I converted the attached coordinates (x,y) into an alphashape. See image.
shp = alphaShape(coordinates(:,1),coordinates(:,2),’HoleThreshold’,50);
What I need are the coordinates at the polygon vertices (shown in red), and importantly, in the proper order shown with the numbers. I would like to be able to use the polyshape function next…
I have toyed with boundaryfacet, delauney but with no luck.
Any suggestions?
Thanks.I converted the attached coordinates (x,y) into an alphashape. See image.
shp = alphaShape(coordinates(:,1),coordinates(:,2),’HoleThreshold’,50);
What I need are the coordinates at the polygon vertices (shown in red), and importantly, in the proper order shown with the numbers. I would like to be able to use the polyshape function next…
I have toyed with boundaryfacet, delauney but with no luck.
Any suggestions?
Thanks. I converted the attached coordinates (x,y) into an alphashape. See image.
shp = alphaShape(coordinates(:,1),coordinates(:,2),’HoleThreshold’,50);
What I need are the coordinates at the polygon vertices (shown in red), and importantly, in the proper order shown with the numbers. I would like to be able to use the polyshape function next…
I have toyed with boundaryfacet, delauney but with no luck.
Any suggestions?
Thanks. alphashape, polyshape, boundary MATLAB Answers — New Questions

​

I tried to obtain the coefficients of B-spline matrix using the following code :
function C_matrix = calculateBsplineCoefficientMatrix(s)
% Number of input samples
N = length(s);
% Construct Alpha Values (Used in the Recursive Calculation)
alpha_values = zeros(1, 10);
alpha_values(1) = -1/4;
for i = 2:10
alpha_values(i) = -1 / (4 + alpha_values(i – 1));
end
alpha = alpha_values(end);
b_i = -alpha / (1 – alpha^2);
% Construct Forward Recursion Matrix (C_plus)
C_plus = zeros(N, N); % Matrix to store forward recursion values
C_plus(:, 1) = 1; % Initialize first column as 1
for k = 2:N
C_plus(k, 🙂 = alpha * C_plus(k-1, :);
C_plus(k, k) = 1; % Set diagonal values to 1 for recursion steps
end
% Construct Backward Recursion Matrix (C_minus)
C_minus = zeros(N, N);
I_N = eye(N); % Identity matrix of size NxN
C_minus(N, 🙂 = b_i * (2 * C_plus(N, 🙂 – I_N(N, :));
for k = N-1:-1:1
C_minus(k, 🙂 = alpha * (C_minus(k+1, 🙂 – C_plus(k, :));
end
% Construct Final B-spline Coefficient Matrix
C_matrix = 6 * C_minus;
end

s here represents the signal or sequence basically this code is created from the paper FastO-line interpolation by Dooley in pdf document(1).
s = rand(1, 4);
C_spline_matrix = calculateBsplineCoefficientMatrix(s);
disp(‘B-Spline Coefficients Matrix Form:’);
disp(C_spline_matrix);

The Solution is B-Spline Coefficients Matrix Form:
B-Spline Coefficients Matrix Form:
1.7327 -0.4665 0.1333 -0.0333
-0.4665 1.7410 -0.4974 0.1244
0.1333 -0.4974 1.8564 -0.4641
-0.0666 0.2487 -0.9282 1.7321

Then I tried to calculate its quasi diagonal matrix using the following code :
M = length(s); % Polynomial order or interpolation order
% Compute Transformation Matrices
T_mu = compute_TD(M); % Transformation matrix T_mu
T_z = calculate_Tz(M); % Transformation matrix T_z
% Ensure numerical stability
T_mu_invT = inv(transpose(T_mu)); % Explicit inverse transpose of T_mu
T_z_inv = inv(T_z); % Explicit inverse of T_z
% Compute the quasi-diagonal matrix C_LCN
C_LCN = T_mu_invT .*C_spline_matrix .* T_z_inv;
% Display the final quasi-diagonal matrix
disp(‘Computed C_LCN Matrix:’);
disp(C_LCN);

The solution of quasi diagonal matrix
Computed C_LCN Matrix:
1.8384 0 0 0
-0.5937 -2.4397 0 0
0.1728 1.1772 0.8176 0
-0.0168 -0.2263 -0.3135 -0.0388

