Hello.
We’re using the export function to run an *.mlx file and then export it to a pdf.
I don’t seem to be able to interact with the OS filesystem or store data in the base workspace when the *.mlx is executed using the export function
export("Analyse_Template",filename,PageSize="A4",Run=true,FigureResolution=800,HideCode=true);

% using assignin does not work
assignin(‘base’,’OutPutTable’,OutPutTable)
% using writetable does not work
writetable(inputTable, FName,"WriteMode","append");
Executing the *.mlx file itself produces the desired results.
Why can I not work outside of the scope of the export function?
Many thanks in advance.Hello.
We’re using the export function to run an *.mlx file and then export it to a pdf.
I don’t seem to be able to interact with the OS filesystem or store data in the base workspace when the *.mlx is executed using the export function
export("Analyse_Template",filename,PageSize="A4",Run=true,FigureResolution=800,HideCode=true);

% using assignin does not work
assignin(‘base’,’OutPutTable’,OutPutTable)
% using writetable does not work
writetable(inputTable, FName,"WriteMode","append");
Executing the *.mlx file itself produces the desired results.
Why can I not work outside of the scope of the export function?
Many thanks in advance. Hello.
We’re using the export function to run an *.mlx file and then export it to a pdf.
I don’t seem to be able to interact with the OS filesystem or store data in the base workspace when the *.mlx is executed using the export function
export("Analyse_Template",filename,PageSize="A4",Run=true,FigureResolution=800,HideCode=true);

% using assignin does not work
assignin(‘base’,’OutPutTable’,OutPutTable)
% using writetable does not work
writetable(inputTable, FName,"WriteMode","append");
Executing the *.mlx file itself produces the desired results.
Why can I not work outside of the scope of the export function?
Many thanks in advance. export, live script, assignin MATLAB Answers — New Questions

​

Hello Everyone,
Anyone can help me.
I have one .dat file as attached. According to the manual, the description of my file is 364 × 364 × 110, integer, 8-bit.
Anyone can help how to open my file.
Then how to modify some infi like the geometry.Hello Everyone,
Anyone can help me.
I have one .dat file as attached. According to the manual, the description of my file is 364 × 364 × 110, integer, 8-bit.
Anyone can help how to open my file.
Then how to modify some infi like the geometry. Hello Everyone,
Anyone can help me.
I have one .dat file as attached. According to the manual, the description of my file is 364 × 364 × 110, integer, 8-bit.
Anyone can help how to open my file.
Then how to modify some infi like the geometry. file, image, image analysis, image acquisition, image segmentation, digital image processing, image processing MATLAB Answers — New Questions

​

Hello everyone.
I’m using the F28379D LaunchPad in Simulink and I want to read some Digital Inputs to read the pulses of a quadrature encoder and make a pulse count. I know that the board has built in pins to read quadrature encoder but I want to try reading it by the digital input pins. The problem is that I put a Scope and a Display and the value doesn’t change. I already checked that the encoder works, I tried input a square signal, even using a push button with pull-down resistor and the value doesn’t change at all. Here is the piece of code of what I am doing.
Hope someone can help me. Thanks in advance.Hello everyone.
I’m using the F28379D LaunchPad in Simulink and I want to read some Digital Inputs to read the pulses of a quadrature encoder and make a pulse count. I know that the board has built in pins to read quadrature encoder but I want to try reading it by the digital input pins. The problem is that I put a Scope and a Display and the value doesn’t change. I already checked that the encoder works, I tried input a square signal, even using a push button with pull-down resistor and the value doesn’t change at all. Here is the piece of code of what I am doing.
Hope someone can help me. Thanks in advance. Hello everyone.
I’m using the F28379D LaunchPad in Simulink and I want to read some Digital Inputs to read the pulses of a quadrature encoder and make a pulse count. I know that the board has built in pins to read quadrature encoder but I want to try reading it by the digital input pins. The problem is that I put a Scope and a Display and the value doesn’t change. I already checked that the encoder works, I tried input a square signal, even using a push button with pull-down resistor and the value doesn’t change at all. Here is the piece of code of what I am doing.
Hope someone can help me. Thanks in advance. simulink MATLAB Answers — New Questions

​

The Original Equation is the following:

Let . This implies that
Then we eewrite it as a System of First-Order Equations
Using the substitution for , the second-order equation can be transformed into the following system of first-order equations:

Exploring the Effect of .

