Category: News
Experimental Design in Matlab
Can anyone help me with run a complete CCD or BBD experimental design in matlab? Actually I use RStudio for run my DoE but i would like to learn how to perform these in Matlab.
I found this page https://it.mathworks.com/help/stats/design-of-experiments-1.html that give me some information but this is not enough…
I would like to know also the script for give the experimental responses, extimate coefficient, perform the ANOVA and response surface and so on..
Thank you so much!Can anyone help me with run a complete CCD or BBD experimental design in matlab? Actually I use RStudio for run my DoE but i would like to learn how to perform these in Matlab.
I found this page https://it.mathworks.com/help/stats/design-of-experiments-1.html that give me some information but this is not enough…
I would like to know also the script for give the experimental responses, extimate coefficient, perform the ANOVA and response surface and so on..
Thank you so much! Can anyone help me with run a complete CCD or BBD experimental design in matlab? Actually I use RStudio for run my DoE but i would like to learn how to perform these in Matlab.
I found this page https://it.mathworks.com/help/stats/design-of-experiments-1.html that give me some information but this is not enough…
I would like to know also the script for give the experimental responses, extimate coefficient, perform the ANOVA and response surface and so on..
Thank you so much! doe, ccd, bbd MATLAB Answers — New Questions
Solving ODE just for one time step
I want to solve some ODE using matlab solver (ode45, ode15s, etc) just for one time step. The time step is decided internally by the solver. Is this possible to do?I want to solve some ODE using matlab solver (ode45, ode15s, etc) just for one time step. The time step is decided internally by the solver. Is this possible to do? I want to solve some ODE using matlab solver (ode45, ode15s, etc) just for one time step. The time step is decided internally by the solver. Is this possible to do? ode, solver MATLAB Answers — New Questions
How do I put more than one patternCustom plot into a figure when it seems like neither subplot or tiledlayout will work?
I am using multiple patternCustom plots that use CoordinateSystem="polar", Slice="theta" that I would like to show in the same figure in a 2×3 format. It seems that neither tiledlayout or subplots will work. I would think there has to be a straightforward way to do this, that someone can show me.I am using multiple patternCustom plots that use CoordinateSystem="polar", Slice="theta" that I would like to show in the same figure in a 2×3 format. It seems that neither tiledlayout or subplots will work. I would think there has to be a straightforward way to do this, that someone can show me. I am using multiple patternCustom plots that use CoordinateSystem="polar", Slice="theta" that I would like to show in the same figure in a 2×3 format. It seems that neither tiledlayout or subplots will work. I would think there has to be a straightforward way to do this, that someone can show me. patterncustom, tiledlayout, subplot, figure, multiple plots MATLAB Answers — New Questions
Intersection points of multiple straight line segments
This feels as though it should be easy, but other than creating loops which for large data sets run for ever I can’t see how to do this
I have mulitple line segments with end points in the form [x11 y11 x12 y12; x21 y21 x22 y22;…] [x13 y13 x14 y14; x23 y23 x24 y24;…]
I am trying to find intersection points between any pair of line segments in the two lists
I know how to find where any specific two intersect, using xy = [x1*y2-x2*y1,x3*y4-x4*y3]/[y2-y1,y4-y3;-(x2-x1),-(x4-x3)]; and looping through each list in turn
but can’t see an efficient way of testing all possible pairs
Any help gratefully receivedThis feels as though it should be easy, but other than creating loops which for large data sets run for ever I can’t see how to do this
I have mulitple line segments with end points in the form [x11 y11 x12 y12; x21 y21 x22 y22;…] [x13 y13 x14 y14; x23 y23 x24 y24;…]
I am trying to find intersection points between any pair of line segments in the two lists
I know how to find where any specific two intersect, using xy = [x1*y2-x2*y1,x3*y4-x4*y3]/[y2-y1,y4-y3;-(x2-x1),-(x4-x3)]; and looping through each list in turn
but can’t see an efficient way of testing all possible pairs
Any help gratefully received This feels as though it should be easy, but other than creating loops which for large data sets run for ever I can’t see how to do this
I have mulitple line segments with end points in the form [x11 y11 x12 y12; x21 y21 x22 y22;…] [x13 y13 x14 y14; x23 y23 x24 y24;…]
I am trying to find intersection points between any pair of line segments in the two lists
I know how to find where any specific two intersect, using xy = [x1*y2-x2*y1,x3*y4-x4*y3]/[y2-y1,y4-y3;-(x2-x1),-(x4-x3)]; and looping through each list in turn
but can’t see an efficient way of testing all possible pairs
Any help gratefully received intersects MATLAB Answers — New Questions
How Microsoft 365 Copilot Tenants Benefit from SharePoint Advanced Management
Ignite Announcement About SAM for Copilot Customers Misinterpreted by Many
At the Ignite 2024 conference, Microsoft announced that “Microsoft 365 Copilot will now include built-in content governance controls and insights provided by SharePoint Advanced Management.” At the time, and still broadly believed, the assumption was that Microsoft would provide customers with Microsoft 365 Copilot licenses with SharePoint Advanced Management (SAM) licenses. Maybe even a single SAM license would be sufficient to license SAM technology alongside Copilot. That’s not the case.
If you’ve been waiting for a SAM license to appear in your tenant, you’ll be disappointed and won’t see SAM listed in the set of tenant subscriptions. Don’t be swayed by the banner in the SharePoint Online admin center to announce that your SharePoint Advanced Management subscription is enabled (Figure 1). It’s not. Access to SAM features is granted through a check enabled in code for the presence of Copilot. The necessary update is now broadly available to customers.
SAM Features for Microsoft 365 Copilot Customers
The facts are laid out in the SAM documentation. Customers with eligible Copilot licenses can use some, but not all, SAM functionality without a SAM license. Here’s the list:
- Site Lifecycle Policy
- Inactive SharePoint sites policy
- Site Ownership Policy
- Data Access Governance (DAG) Insights
- “Everyone Except External Users” (EEEU) insights
- Sharing Links and Sensitivity Labels
- PowerShell: Permission state report for SharePoint and OneDrive Sites, and Files
- Sharing links report
- Site Access Review
- Restricted Content Discovery (RCD – enabled via PowerShell)
- Restricted Access Control (RAC) for SharePoint and OneDrive for Business.