The relevant functions are given below:
function Td1 = compute_Td1(M)
% Create a matrix of binomial coefficients
[I, J] = ndgrid(1:M, 1:M);
binomials = zeros(M, M);
for ii = 1:M
for jj = 1:M
if jj <= ii
binomials(ii, jj) = nchoosek(ii-1, jj-1);
end
end
end
% Compute powers and combine with binomial coefficients
powers = ((- (M – 1) / 2) .^ (I – J)) .* (J <= I);
Td1 = binomials .* powers;
% Debugging: Display Td1 matrix
disp(‘Td1 Matrix:’);
disp(Td1);
end
function Td2 = compute_Td2(M)
% Compute the Td2 matrix
Td2 = eye(M+1);
for n = 2:M+1
for k = 2:n-1
Td2(n, k) = Td2(n-1, k-1) – (n-2) * Td2(n-1, k);
end
end
Td2 = Td2(2:M+1, 2:M+1);
% Debugging: Display Td2 matrix
disp(‘Td2 Matrix:’);
disp(Td2);
end
function TD = compute_TD(M)
Td1 = compute_Td1(M);
Td2 = compute_Td2(M);
% Compute TD
TD = Td1 * Td2;
% Force symmetry in TD
TD = (TD + TD’) / 2;
% Regularization for numerical stability
TD = TD + 1e-8 * eye(size(TD));
% Debugging: Display TD matrix
disp(‘Symmetrized TD Matrix:’);
disp(TD);
end
function Tz = calculate_Tz(M)
% Generate the Tz transformation matrix for Newton interpolation
Tz = zeros(M, M);
for i = 1:M
for j = 1:i
Tz(i, j) = nchoosek(i-1, j-1) * (-1)^(j+1);
end
end
% Regularization for numerical stability
Tz = Tz + 1e-8 * eye(size(Tz));
% Debugging: Display Tz matrix
disp(‘Tz Matrix:’);
disp(Tz);
end
The problem is the it does not match the values as solved in the paper Electronic letters…
The solutions provided are :

Please could any body help me correcting this code to obtain the solution ?I tried to obtain the coefficients of B-spline matrix using the following code :
function C_matrix = calculateBsplineCoefficientMatrix(s)
% Number of input samples
N = length(s);
% Construct Alpha Values (Used in the Recursive Calculation)
alpha_values = zeros(1, 10);
alpha_values(1) = -1/4;
for i = 2:10
alpha_values(i) = -1 / (4 + alpha_values(i – 1));
end
alpha = alpha_values(end);
b_i = -alpha / (1 – alpha^2);
% Construct Forward Recursion Matrix (C_plus)
C_plus = zeros(N, N); % Matrix to store forward recursion values
C_plus(:, 1) = 1; % Initialize first column as 1
for k = 2:N
C_plus(k, 🙂 = alpha * C_plus(k-1, :);
C_plus(k, k) = 1; % Set diagonal values to 1 for recursion steps
end
% Construct Backward Recursion Matrix (C_minus)
C_minus = zeros(N, N);
I_N = eye(N); % Identity matrix of size NxN
C_minus(N, 🙂 = b_i * (2 * C_plus(N, 🙂 – I_N(N, :));
for k = N-1:-1:1
C_minus(k, 🙂 = alpha * (C_minus(k+1, 🙂 – C_plus(k, :));
end
% Construct Final B-spline Coefficient Matrix
C_matrix = 6 * C_minus;
end

s here represents the signal or sequence basically this code is created from the paper FastO-line interpolation by Dooley in pdf document(1).
s = rand(1, 4);
C_spline_matrix = calculateBsplineCoefficientMatrix(s);
disp(‘B-Spline Coefficients Matrix Form:’);
disp(C_spline_matrix);

The Solution is B-Spline Coefficients Matrix Form:
B-Spline Coefficients Matrix Form:
1.7327 -0.4665 0.1333 -0.0333
-0.4665 1.7410 -0.4974 0.1244
0.1333 -0.4974 1.8564 -0.4641
-0.0666 0.2487 -0.9282 1.7321