% Define parameters
gamma = 0.338;
alpha = -1;
beta = 1;
delta = 0.1;
omega = 1.4;

% Define the system of equations
odeSystem = @(t, y) [y(2);
-delta*y(2) – alpha*y(1) – beta*y(1)^3 + gamma*cos(omega*t)];

% Initial conditions
y0 = [0; 0]; % x(0) = 0, v(0) = 0

% Time span
tspan = [0 2000];

% Solve the system
[t, y] = ode45(odeSystem, tspan, y0);

% Plot the results
figure;
plot(t, y(:, 1));
xlabel(‘Time’);
ylabel(‘x(t)’);
title(‘Solution of the nonlinear system’);
grid on;

% Plot the phase portrait
figure;
plot(y(:, 1), y(:, 2));
xlabel(‘x(t)’);
ylabel(‘v(t)’);
title(‘Phase-Plane $$ddot{x}+delta dot{x}+alpha x+beta x^3=0$$ for $$gamma=0.318$$, $$alpha=-1$$,$$beta=1$$,$$delta=0.1$$,$$omega=1.4$$’,’Interpreter’, ‘latex’);
grid on;

% Define the tail (e.g., last 10% of the time interval)
tail_start = floor(0.9 * length(t)); % Starting index for the tail
tail_end = length(t); % Ending index for the tail

% Plot the tail of the solution
figure;
plot(y(tail_start:tail_end, 1), y(tail_start:tail_end, 2), ‘r’, ‘LineWidth’, 1.5);
xlabel(‘x(t)’);
ylabel(‘v(t)’);
title(‘Phase-Plane $$ddot{x}+delta dot{x}+alpha x+beta x^3=0$$ for $$gamma=0.318$$, $$alpha=-1$$,$$beta=1$$,$$delta=0.1$$,$$omega=1.4$$’,’Interpreter’, ‘latex’);
grid on;

ax = gca;
chart = ax.Children(1);
datatip(chart,0.5581,-0.1126);
Then I used but the results were not that I expected for

My code gives me the following. Any suggestion?
% Define parameters
gamma = 0.35;
alpha = -1;
beta = 1;
delta = 0.1;
omega = 1.4;

% Define the system of equations
odeSystem = @(t, y) [y(2);
-delta*y(2) – alpha*y(1) – beta*y(1)^3 + gamma*cos(omega*t)];

% Initial conditions
y0 = [0; 0]; % x(0) = 0, v(0) = 0

% Time span
tspan = [0 3000];

% Solve the system
[t, y] = ode45(odeSystem, tspan, y0);

% Plot the phase portrait
figure;
plot(y(:, 1), y(:, 2));
xlabel(‘x(t)’);
ylabel(‘v(t)’);
title(‘Phase-Plane $$ddot{x}+delta dot{x}+alpha x+beta x^3=0$$ for $$gamma=0.318$$, $$alpha=-1$$,$$beta=1$$,$$delta=0.1$$,$$omega=1.4$$’,’Interpreter’, ‘latex’);
grid on;

% Define the tail (e.g., last 10% of the time interval)
tail_start = floor(0.9 * length(t)); % Starting index for the tail
tail_end = length(t); % Ending index for the tail

% Plot the tail of the solution
figure;
plot(y(tail_start:tail_end, 1), y(tail_start:tail_end, 2), ‘r’, ‘LineWidth’, 1.5);
xlabel(‘x(t)’);
ylabel(‘v(t)’);
title(‘Phase-Plane $$ddot{x}+delta dot{x}+alpha x+beta x^3=0$$ for $$gamma=0.35$$, $$alpha=-1$$,$$beta=1$$,$$delta=0.1$$,$$omega=1.4$$’,’Interpreter’, ‘latex’);
grid on;

ax = gca;
chart = ax.Children(1);
datatip(chart,0.5581,-0.1126);The Original Equation is the following:

Let . This implies that
Then we eewrite it as a System of First-Order Equations
Using the substitution for , the second-order equation can be transformed into the following system of first-order equations:

Exploring the Effect of .