- Recent Admin Actions and Change History
- Block Download Policy
- SharePoint and OneDrive sites
- Teams recordings
There’s some good stuff here, particularly Restricted Content Discovery (RCD), the Site Lifecycle Policy to manage inactive sites, and the Block download policy. Every tenant with Microsoft 365 Copilot should consider enabling RCD to block Copilot access to sites containing sensitive Office and PDF files and sites containing old and obsolete material (the digital rot or debris that clutters up so many tenants).
The problem with Copilot reusing sensitive material in its responses is obvious. The issue with Copilot reusing old, obsolete, and potentially misleading content in its responses is equally problematic, especially if human checks don’t catch errors in responses. Copilot doesn’t know when a Word document written ten years ago is outdated and inaccurate. All Copilot sees is words that can be processed and reused.
When SAM is Needed
All of which brings me to a point where a SAM license is required. In my case, I wanted to test the extend SharePoint protections with a default sensitivity label feature. The idea here is to make sure that unlabeled files receive protection when downloaded by applying a sensitivity label with equivalent rights to those enjoyed by site users. Defining a default sensitivity label for a document library already requires an Office 365 E5 license or equivalent. Why this slight extension wanders into the need to have SAM is another example of bizarre Microsoft licensing.
The documentation notes that Copilot can’t currently open files with sensitivity labels applied in this manner. This means that Copilot cannot extract the protected content to use in its responses because it doesn’t have the right to do so. However, Copilot can search the metadata of labeled files and show that metadata to those who perform searches. Restricted Content Discovery is the right way to block Copilot access to files.
Anyway, without a SAM license, I can’t test. Do I want to pay Microsoft for a license for the privilege of testing their software? I don’t think so.
Copilot in Word for iOS
In closing, I attempted to use a new feature in Word for iOS (and Android) to dictate some notes for this article for Copilot to reason over and produce a draft. The feature is covered in MC1060866 (23 April 2025) and deployment has begun, which is why I guess I could use it. The dictation part worked, even if some of my words were misunderstood (Figure 2). But any attempt to have Copilot do some magic failed utterly. I guess that AI can’t help me…
Support the work of the Office 365 for IT Pros team by subscribing to the Office 365 for IT Pros eBook. Your support pays for the time we need to track, analyze, and document the changing world of Microsoft 365 and Office 365.
I haven’t defined syms x but solve function gives me values depend on x
I am designing a PID controller for which I set the denominator of my closed-loop transfer function equal to the product of the dominant poles and the residue polynomial.
My goal is to get values that depend on Ki, but even though I don’t use the syms x command, my values depend on x and Ki.
clear all;
clc;
syms zeta positive
syms s;
syms Kp Ki Kd;
% Given values
ts = 4; % Settling time
Mp = 0.05; % Maximum overshoot
% Define the overshoot equation
eq1 = exp(-zeta*pi / sqrt(1 – zeta^2)) == Mp;
% Solve for zeta numerically (only real positive solution makes sense)
zeta = double(vpasolve(eq1, zeta, [0 1]));
% Now calculate omega_n using Ts = 4 / (zeta * omega_n)
w_n = 4 / (zeta * ts);
G= (s+1)/(s^3 +6*s^2 +10*s -15)
H= 10/(s+10)
F= (Kd*s^2 + Kp*s + Ki)/s
Tcl= (F*G)/(1 + F*G*H)
[num, den] = numden(simplifyFraction(Tcl));
num = expand(num)
den = expand(den)
num_s = coeffs(num, s, ‘All’)
den_s = coeffs(den, s, ‘All’)
Pd=s^2 +2*zeta*w_n*s+(w_n)^2
syms e0 e1 e2 positive;
Pr= (s+e0)*(s+e1)*(s+e2)
Pe=expand(Pd*Pr)
Pe_s= coeffs(Pe,s,"All")
eq2 = Pe_s == den_s
sol2 = solve(eq2, [Kp, Kd, e0, e1, e2], ‘Real’, true);
% den_s=subs(den_s,[Kd,Kp],[sol2.Kd,sol2.Kp])
% F = subs(F,[Kd,Kp],[sol2.Kd,sol2.Kp])I am designing a PID controller for which I set the denominator of my closed-loop transfer function equal to the product of the dominant poles and the residue polynomial.
My goal is to get values that depend on Ki, but even though I don’t use the syms x command, my values depend on x and Ki.
clear all;
clc;
syms zeta positive
syms s;
syms Kp Ki Kd;
% Given values
ts = 4; % Settling time
Mp = 0.05; % Maximum overshoot
% Define the overshoot equation
eq1 = exp(-zeta*pi / sqrt(1 – zeta^2)) == Mp;
% Solve for zeta numerically (only real positive solution makes sense)
zeta = double(vpasolve(eq1, zeta, [0 1]));
% Now calculate omega_n using Ts = 4 / (zeta * omega_n)
w_n = 4 / (zeta * ts);
G= (s+1)/(s^3 +6*s^2 +10*s -15)
H= 10/(s+10)
F= (Kd*s^2 + Kp*s + Ki)/s
Tcl= (F*G)/(1 + F*G*H)
[num, den] = numden(simplifyFraction(Tcl));
num = expand(num)
den = expand(den)
num_s = coeffs(num, s, ‘All’)
den_s = coeffs(den, s, ‘All’)
Pd=s^2 +2*zeta*w_n*s+(w_n)^2
syms e0 e1 e2 positive;
Pr= (s+e0)*(s+e1)*(s+e2)
Pe=expand(Pd*Pr)
Pe_s= coeffs(Pe,s,"All")
eq2 = Pe_s == den_s
sol2 = solve(eq2, [Kp, Kd, e0, e1, e2], ‘Real’, true);
% den_s=subs(den_s,[Kd,Kp],[sol2.Kd,sol2.Kp])
% F = subs(F,[Kd,Kp],[sol2.Kd,sol2.Kp]) I am designing a PID controller for which I set the denominator of my closed-loop transfer function equal to the product of the dominant poles and the residue polynomial.