Then I tried to calculate its quasi diagonal matrix using the following code :
M = length(s); % Polynomial order or interpolation order
% Compute Transformation Matrices
T_mu = compute_TD(M); % Transformation matrix T_mu
T_z = calculate_Tz(M); % Transformation matrix T_z
% Ensure numerical stability
T_mu_invT = inv(transpose(T_mu)); % Explicit inverse transpose of T_mu
T_z_inv = inv(T_z); % Explicit inverse of T_z
% Compute the quasi-diagonal matrix C_LCN
C_LCN = T_mu_invT .*C_spline_matrix .* T_z_inv;
% Display the final quasi-diagonal matrix
disp(‘Computed C_LCN Matrix:’);
disp(C_LCN);

The solution of quasi diagonal matrix
Computed C_LCN Matrix:
1.8384 0 0 0
-0.5937 -2.4397 0 0
0.1728 1.1772 0.8176 0
-0.0168 -0.2263 -0.3135 -0.0388

The relevant functions are given below:
function Td1 = compute_Td1(M)
% Create a matrix of binomial coefficients
[I, J] = ndgrid(1:M, 1:M);
binomials = zeros(M, M);
for ii = 1:M
for jj = 1:M
if jj <= ii
binomials(ii, jj) = nchoosek(ii-1, jj-1);
end
end
end
% Compute powers and combine with binomial coefficients
powers = ((- (M – 1) / 2) .^ (I – J)) .* (J <= I);
Td1 = binomials .* powers;
% Debugging: Display Td1 matrix
disp(‘Td1 Matrix:’);
disp(Td1);
end
function Td2 = compute_Td2(M)
% Compute the Td2 matrix
Td2 = eye(M+1);
for n = 2:M+1
for k = 2:n-1
Td2(n, k) = Td2(n-1, k-1) – (n-2) * Td2(n-1, k);
end
end
Td2 = Td2(2:M+1, 2:M+1);
% Debugging: Display Td2 matrix
disp(‘Td2 Matrix:’);
disp(Td2);
end
function TD = compute_TD(M)
Td1 = compute_Td1(M);
Td2 = compute_Td2(M);
% Compute TD
TD = Td1 * Td2;
% Force symmetry in TD
TD = (TD + TD’) / 2;
% Regularization for numerical stability
TD = TD + 1e-8 * eye(size(TD));
% Debugging: Display TD matrix
disp(‘Symmetrized TD Matrix:’);
disp(TD);
end
function Tz = calculate_Tz(M)
% Generate the Tz transformation matrix for Newton interpolation
Tz = zeros(M, M);
for i = 1:M
for j = 1:i
Tz(i, j) = nchoosek(i-1, j-1) * (-1)^(j+1);
end
end
% Regularization for numerical stability
Tz = Tz + 1e-8 * eye(size(Tz));
% Debugging: Display Tz matrix
disp(‘Tz Matrix:’);
disp(Tz);
end
The problem is the it does not match the values as solved in the paper Electronic letters…
The solutions provided are :

Please could any body help me correcting this code to obtain the solution ? I tried to obtain the coefficients of B-spline matrix using the following code :
function C_matrix = calculateBsplineCoefficientMatrix(s)
% Number of input samples
N = length(s);
% Construct Alpha Values (Used in the Recursive Calculation)
alpha_values = zeros(1, 10);
alpha_values(1) = -1/4;
for i = 2:10
alpha_values(i) = -1 / (4 + alpha_values(i – 1));
end
alpha = alpha_values(end);
b_i = -alpha / (1 – alpha^2);
% Construct Forward Recursion Matrix (C_plus)
C_plus = zeros(N, N); % Matrix to store forward recursion values
C_plus(:, 1) = 1; % Initialize first column as 1
for k = 2:N
C_plus(k, 🙂 = alpha * C_plus(k-1, :);
C_plus(k, k) = 1; % Set diagonal values to 1 for recursion steps
end
% Construct Backward Recursion Matrix (C_minus)
C_minus = zeros(N, N);
I_N = eye(N); % Identity matrix of size NxN
C_minus(N, 🙂 = b_i * (2 * C_plus(N, 🙂 – I_N(N, :));
for k = N-1:-1:1
C_minus(k, 🙂 = alpha * (C_minus(k+1, 🙂 – C_plus(k, :));
end
% Construct Final B-spline Coefficient Matrix
C_matrix = 6 * C_minus;
end

s here represents the signal or sequence basically this code is created from the paper FastO-line interpolation by Dooley in pdf document(1).
s = rand(1, 4);
C_spline_matrix = calculateBsplineCoefficientMatrix(s);
disp(‘B-Spline Coefficients Matrix Form:’);
disp(C_spline_matrix);