% Define parameters
gamma = 0.338;
alpha = -1;
beta = 1;
delta = 0.1;
omega = 1.4;

% Define the system of equations
odeSystem = @(t, y) [y(2);
-delta*y(2) – alpha*y(1) – beta*y(1)^3 + gamma*cos(omega*t)];

% Initial conditions
y0 = [0; 0]; % x(0) = 0, v(0) = 0

% Time span
tspan = [0 2000];

% Solve the system
[t, y] = ode45(odeSystem, tspan, y0);

% Plot the results
figure;
plot(t, y(:, 1));
xlabel(‘Time’);
ylabel(‘x(t)’);
title(‘Solution of the nonlinear system’);
grid on;

% Plot the phase portrait
figure;
plot(y(:, 1), y(:, 2));
xlabel(‘x(t)’);
ylabel(‘v(t)’);
title(‘Phase-Plane $$ddot{x}+delta dot{x}+alpha x+beta x^3=0$$ for $$gamma=0.318$$, $$alpha=-1$$,$$beta=1$$,$$delta=0.1$$,$$omega=1.4$$’,’Interpreter’, ‘latex’);
grid on;

% Define the tail (e.g., last 10% of the time interval)
tail_start = floor(0.9 * length(t)); % Starting index for the tail
tail_end = length(t); % Ending index for the tail

% Plot the tail of the solution
figure;
plot(y(tail_start:tail_end, 1), y(tail_start:tail_end, 2), ‘r’, ‘LineWidth’, 1.5);
xlabel(‘x(t)’);
ylabel(‘v(t)’);
title(‘Phase-Plane $$ddot{x}+delta dot{x}+alpha x+beta x^3=0$$ for $$gamma=0.318$$, $$alpha=-1$$,$$beta=1$$,$$delta=0.1$$,$$omega=1.4$$’,’Interpreter’, ‘latex’);
grid on;

ax = gca;
chart = ax.Children(1);
datatip(chart,0.5581,-0.1126);
Then I used but the results were not that I expected for

My code gives me the following. Any suggestion?
% Define parameters
gamma = 0.35;
alpha = -1;
beta = 1;
delta = 0.1;
omega = 1.4;

% Define the system of equations
odeSystem = @(t, y) [y(2);
-delta*y(2) – alpha*y(1) – beta*y(1)^3 + gamma*cos(omega*t)];

% Initial conditions
y0 = [0; 0]; % x(0) = 0, v(0) = 0

% Time span
tspan = [0 3000];

% Solve the system
[t, y] = ode45(odeSystem, tspan, y0);

% Plot the phase portrait
figure;
plot(y(:, 1), y(:, 2));
xlabel(‘x(t)’);
ylabel(‘v(t)’);
title(‘Phase-Plane $$ddot{x}+delta dot{x}+alpha x+beta x^3=0$$ for $$gamma=0.318$$, $$alpha=-1$$,$$beta=1$$,$$delta=0.1$$,$$omega=1.4$$’,’Interpreter’, ‘latex’);
grid on;

% Define the tail (e.g., last 10% of the time interval)
tail_start = floor(0.9 * length(t)); % Starting index for the tail
tail_end = length(t); % Ending index for the tail

% Plot the tail of the solution
figure;
plot(y(tail_start:tail_end, 1), y(tail_start:tail_end, 2), ‘r’, ‘LineWidth’, 1.5);
xlabel(‘x(t)’);
ylabel(‘v(t)’);
title(‘Phase-Plane $$ddot{x}+delta dot{x}+alpha x+beta x^3=0$$ for $$gamma=0.35$$, $$alpha=-1$$,$$beta=1$$,$$delta=0.1$$,$$omega=1.4$$’,’Interpreter’, ‘latex’);
grid on;

ax = gca;
chart = ax.Children(1);
datatip(chart,0.5581,-0.1126); The Original Equation is the following:

Let . This implies that
Then we eewrite it as a System of First-Order Equations
Using the substitution for , the second-order equation can be transformed into the following system of first-order equations:

Exploring the Effect of .

