Month: July 2024
Connect to a Linux VM using Bastion and Key Vault without a private key file
In enterprise environments, requirements often demand the use of private networks, resulting in VMs not being assigned Public IP addresses. In such cases, Azure Bastion offers an extremely useful feature for accessing these VMs without complexity. When logging into Linux VMs on Azure via Azure Bastion using an SSH Private Key, the key is often managed as a local file. Managing SSH Private Key files on individual devices poses risks such as potential key leakage. Therefore, centralized management, including proper role assignment, is preferable from a security standpoint. A common need in such scenarios is to manage SSH Private Keys using Key Vault. This article provides such a comprehensive step-by-step guide.
There are two huge benefits here.
SSH Private Key is no longer required to store as a local file Local file by storing it in Azure Key Vault.
Access to Linux VM can be be controlled using RBAC assignment for Azure Key Vault.
Regarding the first point, storing the SSH Private Key locally after creating it through the Azure Portal can be issues such as loss the file or difficulty in identifying which VM it is for. Additionally, it has the benefit of reducing the risk of file leakage if a developer leaves the company or the project.
Regarding the second point, using Key Vault for access management allows you to control VM access through Key Vault. You can manage access control to specific VMs.
Here is the architecture diagram for the current setup. When creating a Linux VM, the SSH Private Key generated through the Azure portal is registered in Key Vault. This SSH Private Key is then used to access the Linux VM via Azure Bastion.
Prerequisites
Before following this article, you have to create Azure resources as follows:
Virtual Network
Key Vault
Azure Bastion
Setup
First, we create a new SSH Key when creating the Linux VM through the Azure portal. While it is possible to create an SSH Private Key when creating a Linux VM through the Azure portal, you can also create it separately. Once you input the parameters and create the Linux VM, you can download the SSH Private Key and save it as a local file.
If you choose to create the SSH Private Key separately, note that you need to select “Reset password” from the VM menu on Azure Portal and use “Add SSH public key” to set the SSH Key information for the Linux VM. The first link in the References section contains relevant information.
Next, you have to acquire proper RBAC role to manage Azure Key Vault. It’s not enough just having subscription Owner role. In this article, we assign Key Vault Administrator role.
We can use Key Vault Secrets Officer role for developers who access your Linux VM. Please refer to the article as follows to assign proper RBA roles.
Provide access to Key Vault keys, certificates, and secrets with an Azure role-based access control
Then, we register the SSH Private Key in the secrets of Azure Key Vault (you cannot select Keys or Certificates with Azure Bastion). There is a tip here that the SSH Private Key contains a multi-line file, but the Key Vault portal does not support the registration of multi-line strings. Therefore, we have to execute this via command line as follows:
az keyvault secret set –name “your-secret-name” –vault-name “your-key-name” –file “my-ssh-private-key.pem”
If the command works well, you can find your SSH Private Key on the Azure Key Vault.
Then, we can access the Linux VM via Azure Bastion. Select the Linux VM, choose “SSH Private Key from Azure Key Vault” from the Authentication Type options, and use the the SSH Private Key you have registered.
Error #1: Register SSH Private Key without RBAC role assignment
You got error as follows if you do not have proper RBAC. I have assigned Key Vault Administrator role to the operator to solve this issue.
Error #2: Access Linux VM without RBAC role assignment
You got error as follows if you do not have proper RBAC. I have assigned Key Vault Secrets Officer role to the operator to solve this issue.
References
ssl – How to use a certificate from a Azure Key Vault to connect to a Azure Virtual Machine through Bastion? – Stack Overflow
az keyvault secret | Microsoft Learn
az sshkey | Microsoft Learn
Connect to a Linux VM using SSH – Azure Bastion | Microsoft Learn
Microsoft Tech Community – Latest Blogs –Read More
Trying to make a 2 value sz vector, get Error using tabular/horzcat All input arguments must be tables.
Seems so simple, I’m trying to set a new table size based on values in a previous table
C = sum(app.UITable2.Data(1,:));
No matter how hard I try I can’t get the sum of a single column!
So I give up and do
sz = [C(1,1) 4];
Get the weird error that all inputs must be tables!
Try making sz a known vector type
make a property sz = [4 4]
then try
app.sz = [C(1,1) 4];
Sorry for all the asks but I’m a very experienced C, C++ (and many other languages) programmer but this .m code is driving me crazy! Help give all kinds of useless details about setting Fonts etc but little about simple syntax for e.g. summing a column.
S = sum(A,1); does sum by columns (vs ,2 for by rows but no way to say just do column 1. Even though C is a 1×4 Can’t seem to get the scaler C(1,1) out of it properly.
C(1,1) or C(1) works in the command line, just not in my app created by app designer.
Ultimately all I really want is to make a new table of the correct size with
app.TrialsTable = table(‘Size’,sz,’VariableTypes’,{‘uint8′,’uint8′,’categorical’,’int16′});Seems so simple, I’m trying to set a new table size based on values in a previous table
C = sum(app.UITable2.Data(1,:));
No matter how hard I try I can’t get the sum of a single column!
So I give up and do
sz = [C(1,1) 4];
Get the weird error that all inputs must be tables!
Try making sz a known vector type
make a property sz = [4 4]
then try
app.sz = [C(1,1) 4];
Sorry for all the asks but I’m a very experienced C, C++ (and many other languages) programmer but this .m code is driving me crazy! Help give all kinds of useless details about setting Fonts etc but little about simple syntax for e.g. summing a column.
S = sum(A,1); does sum by columns (vs ,2 for by rows but no way to say just do column 1. Even though C is a 1×4 Can’t seem to get the scaler C(1,1) out of it properly.
C(1,1) or C(1) works in the command line, just not in my app created by app designer.
Ultimately all I really want is to make a new table of the correct size with
app.TrialsTable = table(‘Size’,sz,’VariableTypes’,{‘uint8′,’uint8′,’categorical’,’int16′}); Seems so simple, I’m trying to set a new table size based on values in a previous table
C = sum(app.UITable2.Data(1,:));
No matter how hard I try I can’t get the sum of a single column!
So I give up and do
sz = [C(1,1) 4];
Get the weird error that all inputs must be tables!
Try making sz a known vector type
make a property sz = [4 4]
then try
app.sz = [C(1,1) 4];
Sorry for all the asks but I’m a very experienced C, C++ (and many other languages) programmer but this .m code is driving me crazy! Help give all kinds of useless details about setting Fonts etc but little about simple syntax for e.g. summing a column.
S = sum(A,1); does sum by columns (vs ,2 for by rows but no way to say just do column 1. Even though C is a 1×4 Can’t seem to get the scaler C(1,1) out of it properly.
C(1,1) or C(1) works in the command line, just not in my app created by app designer.