My goal is to get values that depend on Ki, but even though I don’t use the syms x command, my values depend on x and Ki.
clear all;
clc;
syms zeta positive
syms s;
syms Kp Ki Kd;
% Given values
ts = 4; % Settling time
Mp = 0.05; % Maximum overshoot
% Define the overshoot equation
eq1 = exp(-zeta*pi / sqrt(1 – zeta^2)) == Mp;
% Solve for zeta numerically (only real positive solution makes sense)
zeta = double(vpasolve(eq1, zeta, [0 1]));
% Now calculate omega_n using Ts = 4 / (zeta * omega_n)
w_n = 4 / (zeta * ts);
G= (s+1)/(s^3 +6*s^2 +10*s -15)
H= 10/(s+10)
F= (Kd*s^2 + Kp*s + Ki)/s
Tcl= (F*G)/(1 + F*G*H)
[num, den] = numden(simplifyFraction(Tcl));
num = expand(num)
den = expand(den)
num_s = coeffs(num, s, ‘All’)
den_s = coeffs(den, s, ‘All’)
Pd=s^2 +2*zeta*w_n*s+(w_n)^2
syms e0 e1 e2 positive;
Pr= (s+e0)*(s+e1)*(s+e2)
Pe=expand(Pd*Pr)
Pe_s= coeffs(Pe,s,"All")
eq2 = Pe_s == den_s
sol2 = solve(eq2, [Kp, Kd, e0, e1, e2], ‘Real’, true);
% den_s=subs(den_s,[Kd,Kp],[sol2.Kd,sol2.Kp])
% F = subs(F,[Kd,Kp],[sol2.Kd,sol2.Kp]) solve MATLAB Answers — New Questions
Classification problem parsed as regression problem when Split Criterion is supplied to fitcensemble
Hi
I ran a hyperparameter optimization to find the best parameters for a two-class classification problem using _fitcensemble_. But when I try to use these I get a strange warning:
*Warning: You must pass ‘SplitCriterion’ as a character vector ‘mse’ for regression.*
What is wrong with my code? The warning comes when I use a boosting ensemble as ‘method’. When I remove the ‘SplitCriterion’ everything works fine, but I cannot understand why Matlab somewhere on the line thinks this is a regression problem when I use fit"c"ensemble.
Here is a toy example with arbitrarily chosen settings that you can run to reproduce the Warning/Error.
load carsmall
X = table(Acceleration,Cylinders,Displacement,Horsepower,Mfg,Model_Year,Weight,MPG);
X.Cylinders(X.Cylinders < 8) = 0; % Create two classes in the Cylinders variable
t = templateTree( ‘MaxNumSplits’, 30,…
‘MinLeafSize’, 10,…
‘SplitCriterion’, ‘gdi’);
classificationEnsemble = fitcensemble(X,’Cylinders’,…
‘Method’, ‘LogitBoost’, …
‘NumLearningCycles’,12,…
‘Learners’,t,…
‘KFold’,7,…
‘LearnRate’,0.1);Hi
I ran a hyperparameter optimization to find the best parameters for a two-class classification problem using _fitcensemble_. But when I try to use these I get a strange warning:
*Warning: You must pass ‘SplitCriterion’ as a character vector ‘mse’ for regression.*
What is wrong with my code? The warning comes when I use a boosting ensemble as ‘method’. When I remove the ‘SplitCriterion’ everything works fine, but I cannot understand why Matlab somewhere on the line thinks this is a regression problem when I use fit"c"ensemble.
Here is a toy example with arbitrarily chosen settings that you can run to reproduce the Warning/Error.
load carsmall
X = table(Acceleration,Cylinders,Displacement,Horsepower,Mfg,Model_Year,Weight,MPG);
X.Cylinders(X.Cylinders < 8) = 0; % Create two classes in the Cylinders variable
t = templateTree( ‘MaxNumSplits’, 30,…
‘MinLeafSize’, 10,…
‘SplitCriterion’, ‘gdi’);
classificationEnsemble = fitcensemble(X,’Cylinders’,…
‘Method’, ‘LogitBoost’, …
‘NumLearningCycles’,12,…
‘Learners’,t,…
‘KFold’,7,…
‘LearnRate’,0.1); Hi
I ran a hyperparameter optimization to find the best parameters for a two-class classification problem using _fitcensemble_. But when I try to use these I get a strange warning:
*Warning: You must pass ‘SplitCriterion’ as a character vector ‘mse’ for regression.*
What is wrong with my code? The warning comes when I use a boosting ensemble as ‘method’. When I remove the ‘SplitCriterion’ everything works fine, but I cannot understand why Matlab somewhere on the line thinks this is a regression problem when I use fit"c"ensemble.
Here is a toy example with arbitrarily chosen settings that you can run to reproduce the Warning/Error.
load carsmall
X = table(Acceleration,Cylinders,Displacement,Horsepower,Mfg,Model_Year,Weight,MPG);
X.Cylinders(X.Cylinders < 8) = 0; % Create two classes in the Cylinders variable
t = templateTree( ‘MaxNumSplits’, 30,…
‘MinLeafSize’, 10,…
‘SplitCriterion’, ‘gdi’);
classificationEnsemble = fitcensemble(X,’Cylinders’,…
‘Method’, ‘LogitBoost’, …
‘NumLearningCycles’,12,…
‘Learners’,t,…
‘KFold’,7,…
‘LearnRate’,0.1); fitcensemble, split criterion, classification, regression, hyperparameter, optimization, boost, templatetree MATLAB Answers — New Questions
How do I make a 10×10 showing all numbers 1 to 100?
how do i create a nested loop tht will provide me with a 10×10 matrix with all the numbers 1 to 100 like this
1 2 … 10
11 … … …
… … … …
… … … 100how do i create a nested loop tht will provide me with a 10×10 matrix with all the numbers 1 to 100 like this
1 2 … 10
11 … … …
… … … …
… … … 100 how do i create a nested loop tht will provide me with a 10×10 matrix with all the numbers 1 to 100 like this
1 2 … 10
11 … … …
… … … …
… … … 100 matrix loop nested MATLAB Answers — New Questions
Is it possible to get the version number of a compiled program inside the program? I am using the Application Compiler.
For a project I am working on, I would like the program to output the current version being used in a char array. I have the following code:
if(isdeployed)
version = (Help here);
version_message = [‘Now running program.exe version: ‘ version];
else
version = ‘script’;
version_message = [‘Now running program.m version: ‘ version];
end
disp(version_message);
Is it possible to dynamically get the version number, or do I have to manually set it to ensure they are the same? I know it can be done with App Designer, but this executable does not use App Designer.For a project I am working on, I would like the program to output the current version being used in a char array. I have the following code:
if(isdeployed)
version = (Help here);
version_message = [‘Now running program.exe version: ‘ version];
else
version = ‘script’;
version_message = [‘Now running program.m version: ‘ version];
end
disp(version_message);
Is it possible to dynamically get the version number, or do I have to manually set it to ensure they are the same? I know it can be done with App Designer, but this executable does not use App Designer. For a project I am working on, I would like the program to output the current version being used in a char array. I have the following code:
if(isdeployed)
version = (Help here);
version_message = [‘Now running program.exe version: ‘ version];
else
version = ‘script’;
version_message = [‘Now running program.m version: ‘ version];
end
disp(version_message);
Is it possible to dynamically get the version number, or do I have to manually set it to ensure they are the same? I know it can be done with App Designer, but this executable does not use App Designer. compiler, version number, matlab MATLAB Answers — New Questions
Same script, faster execution when running in a spawned instance (“!matlab …”)?