% Define parameters
gamma = 0.338;
alpha = -1;
beta = 1;
delta = 0.1;
omega = 1.4;

% Define the system of equations
odeSystem = @(t, y) [y(2);
-delta*y(2) – alpha*y(1) – beta*y(1)^3 + gamma*cos(omega*t)];

% Initial conditions
y0 = [0; 0]; % x(0) = 0, v(0) = 0

% Time span
tspan = [0 2000];

% Solve the system
[t, y] = ode45(odeSystem, tspan, y0);

% Plot the results
figure;
plot(t, y(:, 1));
xlabel(‘Time’);
ylabel(‘x(t)’);
title(‘Solution of the nonlinear system’);
grid on;

% Plot the phase portrait
figure;
plot(y(:, 1), y(:, 2));
xlabel(‘x(t)’);
ylabel(‘v(t)’);
title(‘Phase-Plane $$ddot{x}+delta dot{x}+alpha x+beta x^3=0$$ for $$gamma=0.318$$, $$alpha=-1$$,$$beta=1$$,$$delta=0.1$$,$$omega=1.4$$’,’Interpreter’, ‘latex’);
grid on;

% Define the tail (e.g., last 10% of the time interval)
tail_start = floor(0.9 * length(t)); % Starting index for the tail
tail_end = length(t); % Ending index for the tail

% Plot the tail of the solution
figure;
plot(y(tail_start:tail_end, 1), y(tail_start:tail_end, 2), ‘r’, ‘LineWidth’, 1.5);
xlabel(‘x(t)’);
ylabel(‘v(t)’);
title(‘Phase-Plane $$ddot{x}+delta dot{x}+alpha x+beta x^3=0$$ for $$gamma=0.318$$, $$alpha=-1$$,$$beta=1$$,$$delta=0.1$$,$$omega=1.4$$’,’Interpreter’, ‘latex’);
grid on;

ax = gca;
chart = ax.Children(1);
datatip(chart,0.5581,-0.1126);
Then I used but the results were not that I expected for

My code gives me the following. Any suggestion?
% Define parameters
gamma = 0.35;
alpha = -1;
beta = 1;
delta = 0.1;
omega = 1.4;

% Define the system of equations
odeSystem = @(t, y) [y(2);
-delta*y(2) – alpha*y(1) – beta*y(1)^3 + gamma*cos(omega*t)];

% Initial conditions
y0 = [0; 0]; % x(0) = 0, v(0) = 0

% Time span
tspan = [0 3000];

% Solve the system
[t, y] = ode45(odeSystem, tspan, y0);

% Plot the phase portrait
figure;
plot(y(:, 1), y(:, 2));
xlabel(‘x(t)’);
ylabel(‘v(t)’);
title(‘Phase-Plane $$ddot{x}+delta dot{x}+alpha x+beta x^3=0$$ for $$gamma=0.318$$, $$alpha=-1$$,$$beta=1$$,$$delta=0.1$$,$$omega=1.4$$’,’Interpreter’, ‘latex’);
grid on;

% Define the tail (e.g., last 10% of the time interval)
tail_start = floor(0.9 * length(t)); % Starting index for the tail
tail_end = length(t); % Ending index for the tail

% Plot the tail of the solution
figure;
plot(y(tail_start:tail_end, 1), y(tail_start:tail_end, 2), ‘r’, ‘LineWidth’, 1.5);
xlabel(‘x(t)’);
ylabel(‘v(t)’);
title(‘Phase-Plane $$ddot{x}+delta dot{x}+alpha x+beta x^3=0$$ for $$gamma=0.35$$, $$alpha=-1$$,$$beta=1$$,$$delta=0.1$$,$$omega=1.4$$’,’Interpreter’, ‘latex’);
grid on;

ax = gca;
chart = ax.Children(1);
datatip(chart,0.5581,-0.1126); ode45, duffing equation, plot, differential equations, plotting MATLAB Answers — New Questions