Ultimately all I really want is to make a new table of the correct size with
app.TrialsTable = table(‘Size’,sz,’VariableTypes’,{‘uint8′,’uint8′,’categorical’,’int16′}); sum, vector, table size MATLAB Answers — New Questions
Files not showing up in ‘current folder’
Hi all,
I’ve been experiencing a weird problem starting in the last 2 days. When go to a certain directory in Matlab, it doesn’t show any of the files in that directory. Moreover, using the dir() command for that directory doesn’t show any files.
I know the directory exists and there are files in it because I can access it through the normal Windows file explorer (see image below).
A few additional details. The directory I’m trying to access is a network drive. However, I can still access other network drives from Matlab and other PCs on the same network are able to view the directory in Matlab. This makes me think its something specific to Matlab on my PC with this directory (although I’m having the same issue on my personal laptop). I’m using a Windows 10 PC and Matlab 2017a.
I’ve already tried restarting matlab, my pc, disconnecting and reconnecting the drive. I have not tried reinstalling Matlab yet. Any help would be greatly appreciated!
<</matlabcentral/answers/uploaded_files/94148/currentfolder.PNG>>Hi all,
I’ve been experiencing a weird problem starting in the last 2 days. When go to a certain directory in Matlab, it doesn’t show any of the files in that directory. Moreover, using the dir() command for that directory doesn’t show any files.
I know the directory exists and there are files in it because I can access it through the normal Windows file explorer (see image below).
A few additional details. The directory I’m trying to access is a network drive. However, I can still access other network drives from Matlab and other PCs on the same network are able to view the directory in Matlab. This makes me think its something specific to Matlab on my PC with this directory (although I’m having the same issue on my personal laptop). I’m using a Windows 10 PC and Matlab 2017a.
I’ve already tried restarting matlab, my pc, disconnecting and reconnecting the drive. I have not tried reinstalling Matlab yet. Any help would be greatly appreciated!
<</matlabcentral/answers/uploaded_files/94148/currentfolder.PNG>> Hi all,
I’ve been experiencing a weird problem starting in the last 2 days. When go to a certain directory in Matlab, it doesn’t show any of the files in that directory. Moreover, using the dir() command for that directory doesn’t show any files.
I know the directory exists and there are files in it because I can access it through the normal Windows file explorer (see image below).
A few additional details. The directory I’m trying to access is a network drive. However, I can still access other network drives from Matlab and other PCs on the same network are able to view the directory in Matlab. This makes me think its something specific to Matlab on my PC with this directory (although I’m having the same issue on my personal laptop). I’m using a Windows 10 PC and Matlab 2017a.
I’ve already tried restarting matlab, my pc, disconnecting and reconnecting the drive. I have not tried reinstalling Matlab yet. Any help would be greatly appreciated!
<</matlabcentral/answers/uploaded_files/94148/currentfolder.PNG>> current folder, directory MATLAB Answers — New Questions
Code checking error in App Designer, it thinks xxxx.BackgroundColor is a scalar
I’m changing the background color to alert the user about various (async) conditions and button pressed.
if app.STOPTRIALSButton.BackgroundColor == [0.95 0.33 0.10]
app.STOPTRIALSButton.BackgroundColor = [0 1 0];
end
and I get the warning:
Unexpected use of ‘[‘ in a scalar context.
but of course colors are RGB three values, so the code runs fine.
Also while the Component browser shows (another) color as 0.65, 0.65, 0.65 that’s the wrong syntax (no commas please for easier cut and paste) and in the startup code the values are actually [0.651 0.651 0.651] so my compare failed without significant research as to why.
Finally, App Designer isn’t listed as a product below, so I can’t flag that. Just saying MATLAB is the product.I’m changing the background color to alert the user about various (async) conditions and button pressed.
if app.STOPTRIALSButton.BackgroundColor == [0.95 0.33 0.10]
app.STOPTRIALSButton.BackgroundColor = [0 1 0];
end
and I get the warning:
Unexpected use of ‘[‘ in a scalar context.
but of course colors are RGB three values, so the code runs fine.
Also while the Component browser shows (another) color as 0.65, 0.65, 0.65 that’s the wrong syntax (no commas please for easier cut and paste) and in the startup code the values are actually [0.651 0.651 0.651] so my compare failed without significant research as to why.
Finally, App Designer isn’t listed as a product below, so I can’t flag that. Just saying MATLAB is the product. I’m changing the background color to alert the user about various (async) conditions and button pressed.
if app.STOPTRIALSButton.BackgroundColor == [0.95 0.33 0.10]
app.STOPTRIALSButton.BackgroundColor = [0 1 0];
end
and I get the warning:
Unexpected use of ‘[‘ in a scalar context.
but of course colors are RGB three values, so the code runs fine.
Also while the Component browser shows (another) color as 0.65, 0.65, 0.65 that’s the wrong syntax (no commas please for easier cut and paste) and in the startup code the values are actually [0.651 0.651 0.651] so my compare failed without significant research as to why.
Finally, App Designer isn’t listed as a product below, so I can’t flag that. Just saying MATLAB is the product. backgroundcolor, scalar MATLAB Answers — New Questions
Using logicals in arrayfun
Hi,
I have two arrays:
tempA, size(10, 18)
tempB, size(1, 10) (is a column vector),
For each row in tempA, I want to extract the number of columns specified for that row by tempB. However, there are certain rows in tempB that are ‘nan’.
u = arrayfun(@(x,y) x{1}(1:y), tempA, tempB, ‘UniformOutput’, false);
Can I use logical in the arrayfun so that it automatically excludes cases that are ‘nan’?
Thank youHi,
I have two arrays:
tempA, size(10, 18)
tempB, size(1, 10) (is a column vector),
For each row in tempA, I want to extract the number of columns specified for that row by tempB. However, there are certain rows in tempB that are ‘nan’.
u = arrayfun(@(x,y) x{1}(1:y), tempA, tempB, ‘UniformOutput’, false);
Can I use logical in the arrayfun so that it automatically excludes cases that are ‘nan’?
Thank you Hi,
I have two arrays:
tempA, size(10, 18)
tempB, size(1, 10) (is a column vector),
For each row in tempA, I want to extract the number of columns specified for that row by tempB. However, there are certain rows in tempB that are ‘nan’.
u = arrayfun(@(x,y) x{1}(1:y), tempA, tempB, ‘UniformOutput’, false);
Can I use logical in the arrayfun so that it automatically excludes cases that are ‘nan’?
Thank you array function, logicals MATLAB Answers — New Questions
Support for Com4 ManagedInProcessServer and others
Com servers may generally be written in managed or unmanaged code. The COM4 schema extensions include elements such as com4:ManagedInProcessServer which allows for the specification of a COM Clsid that is implemented in managed code.