The Solution is B-Spline Coefficients Matrix Form:
B-Spline Coefficients Matrix Form:
1.7327 -0.4665 0.1333 -0.0333
-0.4665 1.7410 -0.4974 0.1244
0.1333 -0.4974 1.8564 -0.4641
-0.0666 0.2487 -0.9282 1.7321

Then I tried to calculate its quasi diagonal matrix using the following code :
M = length(s); % Polynomial order or interpolation order
% Compute Transformation Matrices
T_mu = compute_TD(M); % Transformation matrix T_mu
T_z = calculate_Tz(M); % Transformation matrix T_z
% Ensure numerical stability
T_mu_invT = inv(transpose(T_mu)); % Explicit inverse transpose of T_mu
T_z_inv = inv(T_z); % Explicit inverse of T_z
% Compute the quasi-diagonal matrix C_LCN
C_LCN = T_mu_invT .*C_spline_matrix .* T_z_inv;
% Display the final quasi-diagonal matrix
disp(‘Computed C_LCN Matrix:’);
disp(C_LCN);

The solution of quasi diagonal matrix
Computed C_LCN Matrix:
1.8384 0 0 0
-0.5937 -2.4397 0 0
0.1728 1.1772 0.8176 0
-0.0168 -0.2263 -0.3135 -0.0388

The relevant functions are given below:
function Td1 = compute_Td1(M)
% Create a matrix of binomial coefficients
[I, J] = ndgrid(1:M, 1:M);
binomials = zeros(M, M);
for ii = 1:M
for jj = 1:M
if jj <= ii
binomials(ii, jj) = nchoosek(ii-1, jj-1);
end
end
end
% Compute powers and combine with binomial coefficients
powers = ((- (M – 1) / 2) .^ (I – J)) .* (J <= I);
Td1 = binomials .* powers;
% Debugging: Display Td1 matrix
disp(‘Td1 Matrix:’);
disp(Td1);
end
function Td2 = compute_Td2(M)
% Compute the Td2 matrix
Td2 = eye(M+1);
for n = 2:M+1
for k = 2:n-1
Td2(n, k) = Td2(n-1, k-1) – (n-2) * Td2(n-1, k);
end
end
Td2 = Td2(2:M+1, 2:M+1);
% Debugging: Display Td2 matrix
disp(‘Td2 Matrix:’);
disp(Td2);
end
function TD = compute_TD(M)
Td1 = compute_Td1(M);
Td2 = compute_Td2(M);
% Compute TD
TD = Td1 * Td2;
% Force symmetry in TD
TD = (TD + TD’) / 2;
% Regularization for numerical stability
TD = TD + 1e-8 * eye(size(TD));
% Debugging: Display TD matrix
disp(‘Symmetrized TD Matrix:’);
disp(TD);
end
function Tz = calculate_Tz(M)
% Generate the Tz transformation matrix for Newton interpolation
Tz = zeros(M, M);
for i = 1:M
for j = 1:i
Tz(i, j) = nchoosek(i-1, j-1) * (-1)^(j+1);
end
end
% Regularization for numerical stability
Tz = Tz + 1e-8 * eye(size(Tz));
% Debugging: Display Tz matrix
disp(‘Tz Matrix:’);
disp(Tz);
end
The problem is the it does not match the values as solved in the paper Electronic letters…
The solutions provided are :

Please could any body help me correcting this code to obtain the solution ? mathematics, digital image processing, digital signal processing, linear algebra MATLAB Answers — New Questions