​

Thank you for taking the time to help me with this. I’m using MATLAB R2024a. Here is my code:
% Create coarse-uneven data points
Xc = [5 7 11 17 27 40 59 82 129 191];
Yc = [-5 -2 2 0 3 -4 0 4 0 7];
% Basic parameters for observer and lines-of-sight
height = 5.2;
distance = 200;
radians = -pi/4:pi/24:0;
% Coordinates for intersecting lines
[X2, Y2] = pol2cart(radians, distance);
Y2 = Y2 + height;
X1 = zeros(size(X2));
Y1 = zeros(size(Y2)) + height;
% Plotting
figure
plot(Xc, Yc, ‘-r’)
hold on
plot([X1; X2], [Y1; Y2], ‘-b’)
hold off

As you can see, the x-component of the terrain is coarse and unevenly spaced. I would like to identify which terrain datapoints (Xc and Yc) are within line-of-sight of each "slice" of the observer’s vision. In the attached picture, there are four data points within the first sector of vision. It is likely that the second data point and most certainly the fourth data point would be obscured by the terrain pattern. How can I caluclate the line-of-sight for this and extract the Xc components that fall within line of sight of the observer who is positioned at x=0 and y=height?

Ideally, the output would be an NxM logical array where N is the number of vision sectors and M is the length of Xc. The values contained along M in the matrix would represent with boolean true/false values the indices of Xc in which (Xc,Yc) is visible by the observer for the Nth vision sector.

From what I’ve seen, los() and los2() take inputs in a much different format than just 2-D cartesian points and lines, so I haven’t been able to figure it out using those functions.

Once again, thank you for your time.Thank you for taking the time to help me with this. I’m using MATLAB R2024a. Here is my code:
% Create coarse-uneven data points
Xc = [5 7 11 17 27 40 59 82 129 191];
Yc = [-5 -2 2 0 3 -4 0 4 0 7];
% Basic parameters for observer and lines-of-sight
height = 5.2;
distance = 200;
radians = -pi/4:pi/24:0;
% Coordinates for intersecting lines
[X2, Y2] = pol2cart(radians, distance);
Y2 = Y2 + height;
X1 = zeros(size(X2));
Y1 = zeros(size(Y2)) + height;
% Plotting
figure
plot(Xc, Yc, ‘-r’)
hold on
plot([X1; X2], [Y1; Y2], ‘-b’)
hold off

As you can see, the x-component of the terrain is coarse and unevenly spaced. I would like to identify which terrain datapoints (Xc and Yc) are within line-of-sight of each "slice" of the observer’s vision. In the attached picture, there are four data points within the first sector of vision. It is likely that the second data point and most certainly the fourth data point would be obscured by the terrain pattern. How can I caluclate the line-of-sight for this and extract the Xc components that fall within line of sight of the observer who is positioned at x=0 and y=height?

Ideally, the output would be an NxM logical array where N is the number of vision sectors and M is the length of Xc. The values contained along M in the matrix would represent with boolean true/false values the indices of Xc in which (Xc,Yc) is visible by the observer for the Nth vision sector.

From what I’ve seen, los() and los2() take inputs in a much different format than just 2-D cartesian points and lines, so I haven’t been able to figure it out using those functions.

Once again, thank you for your time. Thank you for taking the time to help me with this. I’m using MATLAB R2024a. Here is my code:
% Create coarse-uneven data points
Xc = [5 7 11 17 27 40 59 82 129 191];
Yc = [-5 -2 2 0 3 -4 0 4 0 7];
% Basic parameters for observer and lines-of-sight
height = 5.2;
distance = 200;
radians = -pi/4:pi/24:0;
% Coordinates for intersecting lines
[X2, Y2] = pol2cart(radians, distance);
Y2 = Y2 + height;
X1 = zeros(size(X2));
Y1 = zeros(size(Y2)) + height;
% Plotting
figure
plot(Xc, Yc, ‘-r’)
hold on
plot([X1; X2], [Y1; Y2], ‘-b’)
hold off

As you can see, the x-component of the terrain is coarse and unevenly spaced. I would like to identify which terrain datapoints (Xc and Yc) are within line-of-sight of each "slice" of the observer’s vision. In the attached picture, there are four data points within the first sector of vision. It is likely that the second data point and most certainly the fourth data point would be obscured by the terrain pattern. How can I caluclate the line-of-sight for this and extract the Xc components that fall within line of sight of the observer who is positioned at x=0 and y=height?