Dear Community,
I am running Matlab 2024b under Windows 10 / 32GB RAM on a 6-core laptop Intel CPU (hyperthreading deactivated). For the sake of runtime optimisation I have done the following comparison:
case #1: a script running within the current Matlab session (single instance) and
case #2: the same script running in a spawned session (instance), using "!matlab -nosplash -desktop -r "load …" (etc.).
The runtime results are as follows:
script #1 finished in about an hour, total CPU load ~50% at boost clock (single Matlab instance)
script #2 finished in about HALF an hour (!), similar total CPU load ~50% at boost clock; main Matlab instance idle (~0%), spawned instance ~40%
I muss admit I do not understand the results: why does a spawned Matlab instance (i.e. two instances running in parallel: main/idle and active) complete the task in half of the time, compared to a single instance?
What might be the reason for this behavior?
Thanks a lot in advance!
MarekDear Community,
I am running Matlab 2024b under Windows 10 / 32GB RAM on a 6-core laptop Intel CPU (hyperthreading deactivated). For the sake of runtime optimisation I have done the following comparison:
case #1: a script running within the current Matlab session (single instance) and
case #2: the same script running in a spawned session (instance), using "!matlab -nosplash -desktop -r "load …" (etc.).
The runtime results are as follows:
script #1 finished in about an hour, total CPU load ~50% at boost clock (single Matlab instance)
script #2 finished in about HALF an hour (!), similar total CPU load ~50% at boost clock; main Matlab instance idle (~0%), spawned instance ~40%
I muss admit I do not understand the results: why does a spawned Matlab instance (i.e. two instances running in parallel: main/idle and active) complete the task in half of the time, compared to a single instance?
What might be the reason for this behavior?
Thanks a lot in advance!
Marek Dear Community,
I am running Matlab 2024b under Windows 10 / 32GB RAM on a 6-core laptop Intel CPU (hyperthreading deactivated). For the sake of runtime optimisation I have done the following comparison:
case #1: a script running within the current Matlab session (single instance) and
case #2: the same script running in a spawned session (instance), using "!matlab -nosplash -desktop -r "load …" (etc.).
The runtime results are as follows:
script #1 finished in about an hour, total CPU load ~50% at boost clock (single Matlab instance)
script #2 finished in about HALF an hour (!), similar total CPU load ~50% at boost clock; main Matlab instance idle (~0%), spawned instance ~40%
I muss admit I do not understand the results: why does a spawned Matlab instance (i.e. two instances running in parallel: main/idle and active) complete the task in half of the time, compared to a single instance?
What might be the reason for this behavior?
Thanks a lot in advance!
Marek multiple instance spawn MATLAB Answers — New Questions
Microsoft Extends DLP Policy for Copilot to Office Apps
Same DLP Policy for Copilot Used to Block BizChat
On May 1, Microsoft announced that the public preview of the DLP policy for Microsoft 365 Copilot is effective for the Office apps (MC1059677, 21 April 2025, Microsoft 365 roadmap item 423483). The new functionality is an extension of the DLP policy introduced in March 2025. At that time, the policy only covered Microsoft 365 Copilot Chat (BizChat). Its extension to cover the Office apps (desktop and web) is logical, even if the implementation is different. We’ll get to what those differences are shortly.
How the DLP Policy for Copilot Works
As a quick refresher, the DLP policy for Copilot works by checking if a file is assigned a specific sensitivity label. If true, the Copilot functionality built into the app is limited and the content of the file cannot be used in Copilot responses, such as creating a document summary.
Apps are responsible for checking if a DLP policy is active within the tenant and what sensitivity labels are associated with the policy, so the announcement marks the inclusion of the necessary code in the Office apps to check for the DLP policy. I tested with Microsoft 365 Enterprise Apps version 2504 (build 18730.20122).
Like any other DLP policy, the policy can have multiple rules. In this case, rules for the DLP policy for Copilot block access for a sensitivity label, so if you want to block access for multiple sensitivity labels, the DLP policy has a separate rule for each label. If you created the DLP policy for Copilot to use with BizChat, you don’t need to do anything to extend the policy to cover the Office apps.
Using the DLP Policy for Copilot in Word
As an example, I created a Word document and tested that all the Copilot functionality worked as expected. I saved the document and reopened it to force Copilot to generate the automatic summary.
I then applied one of the sensitivity labels covered by a rule in the DLP policy for Copilot and tried out some of the Copilot features. As you can see from Figure 1, the automatic summary was not removed (but the summary cannot be updated), and asking Copilot to explicitly summarize the document fails because “your organization’s policy doesn’t allow it.” However, it looks like Copilot can query the content of the document to answer questions in chat.
In their announcement, Microsoft says that “Copilot actions like summarizing or auto-generating content directly in the canvas are blocked.” They also say that chatting with Copilot is also blocked, but as you can see in Figure 1, Copilot answered a predefined question (“What is the purpose of DLP for M365 Copilot”) quite happily. On the other hand, if you go to the Message Copilot section and input the same question, Copilot refuses to answer. The block on chat worked in the web app but not always in the desktop version of Word (but this is preview software, so some bugs are expected).
Finally, Copilot cannot reference a file protected by one of the sensitivity labels covered by the DLP policy (an action that forces Copilot to extract the content of the referenced document).
Maybe Just Turn Copilot Off
I’ve used Copilot for nearly two years, and I was initially confused by the effect the DLP policy for Copilot has on the Office apps. To me, it would be simpler and more understandable to disable Copilot completely for documents within the scope of the DLP policy. I would remove the Copilot button from the menu bar and make sure that no UI elements that expose any Copilot feature, like the automatic summary appear. Right now, the UI is a confusing mishmash of things that work and stuff that doesn’t that needs to be cleaned up.
Learn how to exploit the data available to Microsoft 365 tenant administrators through the Office 365 for IT Pros eBook. We love figuring out how things work.