Unfortunately, the foundation manifest was not extended to include these new categories when performing the uniqueness test to ensure that when other element references a CLSID that the ID is defined once and only once in the document. This was done when COM2 was added to the schema set but seems to have been missed here.
These are the lines from the latest SDK version of the FoundationManifestSchema.xsd (starting on line 134) which appear problematic to me:
<xs:unique name=”Class_Id”>
<xs:selector xpath=”f:Applications/f:Application/f:Extensions/com:Extension/com:ComServer/com:ExeServer/com:Class | f:Applications/f:Application/f:Extensions/com:Extension/com:ComServer/com:SurrogateServer/com:Class | f:Applications/f:Application/f:Extensions/com:Extension/com:ComServer/com:TreatAsClass | f:Applications/f:Application/f:Extensions/com:Extension/com:ComInterface/com:ProxyStub | f:Extensions/com:Extension/com:ComInterface/com:ProxyStub | f:Applications/f:Application/f:Extensions/com2:Extension/com2:ComServer/com:ExeServer/com:Class | f:Applications/f:Application/f:Extensions/com2:Extension/com2:ComServer/com:SurrogateServer/com:Class | f:Applications/f:Application/f:Extensions/com2:Extension/com2:ComServer/com:TreatAsClass | f:Applications/f:Application/f:Extensions/com2:Extension/com2:ComInterface/com:ProxyStub | f:Extensions/com2:Extension/com2:ComInterface/com:ProxyStub”/>
<xs:field xpath=”@Id”/>
</xs:unique>
This means that while we are able to add the ManagedInProcessServer element to define the com server component (as either a PackageExtension or ApplicationExtension as needed), we cannot reference the Clsid in any other element that would need it, such as a ShellExtension.
Fixing the above in the Foundation (and updating all tooling that verifying the schemas) will be necessary in order to build and package up more modern COM components.
Tagging @Fizza for visibility.
Com servers may generally be written in managed or unmanaged code. The COM4 schema extensions include elements such as com4:ManagedInProcessServer which allows for the specification of a COM Clsid that is implemented in managed code.
Unfortunately, the foundation manifest was not extended to include these new categories when performing the uniqueness test to ensure that when other element references a CLSID that the ID is defined once and only once in the document. This was done when COM2 was added to the schema set but seems to have been missed here.
These are the lines from the latest SDK version of the FoundationManifestSchema.xsd (starting on line 134) which appear problematic to me:
<xs:unique name=”Class_Id”><xs:selector xpath=”f:Applications/f:Application/f:Extensions/com:Extension/com:ComServer/com:ExeServer/com:Class | f:Applications/f:Application/f:Extensions/com:Extension/com:ComServer/com:SurrogateServer/com:Class | f:Applications/f:Application/f:Extensions/com:Extension/com:ComServer/com:TreatAsClass | f:Applications/f:Application/f:Extensions/com:Extension/com:ComInterface/com:ProxyStub | f:Extensions/com:Extension/com:ComInterface/com:ProxyStub | f:Applications/f:Application/f:Extensions/com2:Extension/com2:ComServer/com:ExeServer/com:Class | f:Applications/f:Application/f:Extensions/com2:Extension/com2:ComServer/com:SurrogateServer/com:Class | f:Applications/f:Application/f:Extensions/com2:Extension/com2:ComServer/com:TreatAsClass | f:Applications/f:Application/f:Extensions/com2:Extension/com2:ComInterface/com:ProxyStub | f:Extensions/com2:Extension/com2:ComInterface/com:ProxyStub”/><xs:field xpath=”@Id”/></xs:unique>
This means that while we are able to add the ManagedInProcessServer element to define the com server component (as either a PackageExtension or ApplicationExtension as needed), we cannot reference the Clsid in any other element that would need it, such as a ShellExtension.
Fixing the above in the Foundation (and updating all tooling that verifying the schemas) will be necessary in order to build and package up more modern COM components.
Tagging @Fizza for visibility. Read More
Schedule View By Event
I have a given excel schedule, similar to the “Input” table shown in the attachment. I have a need to organize this table by event and show participants in individual columns. I have this table shown as “Desired Output” in the attachment I also have Power BI available to me, if that provides an easier solution.
Any and all help would be greatly appreciated, thank you.
I have a given excel schedule, similar to the “Input” table shown in the attachment. I have a need to organize this table by event and show participants in individual columns. I have this table shown as “Desired Output” in the attachment I also have Power BI available to me, if that provides an easier solution. Any and all help would be greatly appreciated, thank you. Sheet Read More
Find Selected button in Button Group CallBack
I have a group of 5 buttons (including a dummy one for "none"
In the callback I want to do something based on the selected button.
Here is the default start of the callback:
function ManualRewardsButtonGroupSelectionChanged(app, event)
selectedButton = app.ManualRewardsButtonGroup.SelectedObject;
So why is the selectedButton always 1? I’d like 1 to 5 so I can do the appropriate
switch app.ManualRewardsButtonGroup.NewValue % or what here?
case 1
% Give LA reward
case 2
% Give LB reward
case 3
% Give RA reward
case 4
% Give RA reward
end
Help shows that I could use event.NewValue (since selectedButton doesn’t work) but that also returns 1
https://www.mathworks.com/help/matlab/ref/matlab.ui.container.buttongroup-properties.html
Van I set each button to have a different value? (I clicked on the Value check box in the Component but no opportunity to set it. I’ve even tried assigning Tags though I don’t know what they do.