​

I am an engineer by discipline, so I have a basic understanding of sine waves, but I am unfamiliar with signal processing techniques. I have two sets of noisy, irregularly sampled sinusoids and I have been tasked quantify how similar they are. The code below generates arrays that are similar to the data I am looking at. My data, however, was measured, so I have no prior knowledge of the amplitudes/frequencies of the sinusoids, only 3 arrays: one containing the discrete time points, and the other two containig the discrete values for signal x and signal y.
t = sort(10*rand(10000,1)); % Generate uneven sample times
x = 2*sin(pi*t) + 3*cos(2*pi*t) + rand(size(t)); % Generate signal 1
y = 2.1*sin(pi*t+0.3) + 3.1*cos(2*pi*t+0.3) + rand(size(t)); % Generate signal 2
I’d greatly appreciate any guidance/references to how to accomplish the task. Thank you for your time.I am an engineer by discipline, so I have a basic understanding of sine waves, but I am unfamiliar with signal processing techniques. I have two sets of noisy, irregularly sampled sinusoids and I have been tasked quantify how similar they are. The code below generates arrays that are similar to the data I am looking at. My data, however, was measured, so I have no prior knowledge of the amplitudes/frequencies of the sinusoids, only 3 arrays: one containing the discrete time points, and the other two containig the discrete values for signal x and signal y.
t = sort(10*rand(10000,1)); % Generate uneven sample times
x = 2*sin(pi*t) + 3*cos(2*pi*t) + rand(size(t)); % Generate signal 1
y = 2.1*sin(pi*t+0.3) + 3.1*cos(2*pi*t+0.3) + rand(size(t)); % Generate signal 2
I’d greatly appreciate any guidance/references to how to accomplish the task. Thank you for your time. I am an engineer by discipline, so I have a basic understanding of sine waves, but I am unfamiliar with signal processing techniques. I have two sets of noisy, irregularly sampled sinusoids and I have been tasked quantify how similar they are. The code below generates arrays that are similar to the data I am looking at. My data, however, was measured, so I have no prior knowledge of the amplitudes/frequencies of the sinusoids, only 3 arrays: one containing the discrete time points, and the other two containig the discrete values for signal x and signal y.
t = sort(10*rand(10000,1)); % Generate uneven sample times
x = 2*sin(pi*t) + 3*cos(2*pi*t) + rand(size(t)); % Generate signal 1
y = 2.1*sin(pi*t+0.3) + 3.1*cos(2*pi*t+0.3) + rand(size(t)); % Generate signal 2
I’d greatly appreciate any guidance/references to how to accomplish the task. Thank you for your time. fft, plomb, lomb, scargle, signal processing, discrete, discrete signal processing, dsp, fourier, irregular, noisy, irregularly sampled, noisy data, noisy sinusoid MATLAB Answers — New Questions

​

Hello, I want to model a desorption reactor with a fixed bed containing gold-laden carbon, through which a desorbing solution passes, which extracts extra gold. I am trying to solve it using a PDE system in which I create multiple nodes, assimilating it as if it were solved using the finite difference method. Using a forward difference for the initial node, a central difference for the intermediate nodes, and a backward difference for the final node. These are relative to a distance differential.The equations are as follows.

So i tried finite differences for dc/dz with forward difference for eactor entry , central along the reactor , and backward in the exit, and dc/dt and dq/dt use integrator blocks, I consider Co= 0 kg/m3 solution and q0=4.320 kg/m3 carbon. Just considering 5 nodes , shall be more but first i want to make the first five work fine .

And each node consist on the following layout , where can be seen a time integral block term for q(carbon loading) and c(solution loading , it shows also a length step and inputs from the forwarded and current node soluction concentration(in the case of the first node)

My problem is that I am getting the same values on each node, which I don’t know if it is right the layout approach, since they should be different with relation to time and besides when i increase or decrease the input stream speed , the values in carbon and solution loading not change at all. thanks in advance

The model file is attached next: Reactor processHello, I want to model a desorption reactor with a fixed bed containing gold-laden carbon, through which a desorbing solution passes, which extracts extra gold. I am trying to solve it using a PDE system in which I create multiple nodes, assimilating it as if it were solved using the finite difference method. Using a forward difference for the initial node, a central difference for the intermediate nodes, and a backward difference for the final node. These are relative to a distance differential.The equations are as follows.

So i tried finite differences for dc/dz with forward difference for eactor entry , central along the reactor , and backward in the exit, and dc/dt and dq/dt use integrator blocks, I consider Co= 0 kg/m3 solution and q0=4.320 kg/m3 carbon. Just considering 5 nodes , shall be more but first i want to make the first five work fine .

