6, DisplayName=’H2′); % not working in R2024b?
set(findall(gcf, ‘Type’, ‘axes’), LineWidth=1.5, Box="off", fontsize=12); % not working in R2024b?
% Any correct/equivalent/recommended practice in R2024b?

figure;
pzmap(H, H1, H2);
set(gca, YTick=[], Box="off", LineWidth=1.5, XAxisLocation=’origin’, YAxisLocation=’origin’);
% get an R2024b Error using controllib.chart.PZPlot/set Unrecognized property YTick for class PZPlot.
% Any correct/equivalent/recommended practice in R2024b?

R2024a:

R2024b:

The "Version History" of "BodePlot" and "PZPlot" does point out changes in "gca", but for the "Chart Object" mentioned in it, I Googled it and couldn’t find the relevant documentation, so I don’t know how to get axes from Chart Object. Since the R2024b version is too new, I don’t know where to find the information. graphics, figure, axis, handles MATLAB Answers — New Questions

​

I tried this block "CustomPythonModelPredict" by Statistics and Machine Learning Toolbox.
See below link.
Predict Responses Using Custom Python Model in Simulink

I opened slexCustomPythonModelPredictExample.slx as this documentation.
I executed PredictResponsesUsingCustomPythonModelExample.mlx but
I got this error.
I looked into MATLAB Answer community and found similar posts but I couldn’t figure out the solution.

Also I did this command
system(‘python -m pip install numpy’)
but
/usr/bin/python: No module named pip
shows up. I’m not sure….

In the case of MATLAB Online, how do you solve this issue?

It’d be nice if you could give me any advice.

Here is the result I did pyversion command.
>>pyversion

version: ‘3.10’
executable: ‘/usr/bin/python3’
library: ‘libpython3.10.so.1.0’
home: ‘/usr’
isloaded: 1I tried this block "CustomPythonModelPredict" by Statistics and Machine Learning Toolbox.
See below link.
Predict Responses Using Custom Python Model in Simulink

I opened slexCustomPythonModelPredictExample.slx as this documentation.
I executed PredictResponsesUsingCustomPythonModelExample.mlx but
I got this error.
I looked into MATLAB Answer community and found similar posts but I couldn’t figure out the solution.

Also I did this command
system(‘python -m pip install numpy’)
but
/usr/bin/python: No module named pip
shows up. I’m not sure….

In the case of MATLAB Online, how do you solve this issue?

It’d be nice if you could give me any advice.

Here is the result I did pyversion command.
>>pyversion

version: ‘3.10’
executable: ‘/usr/bin/python3’
library: ‘libpython3.10.so.1.0’
home: ‘/usr’
isloaded: 1 I tried this block "CustomPythonModelPredict" by Statistics and Machine Learning Toolbox.
See below link.
Predict Responses Using Custom Python Model in Simulink

I opened slexCustomPythonModelPredictExample.slx as this documentation.
I executed PredictResponsesUsingCustomPythonModelExample.mlx but
I got this error.
I looked into MATLAB Answer community and found similar posts but I couldn’t figure out the solution.

Also I did this command
system(‘python -m pip install numpy’)
but
/usr/bin/python: No module named pip
shows up. I’m not sure….

In the case of MATLAB Online, how do you solve this issue?

It’d be nice if you could give me any advice.

Here is the result I did pyversion command.
>>pyversion

version: ‘3.10’
executable: ‘/usr/bin/python3’
library: ‘libpython3.10.so.1.0’
home: ‘/usr’
isloaded: 1 python, simulink, machine learning, toolbox MATLAB Answers — New Questions

​

for example -7+13i

how do i calculate and display the magnitude and phase angle of thisfor example -7+13i

how do i calculate and display the magnitude and phase angle of this for example -7+13i

how do i calculate and display the magnitude and phase angle of this complex, number, phase angle, magnitude MATLAB Answers — New Questions

​

Attaced is AAA.mat which contains the matrix that generates the issue, as well as a code that reproduces the problem.
Following, copied from code, shows what the issue is, essentially, if I do max in the first dimension of permute(AAA,[2,1,3,4]) then everything works fine. But if I do max in the second dimension of AAA —which should give exact same answer— then the max-values are reported correctly but the max-indexes are incorrect.
%% The problematic example
load AAA.mat
[vv1,ii1]=max(AAA,[],2);
[vv2,ii2]=max(permute(AAA,[2,1,3,4]),[],1);

squeeze(vv1)-squeeze(vv2) % all zeros as expected
squeeze(ii1)-squeeze(ii2) % many non-zeros, which should not happen

% Note: vv1 is correct, but ii1 is incorrect

% E.g., vv1(1,1,1,2)=-2.6733
% Which is correct, and the same as vv2(1,1,1,2)=-2.6733

% By comparison, ii1(1,1,1,2)=3081
% But the correct answer is ii2(1,1,1,2)=1
% We can see this because AAA(1,3081,1,2)=-5.1175, which is not vv1
% By comparison, AAA(1,1,1,2)=-2.6733, which is exactly what we expect it to be

% Unless I am missing something, [vv1,ii1]=max(AAA,[],2); is returning the incorrect index for the max.