App Designer doesn’t set values and doesn’t let me do it either
% Create ManualRewardsButtonGroup
app.ManualRewardsButtonGroup = uibuttongroup(app.RewardControls);
app.ManualRewardsButtonGroup.AutoResizeChildren = ‘off’;
app.ManualRewardsButtonGroup.SelectionChangedFcn = createCallbackFcn(app, @ManualRewardsButtonGroupSelectionChanged, true);
app.ManualRewardsButtonGroup.TitlePosition = ‘centertop’;
app.ManualRewardsButtonGroup.Title = ‘Manual Rewards’;
app.ManualRewardsButtonGroup.FontSize = 10;
app.ManualRewardsButtonGroup.Position = [38 6 190 95];
% Create Rew_ALeftButton
app.Rew_ALeftButton = uiradiobutton(app.ManualRewardsButtonGroup);
app.Rew_ALeftButton.Tag = ‘1’;
app.Rew_ALeftButton.Text = ‘Rew_A Left’;
app.Rew_ALeftButton.FontSize = 10;
app.Rew_ALeftButton.Position = [11 49 73 22];
% Create Rew_ARightButton
app.Rew_ARightButton = uiradiobutton(app.ManualRewardsButtonGroup);
app.Rew_ARightButton.Tag = ‘2’;
app.Rew_ARightButton.Text = ‘Rew_A Right’;
app.Rew_ARightButton.FontSize = 10;
app.Rew_ARightButton.Position = [101 49 80 22];
% Create Rew_BLeftButton
app.Rew_BLeftButton = uiradiobutton(app.ManualRewardsButtonGroup);
app.Rew_BLeftButton.Tag = ‘3’;
app.Rew_BLeftButton.Text = ‘Rew_B Left’;
app.Rew_BLeftButton.FontSize = 10;
app.Rew_BLeftButton.Position = [11 27 74 22];
% Create Rew_BRightButton
app.Rew_BRightButton = uiradiobutton(app.ManualRewardsButtonGroup);
app.Rew_BRightButton.Tag = ‘4’;
app.Rew_BRightButton.Text = ‘Rew_B Right’;
app.Rew_BRightButton.FontSize = 10;
app.Rew_BRightButton.Position = [101 27 80 22];
app.Rew_BRightButton.Value = true;
% Create OffButton
app.OffButton = uiradiobutton(app.ManualRewardsButtonGroup);
app.OffButton.Text = ‘Off’;
app.OffButton.FontSize = 10;
app.OffButton.Position = [85 5 35 22];% Create ManualRewardsButtonGroup
app.ManualRewardsButtonGroup = uibuttongroup(app.RewardControls);
app.ManualRewardsButtonGroup.AutoResizeChildren = ‘off’;
app.ManualRewardsButtonGroup.SelectionChangedFcn = createCallbackFcn(app, @ManualRewardsButtonGroupSelectionChanged, true);
app.ManualRewardsButtonGroup.TitlePosition = ‘centertop’;
app.ManualRewardsButtonGroup.Title = ‘Manual Rewards’;
app.ManualRewardsButtonGroup.FontSize = 10;
app.ManualRewardsButtonGroup.Position = [38 6 190 95];
% Create Rew_ALeftButton
app.Rew_ALeftButton = uiradiobutton(app.ManualRewardsButtonGroup);
app.Rew_ALeftButton.Tag = ‘1’;
app.Rew_ALeftButton.Text = ‘Rew_A Left’;
app.Rew_ALeftButton.FontSize = 10;
app.Rew_ALeftButton.Position = [11 49 73 22];
% Create Rew_ARightButton
app.Rew_ARightButton = uiradiobutton(app.ManualRewardsButtonGroup);
app.Rew_ARightButton.Tag = ‘2’;
app.Rew_ARightButton.Text = ‘Rew_A Right’;
app.Rew_ARightButton.FontSize = 10;
app.Rew_ARightButton.Position = [101 49 80 22];
% Create Rew_BLeftButton
app.Rew_BLeftButton = uiradiobutton(app.ManualRewardsButtonGroup);
app.Rew_BLeftButton.Tag = ‘3’;
app.Rew_BLeftButton.Text = ‘Rew_B Left’;
app.Rew_BLeftButton.FontSize = 10;
app.Rew_BLeftButton.Position = [11 27 74 22];
% Create Rew_BRightButton
app.Rew_BRightButton = uiradiobutton(app.ManualRewardsButtonGroup);
app.Rew_BRightButton.Tag = ‘4’;
app.Rew_BRightButton.Text = ‘Rew_B Right’;
app.Rew_BRightButton.FontSize = 10;
app.Rew_BRightButton.Position = [101 27 80 22];
app.Rew_BRightButton.Value = true;
% Create OffButton
app.OffButton = uiradiobutton(app.ManualRewardsButtonGroup);
app.OffButton.Text = ‘Off’;
app.OffButton.FontSize = 10;
app.OffButton.Position = [85 5 35 22];I have a group of 5 buttons (including a dummy one for "none"
In the callback I want to do something based on the selected button.
Here is the default start of the callback:
function ManualRewardsButtonGroupSelectionChanged(app, event)
selectedButton = app.ManualRewardsButtonGroup.SelectedObject;
So why is the selectedButton always 1? I’d like 1 to 5 so I can do the appropriate
switch app.ManualRewardsButtonGroup.NewValue % or what here?
case 1
% Give LA reward
case 2
% Give LB reward
case 3
% Give RA reward
case 4
% Give RA reward
end
Help shows that I could use event.NewValue (since selectedButton doesn’t work) but that also returns 1
https://www.mathworks.com/help/matlab/ref/matlab.ui.container.buttongroup-properties.html
Van I set each button to have a different value? (I clicked on the Value check box in the Component but no opportunity to set it. I’ve even tried assigning Tags though I don’t know what they do.
App Designer doesn’t set values and doesn’t let me do it either
% Create ManualRewardsButtonGroup
app.ManualRewardsButtonGroup = uibuttongroup(app.RewardControls);
app.ManualRewardsButtonGroup.AutoResizeChildren = ‘off’;
app.ManualRewardsButtonGroup.SelectionChangedFcn = createCallbackFcn(app, @ManualRewardsButtonGroupSelectionChanged, true);
app.ManualRewardsButtonGroup.TitlePosition = ‘centertop’;
app.ManualRewardsButtonGroup.Title = ‘Manual Rewards’;
app.ManualRewardsButtonGroup.FontSize = 10;
app.ManualRewardsButtonGroup.Position = [38 6 190 95];
% Create Rew_ALeftButton
app.Rew_ALeftButton = uiradiobutton(app.ManualRewardsButtonGroup);
app.Rew_ALeftButton.Tag = ‘1’;
app.Rew_ALeftButton.Text = ‘Rew_A Left’;
app.Rew_ALeftButton.FontSize = 10;
app.Rew_ALeftButton.Position = [11 49 73 22];
% Create Rew_ARightButton
app.Rew_ARightButton = uiradiobutton(app.ManualRewardsButtonGroup);
app.Rew_ARightButton.Tag = ‘2’;
app.Rew_ARightButton.Text = ‘Rew_A Right’;
app.Rew_ARightButton.FontSize = 10;
app.Rew_ARightButton.Position = [101 49 80 22];
% Create Rew_BLeftButton
app.Rew_BLeftButton = uiradiobutton(app.ManualRewardsButtonGroup);
app.Rew_BLeftButton.Tag = ‘3’;
app.Rew_BLeftButton.Text = ‘Rew_B Left’;
app.Rew_BLeftButton.FontSize = 10;
app.Rew_BLeftButton.Position = [11 27 74 22];
% Create Rew_BRightButton
app.Rew_BRightButton = uiradiobutton(app.ManualRewardsButtonGroup);
app.Rew_BRightButton.Tag = ‘4’;
app.Rew_BRightButton.Text = ‘Rew_B Right’;
app.Rew_BRightButton.FontSize = 10;
app.Rew_BRightButton.Position = [101 27 80 22];
app.Rew_BRightButton.Value = true;
% Create OffButton
app.OffButton = uiradiobutton(app.ManualRewardsButtonGroup);
app.OffButton.Text = ‘Off’;
app.OffButton.FontSize = 10;
app.OffButton.Position = [85 5 35 22];% Create ManualRewardsButtonGroup