And each node consist on the following layout , where can be seen a time integral block term for q(carbon loading) and c(solution loading , it shows also a length step and inputs from the forwarded and current node soluction concentration(in the case of the first node)

My problem is that I am getting the same values on each node, which I don’t know if it is right the layout approach, since they should be different with relation to time and besides when i increase or decrease the input stream speed , the values in carbon and solution loading not change at all. thanks in advance

The model file is attached next: Reactor process Hello, I want to model a desorption reactor with a fixed bed containing gold-laden carbon, through which a desorbing solution passes, which extracts extra gold. I am trying to solve it using a PDE system in which I create multiple nodes, assimilating it as if it were solved using the finite difference method. Using a forward difference for the initial node, a central difference for the intermediate nodes, and a backward difference for the final node. These are relative to a distance differential.The equations are as follows.

So i tried finite differences for dc/dz with forward difference for eactor entry , central along the reactor , and backward in the exit, and dc/dt and dq/dt use integrator blocks, I consider Co= 0 kg/m3 solution and q0=4.320 kg/m3 carbon. Just considering 5 nodes , shall be more but first i want to make the first five work fine .

And each node consist on the following layout , where can be seen a time integral block term for q(carbon loading) and c(solution loading , it shows also a length step and inputs from the forwarded and current node soluction concentration(in the case of the first node)

My problem is that I am getting the same values on each node, which I don’t know if it is right the layout approach, since they should be different with relation to time and besides when i increase or decrease the input stream speed , the values in carbon and solution loading not change at all. thanks in advance

The model file is attached next: Reactor process simulink MATLAB Answers — New Questions

​

I am trying to the the hue, saturation and value images from an image (originally colored). I converted the image using rgb2hsv() function but I am confused about how to get the 3 images from there.

Thanks in advance!I am trying to the the hue, saturation and value images from an image (originally colored). I converted the image using rgb2hsv() function but I am confused about how to get the 3 images from there.

Thanks in advance! I am trying to the the hue, saturation and value images from an image (originally colored). I converted the image using rgb2hsv() function but I am confused about how to get the 3 images from there.

Thanks in advance! image processing, hsv, image components MATLAB Answers — New Questions

​

Hi,
I am using the diary function to log my application output. Something like

diary(path_to_logfile)
% all my code with output to the console, matrices written to desk, etc
diary off

When the code fails, the diary is not closed. One option would be to wrap all the code on a try-catch statement and close it on the catch.

Is there another way to avoid the diary being "open" when the app fails?Hi,
I am using the diary function to log my application output. Something like

diary(path_to_logfile)
% all my code with output to the console, matrices written to desk, etc
diary off

When the code fails, the diary is not closed. One option would be to wrap all the code on a try-catch statement and close it on the catch.

Is there another way to avoid the diary being "open" when the app fails? Hi,
I am using the diary function to log my application output. Something like

diary(path_to_logfile)
% all my code with output to the console, matrices written to desk, etc
diary off

When the code fails, the diary is not closed. One option would be to wrap all the code on a try-catch statement and close it on the catch.

Is there another way to avoid the diary being "open" when the app fails? logging, diary, try-catch MATLAB Answers — New Questions

​

I do a lot long running scripts that plot to multiple figure. While these scripts are running, I am unable to use my computer because Matlab steals focus so I can’t do anything else while the script is running. It there a way to keep the plot function from stealing focus?I do a lot long running scripts that plot to multiple figure. While these scripts are running, I am unable to use my computer because Matlab steals focus so I can’t do anything else while the script is running. It there a way to keep the plot function from stealing focus? I do a lot long running scripts that plot to multiple figure. While these scripts are running, I am unable to use my computer because Matlab steals focus so I can’t do anything else while the script is running. It there a way to keep the plot function from stealing focus? focus, stealing MATLAB Answers — New Questions

​

Hi team
I am using EODHD.COM for stock data. I get the following error every now and then. example – if I am iterating over say 50 stocks data, the code sometimes throw an error for 1 stock, then I rerun the full script allover again and it will work. I used webread to start with but webread doesn’t even read the webpage and returns no data (urlread returns data instead).
Error using urlreadwrite_legacy (line 88)
The IP address of "eodhd.com" could not be determined.
Error in urlreadwrite (line 88)
[output,status] = urlreadwrite_legacy(fcn,catchErrors,varargin{:});
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Error in urlread (line 16)
[s,status] = urlreadwrite(mfilename,catchErrors,url,varargin{:});
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Error in NSE_all (line 78)
EOD_data = urlread(URL2); % json data
^^^^^^^^^^^^^