Find a certain number in a vector based on conditions
So I have a vector(A) of distances. I have a vector(B) of ranges. These ranges(B) are not equivalent to the size of the vector(A). I want to find the location of each range in the vector that corresponds to each distance value. Like if a distance is within the range. The range(B) is something like (m by 2) and the distances(A) are (n by 1)So I have a vector(A) of distances. I have a vector(B) of ranges. These ranges(B) are not equivalent to the size of the vector(A). I want to find the location of each range in the vector that corresponds to each distance value. Like if a distance is within the range. The range(B) is something like (m by 2) and the distances(A) are (n by 1) So I have a vector(A) of distances. I have a vector(B) of ranges. These ranges(B) are not equivalent to the size of the vector(A). I want to find the location of each range in the vector that corresponds to each distance value. Like if a distance is within the range. The range(B) is something like (m by 2) and the distances(A) are (n by 1) find, vector MATLAB Answers — New Questions
How to set variable names using Readtable with hierarchical categories (Excel file with merged cells / multiple header rows)
I have an Excel file with a series of header rows containing a hierarchy of descriptors. I would like to use this existing hierarchy to access the data. In the image below, I have data for various engineering majors (electical, mechincal, …) that are split by year of study (freshman, sophomore, …), then by gender, and finally my domestic versus international. This heirachy repeats for each major and there are many rows of data corresponding to academic years (not pictured in the table below) going back a long way. Is there a way to use readtable or a similar function to load that data in such a way that I can retrieve a specific number. For example, can I import the data in such a way that I could retrieve it with something like the following pseudo-code
table = readtable(‘filename’, <other options>)
table(major=electrical, gender=female, <additional options>)
% or even
table(row1=’electrical’, row2=’female’, <additional options>)
Alternatively, can I manipulate the way column names are created? WIthout any tweaking I get names like
‘dometic’, ‘domestic_1’, ‘domestic_2’, …
What I would prefer is to get column names such as
‘electrical-freshman-male-domestic’
‘electrical-freshman-female-domestic’
‘electrical-freshman-unspecified-domestic’
‘electrical-sophomore-male-domestic’
…
and likewise for other combinations so that I drill down to the appropriate column of data.I have an Excel file with a series of header rows containing a hierarchy of descriptors. I would like to use this existing hierarchy to access the data. In the image below, I have data for various engineering majors (electical, mechincal, …) that are split by year of study (freshman, sophomore, …), then by gender, and finally my domestic versus international. This heirachy repeats for each major and there are many rows of data corresponding to academic years (not pictured in the table below) going back a long way. Is there a way to use readtable or a similar function to load that data in such a way that I can retrieve a specific number. For example, can I import the data in such a way that I could retrieve it with something like the following pseudo-code
table = readtable(‘filename’, <other options>)
table(major=electrical, gender=female, <additional options>)
% or even
table(row1=’electrical’, row2=’female’, <additional options>)
Alternatively, can I manipulate the way column names are created? WIthout any tweaking I get names like
‘dometic’, ‘domestic_1’, ‘domestic_2’, …
What I would prefer is to get column names such as
‘electrical-freshman-male-domestic’
‘electrical-freshman-female-domestic’
‘electrical-freshman-unspecified-domestic’
‘electrical-sophomore-male-domestic’
…
and likewise for other combinations so that I drill down to the appropriate column of data. I have an Excel file with a series of header rows containing a hierarchy of descriptors. I would like to use this existing hierarchy to access the data. In the image below, I have data for various engineering majors (electical, mechincal, …) that are split by year of study (freshman, sophomore, …), then by gender, and finally my domestic versus international. This heirachy repeats for each major and there are many rows of data corresponding to academic years (not pictured in the table below) going back a long way. Is there a way to use readtable or a similar function to load that data in such a way that I can retrieve a specific number. For example, can I import the data in such a way that I could retrieve it with something like the following pseudo-code
table = readtable(‘filename’, <other options>)
table(major=electrical, gender=female, <additional options>)
% or even
table(row1=’electrical’, row2=’female’, <additional options>)
Alternatively, can I manipulate the way column names are created? WIthout any tweaking I get names like
‘dometic’, ‘domestic_1’, ‘domestic_2’, …
What I would prefer is to get column names such as
‘electrical-freshman-male-domestic’
‘electrical-freshman-female-domestic’
‘electrical-freshman-unspecified-domestic’
‘electrical-sophomore-male-domestic’
…
and likewise for other combinations so that I drill down to the appropriate column of data. readtable, readcell, table MATLAB Answers — New Questions
Wave simulation
Hello
I’m trying to simulate a single source wave in matlab
does anyone know how to do that? (code tips)
ThanksHello
I’m trying to simulate a single source wave in matlab
does anyone know how to do that? (code tips)
Thanks Hello
I’m trying to simulate a single source wave in matlab
does anyone know how to do that? (code tips)
Thanks wave, matlab, array, plotting, matlab gui, graph, simulation, code generation, 2d MATLAB Answers — New Questions
How to check Matlab code Plagiarism
How can we check MATLAB code plagiarism? Is there any software available?How can we check MATLAB code plagiarism? Is there any software available? How can we check MATLAB code plagiarism? Is there any software available? plagiarism, cheating, copying MATLAB Answers — New Questions