app.ManualRewardsButtonGroup = uibuttongroup(app.RewardControls);
app.ManualRewardsButtonGroup.AutoResizeChildren = ‘off’;
app.ManualRewardsButtonGroup.SelectionChangedFcn = createCallbackFcn(app, @ManualRewardsButtonGroupSelectionChanged, true);
app.ManualRewardsButtonGroup.TitlePosition = ‘centertop’;
app.ManualRewardsButtonGroup.Title = ‘Manual Rewards’;
app.ManualRewardsButtonGroup.FontSize = 10;
app.ManualRewardsButtonGroup.Position = [38 6 190 95];
% Create Rew_ALeftButton
app.Rew_ALeftButton = uiradiobutton(app.ManualRewardsButtonGroup);
app.Rew_ALeftButton.Tag = ‘1’;
app.Rew_ALeftButton.Text = ‘Rew_A Left’;
app.Rew_ALeftButton.FontSize = 10;
app.Rew_ALeftButton.Position = [11 49 73 22];
% Create Rew_ARightButton
app.Rew_ARightButton = uiradiobutton(app.ManualRewardsButtonGroup);
app.Rew_ARightButton.Tag = ‘2’;
app.Rew_ARightButton.Text = ‘Rew_A Right’;
app.Rew_ARightButton.FontSize = 10;
app.Rew_ARightButton.Position = [101 49 80 22];
% Create Rew_BLeftButton
app.Rew_BLeftButton = uiradiobutton(app.ManualRewardsButtonGroup);
app.Rew_BLeftButton.Tag = ‘3’;
app.Rew_BLeftButton.Text = ‘Rew_B Left’;
app.Rew_BLeftButton.FontSize = 10;
app.Rew_BLeftButton.Position = [11 27 74 22];
% Create Rew_BRightButton
app.Rew_BRightButton = uiradiobutton(app.ManualRewardsButtonGroup);
app.Rew_BRightButton.Tag = ‘4’;
app.Rew_BRightButton.Text = ‘Rew_B Right’;
app.Rew_BRightButton.FontSize = 10;
app.Rew_BRightButton.Position = [101 27 80 22];
app.Rew_BRightButton.Value = true;
% Create OffButton
app.OffButton = uiradiobutton(app.ManualRewardsButtonGroup);
app.OffButton.Text = ‘Off’;
app.OffButton.FontSize = 10;
app.OffButton.Position = [85 5 35 22]; I have a group of 5 buttons (including a dummy one for "none"
In the callback I want to do something based on the selected button.
Here is the default start of the callback:
function ManualRewardsButtonGroupSelectionChanged(app, event)
selectedButton = app.ManualRewardsButtonGroup.SelectedObject;
So why is the selectedButton always 1? I’d like 1 to 5 so I can do the appropriate
switch app.ManualRewardsButtonGroup.NewValue % or what here?
case 1
% Give LA reward
case 2
% Give LB reward
case 3
% Give RA reward
case 4
% Give RA reward
end
Help shows that I could use event.NewValue (since selectedButton doesn’t work) but that also returns 1
https://www.mathworks.com/help/matlab/ref/matlab.ui.container.buttongroup-properties.html
Van I set each button to have a different value? (I clicked on the Value check box in the Component but no opportunity to set it. I’ve even tried assigning Tags though I don’t know what they do.
App Designer doesn’t set values and doesn’t let me do it either
% Create ManualRewardsButtonGroup
app.ManualRewardsButtonGroup = uibuttongroup(app.RewardControls);
app.ManualRewardsButtonGroup.AutoResizeChildren = ‘off’;
app.ManualRewardsButtonGroup.SelectionChangedFcn = createCallbackFcn(app, @ManualRewardsButtonGroupSelectionChanged, true);
app.ManualRewardsButtonGroup.TitlePosition = ‘centertop’;
app.ManualRewardsButtonGroup.Title = ‘Manual Rewards’;
app.ManualRewardsButtonGroup.FontSize = 10;
app.ManualRewardsButtonGroup.Position = [38 6 190 95];
% Create Rew_ALeftButton
app.Rew_ALeftButton = uiradiobutton(app.ManualRewardsButtonGroup);
app.Rew_ALeftButton.Tag = ‘1’;
app.Rew_ALeftButton.Text = ‘Rew_A Left’;
app.Rew_ALeftButton.FontSize = 10;
app.Rew_ALeftButton.Position = [11 49 73 22];
% Create Rew_ARightButton
app.Rew_ARightButton = uiradiobutton(app.ManualRewardsButtonGroup);
app.Rew_ARightButton.Tag = ‘2’;
app.Rew_ARightButton.Text = ‘Rew_A Right’;
app.Rew_ARightButton.FontSize = 10;
app.Rew_ARightButton.Position = [101 49 80 22];
% Create Rew_BLeftButton
app.Rew_BLeftButton = uiradiobutton(app.ManualRewardsButtonGroup);
app.Rew_BLeftButton.Tag = ‘3’;
app.Rew_BLeftButton.Text = ‘Rew_B Left’;
app.Rew_BLeftButton.FontSize = 10;
app.Rew_BLeftButton.Position = [11 27 74 22];
% Create Rew_BRightButton
app.Rew_BRightButton = uiradiobutton(app.ManualRewardsButtonGroup);
app.Rew_BRightButton.Tag = ‘4’;
app.Rew_BRightButton.Text = ‘Rew_B Right’;
app.Rew_BRightButton.FontSize = 10;
app.Rew_BRightButton.Position = [101 27 80 22];
app.Rew_BRightButton.Value = true;
% Create OffButton
app.OffButton = uiradiobutton(app.ManualRewardsButtonGroup);
app.OffButton.Text = ‘Off’;
app.OffButton.FontSize = 10;
app.OffButton.Position = [85 5 35 22];% Create ManualRewardsButtonGroup
app.ManualRewardsButtonGroup = uibuttongroup(app.RewardControls);
app.ManualRewardsButtonGroup.AutoResizeChildren = ‘off’;
app.ManualRewardsButtonGroup.SelectionChangedFcn = createCallbackFcn(app, @ManualRewardsButtonGroupSelectionChanged, true);
app.ManualRewardsButtonGroup.TitlePosition = ‘centertop’;
app.ManualRewardsButtonGroup.Title = ‘Manual Rewards’;
app.ManualRewardsButtonGroup.FontSize = 10;
app.ManualRewardsButtonGroup.Position = [38 6 190 95];
% Create Rew_ALeftButton
app.Rew_ALeftButton = uiradiobutton(app.ManualRewardsButtonGroup);
app.Rew_ALeftButton.Tag = ‘1’;
app.Rew_ALeftButton.Text = ‘Rew_A Left’;
app.Rew_ALeftButton.FontSize = 10;
app.Rew_ALeftButton.Position = [11 49 73 22];
% Create Rew_ARightButton
app.Rew_ARightButton = uiradiobutton(app.ManualRewardsButtonGroup);
app.Rew_ARightButton.Tag = ‘2’;
app.Rew_ARightButton.Text = ‘Rew_A Right’;
app.Rew_ARightButton.FontSize = 10;
app.Rew_ARightButton.Position = [101 49 80 22];
% Create Rew_BLeftButton
app.Rew_BLeftButton = uiradiobutton(app.ManualRewardsButtonGroup);
app.Rew_BLeftButton.Tag = ‘3’;
app.Rew_BLeftButton.Text = ‘Rew_B Left’;
app.Rew_BLeftButton.FontSize = 10;
app.Rew_BLeftButton.Position = [11 27 74 22];
% Create Rew_BRightButton
app.Rew_BRightButton = uiradiobutton(app.ManualRewardsButtonGroup);
app.Rew_BRightButton.Tag = ‘4’;
app.Rew_BRightButton.Text = ‘Rew_B Right’;
app.Rew_BRightButton.FontSize = 10;
app.Rew_BRightButton.Position = [101 27 80 22];
app.Rew_BRightButton.Value = true;
% Create OffButton
app.OffButton = uiradiobutton(app.ManualRewardsButtonGroup);
app.OffButton.Text = ‘Off’;
app.OffButton.FontSize = 10;
app.OffButton.Position = [85 5 35 22]; buttongroup, selectedbutton MATLAB Answers — New Questions
How to explain an ANN graph?