Ideally, the output would be an NxM logical array where N is the number of vision sectors and M is the length of Xc. The values contained along M in the matrix would represent with boolean true/false values the indices of Xc in which (Xc,Yc) is visible by the observer for the Nth vision sector.

From what I’ve seen, los() and los2() take inputs in a much different format than just 2-D cartesian points and lines, so I haven’t been able to figure it out using those functions.

Once again, thank you for your time. los, line-of-sight, vectorization, geometry, geography, interpolation, vectors, surface, indexing, nonlinear, find, plot, signal processing MATLAB Answers — New Questions

​

I would like to create Matlab GPU arrays that have a number of elements exceeding the max value of int32 (~2e9). In Matlab, when I try to do this, I get a "Max variable size exceeded" error. Other answers on this forum point to this limitation in GPU array size being caused by limitations in the CUDA API. However, I don’t believe this is accurate. Other languages, like Python 3.9 using theTensorflow 2.11 library for example, that are built on CUDA/CUBLAS allow for arrays that have a number of elements that exceed the MATLAB limitation.
Why is this the case, and why does Matlab seem limited where other languages are not?
The version of Matlab that I am using is 2024a, tested on both Win11 and Ubuntu 20.04LTS with a Nvidia Ada A6000 GPU (VRAM 48GB).
Matlab Code to reproduce:
g = gpuDevice();
arr = ones(1e4, 1e4, 60, ‘single’);
gpu_arr = gpuArray(arr);
Matlab ERROR:
Error using gpuArray
Maximum variable size allowed on the device is exceeded.
Python Code to reproduce:
import tensorflow as tf
arr = tf.ones([10000, 10000, 60])I would like to create Matlab GPU arrays that have a number of elements exceeding the max value of int32 (~2e9). In Matlab, when I try to do this, I get a "Max variable size exceeded" error. Other answers on this forum point to this limitation in GPU array size being caused by limitations in the CUDA API. However, I don’t believe this is accurate. Other languages, like Python 3.9 using theTensorflow 2.11 library for example, that are built on CUDA/CUBLAS allow for arrays that have a number of elements that exceed the MATLAB limitation.
Why is this the case, and why does Matlab seem limited where other languages are not?
The version of Matlab that I am using is 2024a, tested on both Win11 and Ubuntu 20.04LTS with a Nvidia Ada A6000 GPU (VRAM 48GB).
Matlab Code to reproduce:
g = gpuDevice();
arr = ones(1e4, 1e4, 60, ‘single’);
gpu_arr = gpuArray(arr);
Matlab ERROR:
Error using gpuArray
Maximum variable size allowed on the device is exceeded.
Python Code to reproduce:
import tensorflow as tf
arr = tf.ones([10000, 10000, 60]) I would like to create Matlab GPU arrays that have a number of elements exceeding the max value of int32 (~2e9). In Matlab, when I try to do this, I get a "Max variable size exceeded" error. Other answers on this forum point to this limitation in GPU array size being caused by limitations in the CUDA API. However, I don’t believe this is accurate. Other languages, like Python 3.9 using theTensorflow 2.11 library for example, that are built on CUDA/CUBLAS allow for arrays that have a number of elements that exceed the MATLAB limitation.
Why is this the case, and why does Matlab seem limited where other languages are not?
The version of Matlab that I am using is 2024a, tested on both Win11 and Ubuntu 20.04LTS with a Nvidia Ada A6000 GPU (VRAM 48GB).
Matlab Code to reproduce:
g = gpuDevice();
arr = ones(1e4, 1e4, 60, ‘single’);
gpu_arr = gpuArray(arr);
Matlab ERROR:
Error using gpuArray
Maximum variable size allowed on the device is exceeded.
Python Code to reproduce:
import tensorflow as tf
arr = tf.ones([10000, 10000, 60]) gpu, parallel computing toolbox, gpuarray, memory, vram, cuda, python MATLAB Answers — New Questions