any one help me to correct this code or help me my I have a graduation project coming soon
clear; close all; clc;
warning(‘off’, ‘all’);
[contentFile, contentPath] = uigetfile({‘*.bmp;*.jpg;*.png;*.tif’, ‘ملفات الصور’}, ‘اختر صورة المحتوى’);
if isequal(contentFile, 0)
error(‘لم يتم اختيار صورة محتوى’);
end
contentImage = imread(fullfile(contentPath, contentFile));
[styleFile, stylePath] = uigetfile({‘*.bmp;*.jpg;*.png;*.tif’, ‘ملفات الصور’}, ‘اختر صورة النمط’);
if isequal(styleFile, 0)
error(‘لم يتم اختيار صورة نمط’);
end
styleImage = imread(fullfile(stylePath, styleFile));
contentImage = im2double(contentImage);
styleImage = im2double(styleImage);
if size(contentImage,3) == 1
contentImage = repmat(contentImage, [1 1 3]);
end
if size(styleImage,3) == 1
styleImage = repmat(styleImage, [1 1 3]);
end
styleImage = imadjust(styleImage, [0.1 0.9], []);
styleImage = imgaussfilt(styleImage, 1);
net = vgg19();
inputSize = net.Layers(1).InputSize(1:2);
% تغيير حجم الصور مع الحفاظ على التناسب
contentImage = imresize(contentImage, inputSize);
styleImage = imresize(styleImage, inputSize);
%% 4. تحديد الطبقات الصحيحة
contentLayer = ‘conv4_2’;
styleLayers = {‘conv1_1’, ‘conv2_1’, ‘conv3_1’, ‘conv4_1’, ‘conv5_1’};
styleFeatures = getStyleFeatures(styleImage, net, styleLayers);
numIterations = 500;
learningRate = 0.01;
styleWeight = 1e6;
dlContent = dlarray(contentImage, ‘SSC’);
dlTransform = dlContent;
figure;
for iter = 1:numIterations
[grad, loss] = dlfeval(@computeGradients, dlTransform, dlContent, net, …
contentLayer, styleLayers, styleFeatures, styleWeight);
dlTransform = dlTransform – learningRate * grad;
if mod(iter,50) == 0 || iter == 1
fprintf(‘التكرار %d: الخسارة الكلية=%.2f, المحتوى=%.2f, النمط=%.2fn’, …
iter, loss.total, loss.content, loss.style);
% عرض التقدم
currentImg = uint8(extractdata(dlTransform)*255);
imshow(currentImg);
title(sprintf(‘التكرار %d/%d’, iter, numIterations));
drawnow;
end
end
outputImage = uint8(extractdata(dlTransform)*255);
[~,name,ext] = fileparts(contentFile);
outputFile = fullfile(pwd, [name ‘_styled’ ext]);
imwrite(outputImage, outputFile);
fprintf(‘تم حفظ الصورة الناتجة بنجاح في: %sn’, outputFile);
function features = getStyleFeatures(styleImg, net, styleLayers)
if ~ismatrix(styleImg) && ~(ndims(styleImg)==3)
error(‘يجب أن تكون صورة النمط مصفوفة 2D أو 3D’);
end
if size(styleImg,3) ~= 3
error(‘يجب أن تحتوي صورة النمط على 3 قنوات لونية (RGB)’);
end
try
dlStyle = dlarray(styleImg, ‘SSC’);
catch
error(‘فشل تحويل صورة النمط إلى dlarray’);
end
features = struct();
for i = 1:length(styleLayers)
layer = styleLayers{i};
try
dlFeatures = activations(net, dlStyle, layer);
features.(layer) = computeGramMatrix(dlFeatures);
catch ME
error(‘فشل في استخراج خصائص الطبقة %s: %s’, layer, ME.message);
end
end
end
function gramMatrix = computeGramMatrix(features)
[H,W,C] = size(features);
reshaped = reshape(features, H*W, C);
gramMatrix = reshaped’ * reshaped / (H*W*C);
end
function [gradients, loss] = computeGradients(dlTransform, dlContent, net, …
contentLayer, styleLayers, styleFeatures, styleWeight)
contentFeatures = activations(net, dlContent, contentLayer);
transformContentFeatures = activations(net, dlTransform, contentLayer);
contentLoss = mean((transformContentFeatures – contentFeatures).^2);
styleLoss = 0;
for i = 1:length(styleLayers)
layer = styleLayers{i};
transformFeatures = activations(net, dlTransform, layer);
gramTransform = computeGramMatrix(transformFeatures);
gramStyle = styleFeatures.(layer);
styleLoss = styleLoss + mean((gramTransform – gramStyle).^2);
end
styleLoss = styleLoss / length(styleLayers);
totalLoss = contentLoss + styleWeight * styleLoss;
gradients = dlgradient(totalLoss, dlTransform);
loss.total = double(totalLoss);
loss.content = double(contentLoss);
loss.style = double(styleLoss);
end
%%%%
erorr
rror using styletransfer>getStyleFeatures (line 111)
فشل في استخراج خصائص الطبقة conv1_1: Invalid 2-D image data. Specify image data as a 3-D numeric array containing a single image, a
4-D numeric array containing multiple images, a datastore, or a table containing image file paths or images in the first column.
Error in styletransfer (line 48)
styleFeatures = getStyleFeatures(styleImage, net, styleLayers);
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^clear; close all; clc;
warning(‘off’, ‘all’);
[contentFile, contentPath] = uigetfile({‘*.bmp;*.jpg;*.png;*.tif’, ‘ملفات الصور’}, ‘اختر صورة المحتوى’);
if isequal(contentFile, 0)
error(‘لم يتم اختيار صورة محتوى’);
end
contentImage = imread(fullfile(contentPath, contentFile));
[styleFile, stylePath] = uigetfile({‘*.bmp;*.jpg;*.png;*.tif’, ‘ملفات الصور’}, ‘اختر صورة النمط’);
if isequal(styleFile, 0)
error(‘لم يتم اختيار صورة نمط’);
end
styleImage = imread(fullfile(stylePath, styleFile));
contentImage = im2double(contentImage);
styleImage = im2double(styleImage);
if size(contentImage,3) == 1
contentImage = repmat(contentImage, [1 1 3]);
end
if size(styleImage,3) == 1
styleImage = repmat(styleImage, [1 1 3]);
end
styleImage = imadjust(styleImage, [0.1 0.9], []);
styleImage = imgaussfilt(styleImage, 1);
net = vgg19();
inputSize = net.Layers(1).InputSize(1:2);
% تغيير حجم الصور مع الحفاظ على التناسب
contentImage = imresize(contentImage, inputSize);
styleImage = imresize(styleImage, inputSize);
%% 4. تحديد الطبقات الصحيحة
contentLayer = ‘conv4_2’;
styleLayers = {‘conv1_1’, ‘conv2_1’, ‘conv3_1’, ‘conv4_1’, ‘conv5_1’};
styleFeatures = getStyleFeatures(styleImage, net, styleLayers);
numIterations = 500;
learningRate = 0.01;
styleWeight = 1e6;
dlContent = dlarray(contentImage, ‘SSC’);
dlTransform = dlContent;
figure;
for iter = 1:numIterations
[grad, loss] = dlfeval(@computeGradients, dlTransform, dlContent, net, …
contentLayer, styleLayers, styleFeatures, styleWeight);
dlTransform = dlTransform – learningRate * grad;
if mod(iter,50) == 0 || iter == 1
fprintf(‘التكرار %d: الخسارة الكلية=%.2f, المحتوى=%.2f, النمط=%.2fn’, …
iter, loss.total, loss.content, loss.style);
% عرض التقدم
currentImg = uint8(extractdata(dlTransform)*255);
imshow(currentImg);
title(sprintf(‘التكرار %d/%d’, iter, numIterations));
drawnow;
end
end
outputImage = uint8(extractdata(dlTransform)*255);
[~,name,ext] = fileparts(contentFile);
outputFile = fullfile(pwd, [name ‘_styled’ ext]);