Hello, I’m trying to interpret an ANN graph for a paper, but I’m not sure where to start. Should I explain the detailed computations for each data point, or is there a general approach I can use to highlight the overall differences between the two groups? How do I know if it’s considered a good result? Here’s a sample figure
y=root mean square error, x=number of neurons
y=displacement, x=time pointsHello, I’m trying to interpret an ANN graph for a paper, but I’m not sure where to start. Should I explain the detailed computations for each data point, or is there a general approach I can use to highlight the overall differences between the two groups? How do I know if it’s considered a good result? Here’s a sample figure
y=root mean square error, x=number of neurons
y=displacement, x=time points Hello, I’m trying to interpret an ANN graph for a paper, but I’m not sure where to start. Should I explain the detailed computations for each data point, or is there a general approach I can use to highlight the overall differences between the two groups? How do I know if it’s considered a good result? Here’s a sample figure
y=root mean square error, x=number of neurons
y=displacement, x=time points graph, neural network MATLAB Answers — New Questions
Plotted data changes when applying custom colormap
Question:
I am working with a very specific custom color map, that i apply to a matrix using imagesc(). When plotting my data with a built in colormap, everything plots as expected. When im using my own colormap, the image itself plots and the colormap works wonderfully. However: The y-coordinates of my matrix shift! The x coordinates are unaffected by this. Everything is plotted within the same figure, the only difference between the figures is the colormap() command. I hope this is a well known problem, however my internet search didnt resolve the issue.
Thanks in advance for anyone who answers!
Appendix:
Plot with built in colormap: Note that the top-right most pixel of my data is at [30, 60]
imagesc(Vertical_Time_Matrix);
Plot with my Colormap: Note that the top-right most pixel of my data is at [30, 78]
imagesc(Vertical_Time_Matrix);
colormap(myColormap);
A few words regarding my colormap and the data. The data is a 220×30 matrix containing numbers ("States") from 1-28. Depending on the state a color is chosen from a color map. Hence, the colormap is 28 entries long.Question:
I am working with a very specific custom color map, that i apply to a matrix using imagesc(). When plotting my data with a built in colormap, everything plots as expected. When im using my own colormap, the image itself plots and the colormap works wonderfully. However: The y-coordinates of my matrix shift! The x coordinates are unaffected by this. Everything is plotted within the same figure, the only difference between the figures is the colormap() command. I hope this is a well known problem, however my internet search didnt resolve the issue.
Thanks in advance for anyone who answers!
Appendix:
Plot with built in colormap: Note that the top-right most pixel of my data is at [30, 60]
imagesc(Vertical_Time_Matrix);
Plot with my Colormap: Note that the top-right most pixel of my data is at [30, 78]
imagesc(Vertical_Time_Matrix);
colormap(myColormap);
A few words regarding my colormap and the data. The data is a 220×30 matrix containing numbers ("States") from 1-28. Depending on the state a color is chosen from a color map. Hence, the colormap is 28 entries long. Question:
I am working with a very specific custom color map, that i apply to a matrix using imagesc(). When plotting my data with a built in colormap, everything plots as expected. When im using my own colormap, the image itself plots and the colormap works wonderfully. However: The y-coordinates of my matrix shift! The x coordinates are unaffected by this. Everything is plotted within the same figure, the only difference between the figures is the colormap() command. I hope this is a well known problem, however my internet search didnt resolve the issue.
Thanks in advance for anyone who answers!
Appendix:
Plot with built in colormap: Note that the top-right most pixel of my data is at [30, 60]
imagesc(Vertical_Time_Matrix);
Plot with my Colormap: Note that the top-right most pixel of my data is at [30, 78]
imagesc(Vertical_Time_Matrix);
colormap(myColormap);
A few words regarding my colormap and the data. The data is a 220×30 matrix containing numbers ("States") from 1-28. Depending on the state a color is chosen from a color map. Hence, the colormap is 28 entries long. colormap, figure MATLAB Answers — New Questions
Hotmail is conflicting with OneDrive through my Microsoft 365 account
I’ve been struggling for about one month, thus far, with my OneDrive through my Microsoft 365 account. Suddenly, my 40-year-old Hotmail account is frozen, and I cannot send, nor receive email because, the system indicates, my OneDrive is full. When I click on that, to be led to my OneDrive, it shows that I have used less than 0.2 GB from the 15 GB allocated space. In the meantime, I cannot use my Hotmail, where I have deleted about 10 GB of files. This has not helped me, and I cannot reach someone from Microsoft for assistance. Does anyone have any clue about how to resolve this mystery?