​

I am working on an app that ultimately produces 4 separate tables, each containing 4 columns of data. For simplicity, I am only including a single example, as I would think the process should be identical for all 4 tables, aside from the name of each table. All of the tables currently only show 4 decimal places. I would like to show more than this. I tried putting the following code in my startup function:
function startupFcn(app)
clc
close all

app.UITable.ColumnFormat = {‘long’,’long’,’long’,’long’}; %4x ‘long’ – 1 for each column?
This does not seem to work, as all of the tables still only display 4 decimal places. Any help would much appreciated. Thank you.I am working on an app that ultimately produces 4 separate tables, each containing 4 columns of data. For simplicity, I am only including a single example, as I would think the process should be identical for all 4 tables, aside from the name of each table. All of the tables currently only show 4 decimal places. I would like to show more than this. I tried putting the following code in my startup function:
function startupFcn(app)
clc
close all

app.UITable.ColumnFormat = {‘long’,’long’,’long’,’long’}; %4x ‘long’ – 1 for each column?
This does not seem to work, as all of the tables still only display 4 decimal places. Any help would much appreciated. Thank you. I am working on an app that ultimately produces 4 separate tables, each containing 4 columns of data. For simplicity, I am only including a single example, as I would think the process should be identical for all 4 tables, aside from the name of each table. All of the tables currently only show 4 decimal places. I would like to show more than this. I tried putting the following code in my startup function:
function startupFcn(app)
clc
close all

app.UITable.ColumnFormat = {‘long’,’long’,’long’,’long’}; %4x ‘long’ – 1 for each column?
This does not seem to work, as all of the tables still only display 4 decimal places. Any help would much appreciated. Thank you. appdesigner, uitable MATLAB Answers — New Questions

​

When I type into command window, license(‘test’, ‘image_toolbox’), it returns 1.
My code reads and displays images by imread, imshow. But it when I use mat2gray in the next line, there’s an error "’mat2gray’ requires Image Processing Toolbox."When I type into command window, license(‘test’, ‘image_toolbox’), it returns 1.
My code reads and displays images by imread, imshow. But it when I use mat2gray in the next line, there’s an error "’mat2gray’ requires Image Processing Toolbox." When I type into command window, license(‘test’, ‘image_toolbox’), it returns 1.
My code reads and displays images by imread, imshow. But it when I use mat2gray in the next line, there’s an error "’mat2gray’ requires Image Processing Toolbox." image-processing toolbox MATLAB Answers — New Questions

​

Hello.
I have a set of variables that i want to export as one .nc file so that i can use the dataset later on. I have tried using the instructions listed in this thread but it keeps returning an error. I’ll include my code and the variables that i use below. And sorry for the rudamentary code, I’m still new to matlab
clear
clc

%masukin waktu
waktu= ncread(‘FABMSAT.nc’,’time’);
waktu= waktu’;

%masukin fito layer 1
fito= ncread(‘FABMSAT.nc’,’npzd_phy’);
fito= fito(:,1,:);

%masukin pasut
pasut= ncread(‘FABMSAT.nc’,’zeta’);

% masukin vector u
u= ncread(‘FABMSAT.nc’,’u’);
u= u(:,1,:);

%Masukin vector v
v= ncread(‘FABMSAT.nc’,’u’);
v= v(:,1,:);

%CUT SUPAYA MULAI 1 FEB 2022
waktu_feb= waktu;
waktu_feb(:,[1:168])=[];
fito_feb= fito;
fito_feb(:,[1:168])=[];
pasut_feb= pasut;
pasut_feb([2:8307],:)=[];
u_feb= u;
u_feb(:,[1:168])=[];
v_feb= v;
v_feb(:,[1:168])=[];

%define dimensions
netcdf.setDefaultFormat(‘NC_FORMAT_CLASSIC’);
ncid = netcdf.create(‘test.nc’,’NC_WRITE’);
dimidfito = netcdf.defDim(ncid,’fito_feb’,fito_feb);

%Define the name of variable
var_fito = netcdf.defVar(ncid, ‘fito_feb’,’NC_FLOAT’,[dimidfito]);

%variables into .nc file
netcdf.endDef(ncid);
netcdf.putVar(ncid,var_fito,fito_feb);
netcdf.close(ncid);Hello.
I have a set of variables that i want to export as one .nc file so that i can use the dataset later on. I have tried using the instructions listed in this thread but it keeps returning an error. I’ll include my code and the variables that i use below. And sorry for the rudamentary code, I’m still new to matlab
clear
clc