My codes are:
url2A = ‘https://eodhd.com/api/eod/’;
url2B = ‘?from=’;
url2C = ‘&period=d&api_token=’;
url2D = ‘&fmt=json’;
From = datetime("today") – (365 * 20);
dt_str = char(From);
URL2 = url2A + company + url2B + dt_str + url2C + token + url2D;
EOD_data = urlread(URL2);
*please note company is a name of the stock ticker I take from the list for the For loop so I iterate through multiple companies stocks.
Is it possible to tell urlread to wait longer and try more before throwing an error? or even keep trying for the specified time? I want urlread to not stop my script at any random time. I am new to Matlab so if anyone can help it would be great.Hi team
I am using EODHD.COM for stock data. I get the following error every now and then. example – if I am iterating over say 50 stocks data, the code sometimes throw an error for 1 stock, then I rerun the full script allover again and it will work. I used webread to start with but webread doesn’t even read the webpage and returns no data (urlread returns data instead).
Error using urlreadwrite_legacy (line 88)
The IP address of "eodhd.com" could not be determined.
Error in urlreadwrite (line 88)
[output,status] = urlreadwrite_legacy(fcn,catchErrors,varargin{:});
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Error in urlread (line 16)
[s,status] = urlreadwrite(mfilename,catchErrors,url,varargin{:});
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Error in NSE_all (line 78)
EOD_data = urlread(URL2); % json data
^^^^^^^^^^^^^

My codes are:
url2A = ‘https://eodhd.com/api/eod/’;
url2B = ‘?from=’;
url2C = ‘&period=d&api_token=’;
url2D = ‘&fmt=json’;
From = datetime("today") – (365 * 20);
dt_str = char(From);
URL2 = url2A + company + url2B + dt_str + url2C + token + url2D;
EOD_data = urlread(URL2);
*please note company is a name of the stock ticker I take from the list for the For loop so I iterate through multiple companies stocks.
Is it possible to tell urlread to wait longer and try more before throwing an error? or even keep trying for the specified time? I want urlread to not stop my script at any random time. I am new to Matlab so if anyone can help it would be great. Hi team
I am using EODHD.COM for stock data. I get the following error every now and then. example – if I am iterating over say 50 stocks data, the code sometimes throw an error for 1 stock, then I rerun the full script allover again and it will work. I used webread to start with but webread doesn’t even read the webpage and returns no data (urlread returns data instead).
Error using urlreadwrite_legacy (line 88)
The IP address of "eodhd.com" could not be determined.
Error in urlreadwrite (line 88)
[output,status] = urlreadwrite_legacy(fcn,catchErrors,varargin{:});
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Error in urlread (line 16)
[s,status] = urlreadwrite(mfilename,catchErrors,url,varargin{:});
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Error in NSE_all (line 78)
EOD_data = urlread(URL2); % json data
^^^^^^^^^^^^^

My codes are:
url2A = ‘https://eodhd.com/api/eod/’;
url2B = ‘?from=’;
url2C = ‘&period=d&api_token=’;
url2D = ‘&fmt=json’;
From = datetime("today") – (365 * 20);
dt_str = char(From);
URL2 = url2A + company + url2B + dt_str + url2C + token + url2D;
EOD_data = urlread(URL2);
*please note company is a name of the stock ticker I take from the list for the For loop so I iterate through multiple companies stocks.
Is it possible to tell urlread to wait longer and try more before throwing an error? or even keep trying for the specified time? I want urlread to not stop my script at any random time. I am new to Matlab so if anyone can help it would be great. urlread ip_address, webread MATLAB Answers — New Questions

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Hi,
I tried to train SVM on cell array which each element is 20×120 double, How to train SVM fitcecoc on cell which each cell element contains 20×120 array features?Hi,
I tried to train SVM on cell array which each element is 20×120 double, How to train SVM fitcecoc on cell which each cell element contains 20×120 array features? Hi,
I tried to train SVM on cell array which each element is 20×120 double, How to train SVM fitcecoc on cell which each cell element contains 20×120 array features? svm, machine learning MATLAB Answers — New Questions

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