I’ve been struggling for about one month, thus far, with my OneDrive through my Microsoft 365 account. Suddenly, my 40-year-old Hotmail account is frozen, and I cannot send, nor receive email because, the system indicates, my OneDrive is full. When I click on that, to be led to my OneDrive, it shows that I have used less than 0.2 GB from the 15 GB allocated space. In the meantime, I cannot use my Hotmail, where I have deleted about 10 GB of files. This has not helped me, and I cannot reach someone from Microsoft for assistance. Does anyone have any clue about how to resolve this mystery? Read More
Generating data for a scheduling problem that deals with setup families.
clc
%clear all
x1 = 1:30;
x11 = 1:10;
x2 = 0.2:0.2:1;
x12 = 0.2:0.2:4;
%a = 6:100;%b = 125:25:1000;%c = 1050:50:2500;%cc = 3000:500:10000;%cc1 = 15000:5000:30000;%xx = [a b c cc cc1];%[A, B] = size(xx);%mx = max(xx);
n=input(‘n=’);
data = cell(n, 10, 3);
%families = 1:6;
familiesNumber = [2, 4, 6];
familiesSetup = 1:6;
for F = 1:3
% for j = 1:B
% n = xx(j);
% for ex = 1:10
releaseDate = randsrc(1, n, x1);
processingTime = randsrc(1, n, x11);
tf = randsrc(1, 1, x12);
ddr = randsrc(1, 1, x2);
g = sum(releaseDate);
min_due = tf * (1 – g – ddr / 2);
max_due = tf * (1 – g + ddr / 2);
if min_due < 1, min_due = 1; end % Ensure minimum due date is at least 1
if max_due < min_due, max_due = min_due + 1; end % Ensure range is valid
x = min_due:max_due;%linspace(min_due, max_due, n); % Create a range with n points
dueDate = randsrc(1, n, x); % Generate dueDate within the range
% Ensure dueDate is valid
for i = 1:n
if dueDate(i) < processingTime(i)
dueDate(i) = processingTime(i) + 1;
end
end
f = familiesNumber(F); disp(f);
m = mod(n, f);
nn = n – m;
setupTime = zeros(1, f);
for s = 1:f
setupTime = familiesSetup;
end
disp(setupTime);
nf = zeros(1, f);
if m == 0
for i = 1:f
nf(i) = n / f;
end
else
for i = 1:m
nf(i) = (nn / f) + 1;
end
for ii = m+1:f
nf(ii) = (nn / f);
end
end
disp(nf);
da = [releaseDate; processingTime; dueDate]; %disp(da);
data{n, ex, f} = da;
% end
% end
end
savefile = ‘a_data.mat’;
save(savefile, ‘data’);clc
%clear all
x1 = 1:30;
x11 = 1:10;
x2 = 0.2:0.2:1;
x12 = 0.2:0.2:4;
%a = 6:100;%b = 125:25:1000;%c = 1050:50:2500;%cc = 3000:500:10000;%cc1 = 15000:5000:30000;%xx = [a b c cc cc1];%[A, B] = size(xx);%mx = max(xx);
n=input(‘n=’);
data = cell(n, 10, 3);
%families = 1:6;
familiesNumber = [2, 4, 6];
familiesSetup = 1:6;
for F = 1:3
% for j = 1:B
% n = xx(j);
% for ex = 1:10
releaseDate = randsrc(1, n, x1);
processingTime = randsrc(1, n, x11);
tf = randsrc(1, 1, x12);
ddr = randsrc(1, 1, x2);
g = sum(releaseDate);
min_due = tf * (1 – g – ddr / 2);
max_due = tf * (1 – g + ddr / 2);
if min_due < 1, min_due = 1; end % Ensure minimum due date is at least 1
if max_due < min_due, max_due = min_due + 1; end % Ensure range is valid
x = min_due:max_due;%linspace(min_due, max_due, n); % Create a range with n points
dueDate = randsrc(1, n, x); % Generate dueDate within the range
% Ensure dueDate is valid
for i = 1:n
if dueDate(i) < processingTime(i)
dueDate(i) = processingTime(i) + 1;
end
end
f = familiesNumber(F); disp(f);
m = mod(n, f);
nn = n – m;
setupTime = zeros(1, f);
for s = 1:f
setupTime = familiesSetup;
end
disp(setupTime);
nf = zeros(1, f);
if m == 0
for i = 1:f
nf(i) = n / f;
end
else
for i = 1:m
nf(i) = (nn / f) + 1;
end
for ii = m+1:f
nf(ii) = (nn / f);
end
end
disp(nf);
da = [releaseDate; processingTime; dueDate]; %disp(da);
data{n, ex, f} = da;
% end
% end
end
savefile = ‘a_data.mat’;
save(savefile, ‘data’); clc
%clear all
x1 = 1:30;
x11 = 1:10;
x2 = 0.2:0.2:1;
x12 = 0.2:0.2:4;
%a = 6:100;%b = 125:25:1000;%c = 1050:50:2500;%cc = 3000:500:10000;%cc1 = 15000:5000:30000;%xx = [a b c cc cc1];%[A, B] = size(xx);%mx = max(xx);
n=input(‘n=’);
data = cell(n, 10, 3);
%families = 1:6;
familiesNumber = [2, 4, 6];
familiesSetup = 1:6;
for F = 1:3
% for j = 1:B
% n = xx(j);
% for ex = 1:10
releaseDate = randsrc(1, n, x1);
processingTime = randsrc(1, n, x11);
tf = randsrc(1, 1, x12);
ddr = randsrc(1, 1, x2);
g = sum(releaseDate);
min_due = tf * (1 – g – ddr / 2);
max_due = tf * (1 – g + ddr / 2);
if min_due < 1, min_due = 1; end % Ensure minimum due date is at least 1
if max_due < min_due, max_due = min_due + 1; end % Ensure range is valid
x = min_due:max_due;%linspace(min_due, max_due, n); % Create a range with n points
dueDate = randsrc(1, n, x); % Generate dueDate within the range
% Ensure dueDate is valid
for i = 1:n
if dueDate(i) < processingTime(i)
dueDate(i) = processingTime(i) + 1;
end
end
f = familiesNumber(F); disp(f);
m = mod(n, f);
nn = n – m;
setupTime = zeros(1, f);
for s = 1:f
setupTime = familiesSetup;
end
disp(setupTime);
nf = zeros(1, f);
if m == 0
for i = 1:f
nf(i) = n / f;
end
else
for i = 1:m
nf(i) = (nn / f) + 1;
end
for ii = m+1:f
nf(ii) = (nn / f);
end
end
disp(nf);
da = [releaseDate; processingTime; dueDate]; %disp(da);
data{n, ex, f} = da;
% end
% end
end
savefile = ‘a_data.mat’;
save(savefile, ‘data’); data generating, setup families, matlab, single machine scheduling problem MATLAB Answers — New Questions
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I am encountering an issue while trying to install WSL on my Windows 11 Insider Program 24H2. The installation process fails with the following error message:
“`
Installing, this may take a few minutes…
WslRegisterDistribution failed with error: 0x80070003
Error: 0x80070003 The system cannot find the path specified.