imwrite(outputImage, outputFile);
fprintf(‘تم حفظ الصورة الناتجة بنجاح في: %sn’, outputFile);
function features = getStyleFeatures(styleImg, net, styleLayers)
if ~ismatrix(styleImg) && ~(ndims(styleImg)==3)
error(‘يجب أن تكون صورة النمط مصفوفة 2D أو 3D’);
end
if size(styleImg,3) ~= 3
error(‘يجب أن تحتوي صورة النمط على 3 قنوات لونية (RGB)’);
end
try
dlStyle = dlarray(styleImg, ‘SSC’);
catch
error(‘فشل تحويل صورة النمط إلى dlarray’);
end
features = struct();
for i = 1:length(styleLayers)
layer = styleLayers{i};
try
dlFeatures = activations(net, dlStyle, layer);
features.(layer) = computeGramMatrix(dlFeatures);
catch ME
error(‘فشل في استخراج خصائص الطبقة %s: %s’, layer, ME.message);
end
end
end
function gramMatrix = computeGramMatrix(features)
[H,W,C] = size(features);
reshaped = reshape(features, H*W, C);
gramMatrix = reshaped’ * reshaped / (H*W*C);
end
function [gradients, loss] = computeGradients(dlTransform, dlContent, net, …
contentLayer, styleLayers, styleFeatures, styleWeight)
contentFeatures = activations(net, dlContent, contentLayer);
transformContentFeatures = activations(net, dlTransform, contentLayer);
contentLoss = mean((transformContentFeatures – contentFeatures).^2);
styleLoss = 0;
for i = 1:length(styleLayers)
layer = styleLayers{i};
transformFeatures = activations(net, dlTransform, layer);
gramTransform = computeGramMatrix(transformFeatures);
gramStyle = styleFeatures.(layer);
styleLoss = styleLoss + mean((gramTransform – gramStyle).^2);
end
styleLoss = styleLoss / length(styleLayers);
totalLoss = contentLoss + styleWeight * styleLoss;
gradients = dlgradient(totalLoss, dlTransform);
loss.total = double(totalLoss);
loss.content = double(contentLoss);
loss.style = double(styleLoss);
end
%%%%
erorr
rror using styletransfer>getStyleFeatures (line 111)
فشل في استخراج خصائص الطبقة conv1_1: Invalid 2-D image data. Specify image data as a 3-D numeric array containing a single image, a
4-D numeric array containing multiple images, a datastore, or a table containing image file paths or images in the first column.
Error in styletransfer (line 48)
styleFeatures = getStyleFeatures(styleImage, net, styleLayers);
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ clear; close all; clc;
warning(‘off’, ‘all’);
[contentFile, contentPath] = uigetfile({‘*.bmp;*.jpg;*.png;*.tif’, ‘ملفات الصور’}, ‘اختر صورة المحتوى’);
if isequal(contentFile, 0)
error(‘لم يتم اختيار صورة محتوى’);
end
contentImage = imread(fullfile(contentPath, contentFile));
[styleFile, stylePath] = uigetfile({‘*.bmp;*.jpg;*.png;*.tif’, ‘ملفات الصور’}, ‘اختر صورة النمط’);
if isequal(styleFile, 0)
error(‘لم يتم اختيار صورة نمط’);
end
styleImage = imread(fullfile(stylePath, styleFile));
contentImage = im2double(contentImage);
styleImage = im2double(styleImage);
if size(contentImage,3) == 1
contentImage = repmat(contentImage, [1 1 3]);
end
if size(styleImage,3) == 1
styleImage = repmat(styleImage, [1 1 3]);
end
styleImage = imadjust(styleImage, [0.1 0.9], []);
styleImage = imgaussfilt(styleImage, 1);
net = vgg19();
inputSize = net.Layers(1).InputSize(1:2);
% تغيير حجم الصور مع الحفاظ على التناسب
contentImage = imresize(contentImage, inputSize);
styleImage = imresize(styleImage, inputSize);
%% 4. تحديد الطبقات الصحيحة
contentLayer = ‘conv4_2’;
styleLayers = {‘conv1_1’, ‘conv2_1’, ‘conv3_1’, ‘conv4_1’, ‘conv5_1’};
styleFeatures = getStyleFeatures(styleImage, net, styleLayers);
numIterations = 500;
learningRate = 0.01;
styleWeight = 1e6;
dlContent = dlarray(contentImage, ‘SSC’);
dlTransform = dlContent;
figure;
for iter = 1:numIterations
[grad, loss] = dlfeval(@computeGradients, dlTransform, dlContent, net, …
contentLayer, styleLayers, styleFeatures, styleWeight);
dlTransform = dlTransform – learningRate * grad;
if mod(iter,50) == 0 || iter == 1
fprintf(‘التكرار %d: الخسارة الكلية=%.2f, المحتوى=%.2f, النمط=%.2fn’, …
iter, loss.total, loss.content, loss.style);
% عرض التقدم
currentImg = uint8(extractdata(dlTransform)*255);
imshow(currentImg);
title(sprintf(‘التكرار %d/%d’, iter, numIterations));
drawnow;
end
end
outputImage = uint8(extractdata(dlTransform)*255);
[~,name,ext] = fileparts(contentFile);
outputFile = fullfile(pwd, [name ‘_styled’ ext]);
imwrite(outputImage, outputFile);
fprintf(‘تم حفظ الصورة الناتجة بنجاح في: %sn’, outputFile);
function features = getStyleFeatures(styleImg, net, styleLayers)
if ~ismatrix(styleImg) && ~(ndims(styleImg)==3)
error(‘يجب أن تكون صورة النمط مصفوفة 2D أو 3D’);
end
if size(styleImg,3) ~= 3
error(‘يجب أن تحتوي صورة النمط على 3 قنوات لونية (RGB)’);
end
try
dlStyle = dlarray(styleImg, ‘SSC’);
catch
error(‘فشل تحويل صورة النمط إلى dlarray’);
end
features = struct();
for i = 1:length(styleLayers)
layer = styleLayers{i};
try
dlFeatures = activations(net, dlStyle, layer);
features.(layer) = computeGramMatrix(dlFeatures);
catch ME
error(‘فشل في استخراج خصائص الطبقة %s: %s’, layer, ME.message);
end
end
end
function gramMatrix = computeGramMatrix(features)
[H,W,C] = size(features);
reshaped = reshape(features, H*W, C);
gramMatrix = reshaped’ * reshaped / (H*W*C);
end
function [gradients, loss] = computeGradients(dlTransform, dlContent, net, …
contentLayer, styleLayers, styleFeatures, styleWeight)
contentFeatures = activations(net, dlContent, contentLayer);
transformContentFeatures = activations(net, dlTransform, contentLayer);
contentLoss = mean((transformContentFeatures – contentFeatures).^2);
styleLoss = 0;
for i = 1:length(styleLayers)
layer = styleLayers{i};
transformFeatures = activations(net, dlTransform, layer);
gramTransform = computeGramMatrix(transformFeatures);
gramStyle = styleFeatures.(layer);
styleLoss = styleLoss + mean((gramTransform – gramStyle).^2);
end
styleLoss = styleLoss / length(styleLayers);
totalLoss = contentLoss + styleWeight * styleLoss;
gradients = dlgradient(totalLoss, dlTransform);
loss.total = double(totalLoss);
loss.content = double(contentLoss);
loss.style = double(styleLoss);
end
%%%%
erorr
rror using styletransfer>getStyleFeatures (line 111)
فشل في استخراج خصائص الطبقة conv1_1: Invalid 2-D image data. Specify image data as a 3-D numeric array containing a single image, a
4-D numeric array containing multiple images, a datastore, or a table containing image file paths or images in the first column.