%masukin waktu
waktu= ncread(‘FABMSAT.nc’,’time’);
waktu= waktu’;

%masukin fito layer 1
fito= ncread(‘FABMSAT.nc’,’npzd_phy’);
fito= fito(:,1,:);

%masukin pasut
pasut= ncread(‘FABMSAT.nc’,’zeta’);

% masukin vector u
u= ncread(‘FABMSAT.nc’,’u’);
u= u(:,1,:);

%Masukin vector v
v= ncread(‘FABMSAT.nc’,’u’);
v= v(:,1,:);

%CUT SUPAYA MULAI 1 FEB 2022
waktu_feb= waktu;
waktu_feb(:,[1:168])=[];
fito_feb= fito;
fito_feb(:,[1:168])=[];
pasut_feb= pasut;
pasut_feb([2:8307],:)=[];
u_feb= u;
u_feb(:,[1:168])=[];
v_feb= v;
v_feb(:,[1:168])=[];

%define dimensions
netcdf.setDefaultFormat(‘NC_FORMAT_CLASSIC’);
ncid = netcdf.create(‘test.nc’,’NC_WRITE’);
dimidfito = netcdf.defDim(ncid,’fito_feb’,fito_feb);

%Define the name of variable
var_fito = netcdf.defVar(ncid, ‘fito_feb’,’NC_FLOAT’,[dimidfito]);

%variables into .nc file
netcdf.endDef(ncid);
netcdf.putVar(ncid,var_fito,fito_feb);
netcdf.close(ncid); Hello.
I have a set of variables that i want to export as one .nc file so that i can use the dataset later on. I have tried using the instructions listed in this thread but it keeps returning an error. I’ll include my code and the variables that i use below. And sorry for the rudamentary code, I’m still new to matlab
clear
clc

%masukin waktu
waktu= ncread(‘FABMSAT.nc’,’time’);
waktu= waktu’;

%masukin fito layer 1
fito= ncread(‘FABMSAT.nc’,’npzd_phy’);
fito= fito(:,1,:);

%masukin pasut
pasut= ncread(‘FABMSAT.nc’,’zeta’);

% masukin vector u
u= ncread(‘FABMSAT.nc’,’u’);
u= u(:,1,:);

%Masukin vector v
v= ncread(‘FABMSAT.nc’,’u’);
v= v(:,1,:);

%CUT SUPAYA MULAI 1 FEB 2022
waktu_feb= waktu;
waktu_feb(:,[1:168])=[];
fito_feb= fito;
fito_feb(:,[1:168])=[];
pasut_feb= pasut;
pasut_feb([2:8307],:)=[];
u_feb= u;
u_feb(:,[1:168])=[];
v_feb= v;
v_feb(:,[1:168])=[];

%define dimensions
netcdf.setDefaultFormat(‘NC_FORMAT_CLASSIC’);
ncid = netcdf.create(‘test.nc’,’NC_WRITE’);
dimidfito = netcdf.defDim(ncid,’fito_feb’,fito_feb);

%Define the name of variable
var_fito = netcdf.defVar(ncid, ‘fito_feb’,’NC_FLOAT’,[dimidfito]);

%variables into .nc file
netcdf.endDef(ncid);
netcdf.putVar(ncid,var_fito,fito_feb);
netcdf.close(ncid); export, netcdf MATLAB Answers — New Questions

​

Derivative of state ‘1’ in block ‘Fccu_matlabFcn/Integrator2’ at time 0.042441271689357414 is not finite. The simulation will be stopped. There may be a singularity in the solution. If not, try reducing the step size (either by reducing the fixed step size or by tightening the error tolerances)Derivative of state ‘1’ in block ‘Fccu_matlabFcn/Integrator2’ at time 0.042441271689357414 is not finite. The simulation will be stopped. There may be a singularity in the solution. If not, try reducing the step size (either by reducing the fixed step size or by tightening the error tolerances) Derivative of state ‘1’ in block ‘Fccu_matlabFcn/Integrator2’ at time 0.042441271689357414 is not finite. The simulation will be stopped. There may be a singularity in the solution. If not, try reducing the step size (either by reducing the fixed step size or by tightening the error tolerances) error tolerances MATLAB Answers — New Questions

​