Press any key to continue…
“`
This problem started after I accidentally deleted the WSL folder from the Program Files directory. Since then, I have been unable to install WSL (Ubuntu) on my system.
Could someone please guide me on how to resolve this issue?
Thank you in advance for your help!
I am encountering an issue while trying to install WSL on my Windows 11 Insider Program 24H2. The installation process fails with the following error message:“`Installing, this may take a few minutes…WslRegisterDistribution failed with error: 0x80070003Error: 0x80070003 The system cannot find the path specified.Press any key to continue…“`This problem started after I accidentally deleted the WSL folder from the Program Files directory. Since then, I have been unable to install WSL (Ubuntu) on my system.Could someone please guide me on how to resolve this issue?Thank you in advance for your help! Read More
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Thank you. Hello, guys. I have some data points. When I plot these points in Matlab, it connects data points by line. However, I want to connect these points using a curve instead of a line graph for good representation. So please suggest to me which function we should use.
Thank you. plotting, curve, matlab, expert MATLAB Answers — New Questions
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We have 4 servers, 2 exchange 2013 servers and 2 exchange 2016 servers, if all the servers are on ECP porta is accessible, but if we turn off 2 exchange 2013 servers, then ECP portal is not accessible, checked all the internal & External URL’s for ECP and OWA, checked the bindings, checked, SPN’s, checked DNS and all the configurations are Good, nothing we are able to get from Event logs also
Can anyone help on this… we want to get rid of exchange 2013 and run with exchange 2016 servers
We have 4 servers, 2 exchange 2013 servers and 2 exchange 2016 servers, if all the servers are on ECP porta is accessible, but if we turn off 2 exchange 2013 servers, then ECP portal is not accessible, checked all the internal & External URL’s for ECP and OWA, checked the bindings, checked, SPN’s, checked DNS and all the configurations are Good, nothing we are able to get from Event logs also Can anyone help on this… we want to get rid of exchange 2013 and run with exchange 2016 servers Read More
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Hi all,
Just updated today to Windows 11. Upon my computer restarting, I got a ‘sign in’ button beneath my account instead of a box to enter my pin. When I sign in, it takes me to Microsoft account sign in and says “That Microsoft account doesn’t exist”. Before updating my PC, I changed the account associated with my Microsoft account to an Outlook one from the Gmail one I had been using, which is what comes up when I initially go to log in.
I go through the sign in process with my new email and it takes me to the Create a Pin screen. I click next and it says “Something went wrong. We weren’t able to set up your PIN.” It gives me the option to retry, which just takes me back to the “Something went wrong.”, or “Skip for now”, which takes me back to my computer’s lock screen and makes me go through the whole process again.
What can I do to get back into my computer? I’m not reliant on it for work, but I do have important files and some games installed on there.
Thanks in advance.
Hi all, Just updated today to Windows 11. Upon my computer restarting, I got a ‘sign in’ button beneath my account instead of a box to enter my pin. When I sign in, it takes me to Microsoft account sign in and says “That Microsoft account doesn’t exist”. Before updating my PC, I changed the account associated with my Microsoft account to an Outlook one from the Gmail one I had been using, which is what comes up when I initially go to log in. I go through the sign in process with my new email and it takes me to the Create a Pin screen. I click next and it says “Something went wrong. We weren’t able to set up your PIN.” It gives me the option to retry, which just takes me back to the “Something went wrong.”, or “Skip for now”, which takes me back to my computer’s lock screen and makes me go through the whole process again. What can I do to get back into my computer? I’m not reliant on it for work, but I do have important files and some games installed on there. Thanks in advance. Read More
Unraveling the future of POP3 on Outlook email accounts
Does 365 allow POP3 email?
For email, I use the Outlook 2016 desktop client with POP3. This client software reportedly ends in about 15 months. Is there any way I could stay with both POP3 and Outlook email after next year, other than by using third party clients like Thunderbird or Em-Client? Is it correct that 365 Outlook does not, or soon will not, support POP3 for Outlook email? [I assume outgoing Windows Mail and now Outlook for Windows excludes POP3.]
Does 365 allow POP3 email?For email, I use the Outlook 2016 desktop client with POP3. This client software reportedly ends in about 15 months. Is there any way I could stay with both POP3 and Outlook email after next year, other than by using third party clients like Thunderbird or Em-Client? Is it correct that 365 Outlook does not, or soon will not, support POP3 for Outlook email? [I assume outgoing Windows Mail and now Outlook for Windows excludes POP3.] Read More
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i am doing my final year project on simulation software for drying grains using mat lab but i do not have an idea on how to start on my methologyi am doing my final year project on simulation software for drying grains using mat lab but i do not have an idea on how to start on my methology i am doing my final year project on simulation software for drying grains using mat lab but i do not have an idea on how to start on my methology simulation, simulation-of-drying MATLAB Answers — New Questions
Unrecognized function or variable ‘efficiency’
I am getting the following error but can’t figure out why.
Unrecognized function or variable ‘efficiency’.
I am using a full version of MATLAB 2024a with an academic license and all of the toolboxes installed on April 3.
I regularly use the functions from the Antenna Toolbox so I am confident that it is installed correctly. I’ve also tried copying the text from the Help Center article to make sure there weren’t any typos.
‘doc efficiency’ and ‘help efficiency’ in the command window work as expected.
Is the ‘efficiency’ function still available in 2024a? Is there any other reason why it might not be recognized?I am getting the following error but can’t figure out why.
Unrecognized function or variable ‘efficiency’.
I am using a full version of MATLAB 2024a with an academic license and all of the toolboxes installed on April 3.
I regularly use the functions from the Antenna Toolbox so I am confident that it is installed correctly. I’ve also tried copying the text from the Help Center article to make sure there weren’t any typos.
‘doc efficiency’ and ‘help efficiency’ in the command window work as expected.
Is the ‘efficiency’ function still available in 2024a? Is there any other reason why it might not be recognized? I am getting the following error but can’t figure out why.
Unrecognized function or variable ‘efficiency’.
I am using a full version of MATLAB 2024a with an academic license and all of the toolboxes installed on April 3.
I regularly use the functions from the Antenna Toolbox so I am confident that it is installed correctly. I’ve also tried copying the text from the Help Center article to make sure there weren’t any typos.
‘doc efficiency’ and ‘help efficiency’ in the command window work as expected.
Is the ‘efficiency’ function still available in 2024a? Is there any other reason why it might not be recognized? antenna toolbox, efficiency MATLAB Answers — New Questions