Error in styletransfer (line 48)
styleFeatures = getStyleFeatures(styleImage, net, styleLayers);
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ai, fast style transfer, deep learning, transfer learning MATLAB Answers — New Questions
coneprog 代码生成过程中报错：’secondordercone’ not supported for code generation.
coneprog 代码生成coneprog 代码生成 coneprog 代码生成 transferred MATLAB Answers — New Questions
Has the MEX function (mxGetComplexDoubles) been deprecated in R2024b?
I am preparing a MEX file that includes the function: (mxGetComplexDoubles).
I tried to compile and link it to get the executable file (*. mexw64) corresponding to the source file: (*.F) using either of the following commands:
mex -R2018a filename.F
mex -R2018a -I"C:Program FilesMATLABR2024bexterninclude" -L"C:Program FilesMATLABR2024bexternlibwin64microsoft" -llibmex -llibmx filename.F
The compilation was successful; however, the linking was not because:
unresolved external symbol mxgetcomplexdoubles800
This happened despite my file containing the command (mxGetComplexDoubles) rather than the command (mxGetComplexDoubles800).
Surprisingly, upon searching for (mxGetComplexDoubles), I found that it is not there in:
"C:Program FilesMATLABR2024bbinwin64libmex.dll"
"C:Program FilesMATLABR2024bbinwin64libmat.dll"
My questions:
Has the function (mxGetComplexDoubles) been deprecated in R2024b?
What should I do in that case?I am preparing a MEX file that includes the function: (mxGetComplexDoubles).
I tried to compile and link it to get the executable file (*. mexw64) corresponding to the source file: (*.F) using either of the following commands:
mex -R2018a filename.F
mex -R2018a -I"C:Program FilesMATLABR2024bexterninclude" -L"C:Program FilesMATLABR2024bexternlibwin64microsoft" -llibmex -llibmx filename.F
The compilation was successful; however, the linking was not because:
unresolved external symbol mxgetcomplexdoubles800
This happened despite my file containing the command (mxGetComplexDoubles) rather than the command (mxGetComplexDoubles800).
Surprisingly, upon searching for (mxGetComplexDoubles), I found that it is not there in:
"C:Program FilesMATLABR2024bbinwin64libmex.dll"
"C:Program FilesMATLABR2024bbinwin64libmat.dll"
My questions:
Has the function (mxGetComplexDoubles) been deprecated in R2024b?
What should I do in that case? I am preparing a MEX file that includes the function: (mxGetComplexDoubles).
I tried to compile and link it to get the executable file (*. mexw64) corresponding to the source file: (*.F) using either of the following commands:
mex -R2018a filename.F
mex -R2018a -I"C:Program FilesMATLABR2024bexterninclude" -L"C:Program FilesMATLABR2024bexternlibwin64microsoft" -llibmex -llibmx filename.F
The compilation was successful; however, the linking was not because:
unresolved external symbol mxgetcomplexdoubles800
This happened despite my file containing the command (mxGetComplexDoubles) rather than the command (mxGetComplexDoubles800).
Surprisingly, upon searching for (mxGetComplexDoubles), I found that it is not there in:
"C:Program FilesMATLABR2024bbinwin64libmex.dll"
"C:Program FilesMATLABR2024bbinwin64libmat.dll"
My questions:
Has the function (mxGetComplexDoubles) been deprecated in R2024b?
What should I do in that case? mex function mxgetcomplexdoubles MATLAB Answers — New Questions
Gradient based sharpness identifictaion in an image
Hi,
I am trying to identify sharply focused portions in the attached image using imgradient operator in the matlab. I have used "sobel" method.
I have to do this operation on many images, I am wondering is there any way to apply some thresolding while implementing imgradient or (any similar operator) in the matlab.
load matlab; imshow(II,[])Hi,
I am trying to identify sharply focused portions in the attached image using imgradient operator in the matlab. I have used "sobel" method.
I have to do this operation on many images, I am wondering is there any way to apply some thresolding while implementing imgradient or (any similar operator) in the matlab.
load matlab; imshow(II,[]) Hi,
I am trying to identify sharply focused portions in the attached image using imgradient operator in the matlab. I have used "sobel" method.
I have to do this operation on many images, I am wondering is there any way to apply some thresolding while implementing imgradient or (any similar operator) in the matlab.
load matlab; imshow(II,[]) edge detection, image processing, imgradient, sobel MATLAB Answers — New Questions
How to save the proceesed image in the same size as input image
Hi,
I need to save the proceesed image in the same size (width and height) as the input image. I tried below script, but it still saves teh image in the smaller size than the input same.
width=2472;
height=2062;
set(gcf,’position’,[x0,y0,width,height])Hi,
I need to save the proceesed image in the same size (width and height) as the input image. I tried below script, but it still saves teh image in the smaller size than the input same.
width=2472;
height=2062;
set(gcf,’position’,[x0,y0,width,height]) Hi,
I need to save the proceesed image in the same size (width and height) as the input image. I tried below script, but it still saves teh image in the smaller size than the input same.
width=2472;
height=2062;
set(gcf,’position’,[x0,y0,width,height]) image processing, image saving, image size MATLAB Answers — New Questions