% Note: (:,:,1,1) gets the indexes correct, but the indexes corresponding to most other 3rd and 4th
% dimension points, like (:,:,1,5) and (:,:,5,1) are wrong.

I don’t know what about AAA causes this error to occur. The "max2derror.m" tries a bunch of things I thought of but they all work correctly.
Pretty sure I am getting this right 😐
I ran this in R2023b.Attaced is AAA.mat which contains the matrix that generates the issue, as well as a code that reproduces the problem.
Following, copied from code, shows what the issue is, essentially, if I do max in the first dimension of permute(AAA,[2,1,3,4]) then everything works fine. But if I do max in the second dimension of AAA —which should give exact same answer— then the max-values are reported correctly but the max-indexes are incorrect.
%% The problematic example
load AAA.mat
[vv1,ii1]=max(AAA,[],2);
[vv2,ii2]=max(permute(AAA,[2,1,3,4]),[],1);

squeeze(vv1)-squeeze(vv2) % all zeros as expected
squeeze(ii1)-squeeze(ii2) % many non-zeros, which should not happen

% Note: vv1 is correct, but ii1 is incorrect

% E.g., vv1(1,1,1,2)=-2.6733
% Which is correct, and the same as vv2(1,1,1,2)=-2.6733

% By comparison, ii1(1,1,1,2)=3081
% But the correct answer is ii2(1,1,1,2)=1
% We can see this because AAA(1,3081,1,2)=-5.1175, which is not vv1
% By comparison, AAA(1,1,1,2)=-2.6733, which is exactly what we expect it to be

% Unless I am missing something, [vv1,ii1]=max(AAA,[],2); is returning the incorrect index for the max.

% Note: (:,:,1,1) gets the indexes correct, but the indexes corresponding to most other 3rd and 4th
% dimension points, like (:,:,1,5) and (:,:,5,1) are wrong.

I don’t know what about AAA causes this error to occur. The "max2derror.m" tries a bunch of things I thought of but they all work correctly.
Pretty sure I am getting this right 😐
I ran this in R2023b. Attaced is AAA.mat which contains the matrix that generates the issue, as well as a code that reproduces the problem.
Following, copied from code, shows what the issue is, essentially, if I do max in the first dimension of permute(AAA,[2,1,3,4]) then everything works fine. But if I do max in the second dimension of AAA —which should give exact same answer— then the max-values are reported correctly but the max-indexes are incorrect.
%% The problematic example
load AAA.mat
[vv1,ii1]=max(AAA,[],2);
[vv2,ii2]=max(permute(AAA,[2,1,3,4]),[],1);

squeeze(vv1)-squeeze(vv2) % all zeros as expected
squeeze(ii1)-squeeze(ii2) % many non-zeros, which should not happen

% Note: vv1 is correct, but ii1 is incorrect

% E.g., vv1(1,1,1,2)=-2.6733
% Which is correct, and the same as vv2(1,1,1,2)=-2.6733

% By comparison, ii1(1,1,1,2)=3081
% But the correct answer is ii2(1,1,1,2)=1
% We can see this because AAA(1,3081,1,2)=-5.1175, which is not vv1
% By comparison, AAA(1,1,1,2)=-2.6733, which is exactly what we expect it to be

% Unless I am missing something, [vv1,ii1]=max(AAA,[],2); is returning the incorrect index for the max.

% Note: (:,:,1,1) gets the indexes correct, but the indexes corresponding to most other 3rd and 4th
% dimension points, like (:,:,1,5) and (:,:,5,1) are wrong.

I don’t know what about AAA causes this error to occur. The "max2derror.m" tries a bunch of things I thought of but they all work correctly.
Pretty sure I am getting this right 😐
I ran this in R2023b. max MATLAB Answers — New Questions

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Hello everyone,
I am now doing a project about the stock .And now I need to plot a candle line,
I plotted it successesfully,However ,because I have about 300 days to plot into one figure,It is too small to regonize the details.I am know wondering if I can plot two figures with successive time,
ex:the time of first figure is from 01/01 ~05/03 ,and the other one is from 05/03~07/07
but I have no idea how to do it? any idea?
My candle code is simple :candle(DATA(:,2),DATA(:,3),DATA(:,4),DATA(:,1),’b’)Hello everyone,
I am now doing a project about the stock .And now I need to plot a candle line,
I plotted it successesfully,However ,because I have about 300 days to plot into one figure,It is too small to regonize the details.I am know wondering if I can plot two figures with successive time,
ex:the time of first figure is from 01/01 ~05/03 ,and the other one is from 05/03~07/07
but I have no idea how to do it? any idea?
My candle code is simple :candle(DATA(:,2),DATA(:,3),DATA(:,4),DATA(:,1),’b’) Hello everyone,
I am now doing a project about the stock .And now I need to plot a candle line,
I plotted it successesfully,However ,because I have about 300 days to plot into one figure,It is too small to regonize the details.I am know wondering if I can plot two figures with successive time,
ex:the time of first figure is from 01/01 ~05/03 ,and the other one is from 05/03~07/07
but I have no idea how to do it? any idea?
My candle code is simple :candle(DATA(:,2),DATA(:,3),DATA(:,4),DATA(:,1),’b’) plot, financial, graph MATLAB Answers — New Questions

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