Month: June 2024
Recents list keeps being cleared
I use Office 365 on Windows-10. Several times in the last year or so when I opened Word or other Office app, I have found that the “Recents” list has been cleared, though I took no action to do so. I suspect this may be happening when Office is updated, but I’m not sure. Can I prevent the clearing from happening? It is causing me a considerable amount of work to track down the documents that were on the “Recents” list.
I use Office 365 on Windows-10. Several times in the last year or so when I opened Word or other Office app, I have found that the “Recents” list has been cleared, though I took no action to do so. I suspect this may be happening when Office is updated, but I’m not sure. Can I prevent the clearing from happening? It is causing me a considerable amount of work to track down the documents that were on the “Recents” list. Read More
How to hide lines in pivot table, that have no value in a certain sum-value-column?
Hello everyone,
I hope someone is willing to help me with my Pivot Challenge. The actual question is below.
I use a Pivot Table on the following datamodel:
In my Pivot Table I see the spent time on a project. This works fine:
People from Team “Malt” worked on Tasks which belong to Project AAA-11111 and spent 2.13 days. They also worked on Project AAA-11202 and spent 9.88 days.
I like to compare those spent days with the estimations given for project AAA-11111. But if I add the estimation to the values I see all projects and not only those team Malt worked on:
For Project AAA-11111 and AAA-11202 the entries are correct:
How can I make Excel continue to filter on the SPENT entries but still show the ESTIMATE sums?
I understand why excel behaves as it does.
You can download the file I used in the examples from Dropbox.
If you like to share solutions 🙂 you should be able to upload them into a dropbox folder I created.
Thank you for your time reading and thinking about this.
Oliver
Hello everyone,I hope someone is willing to help me with my Pivot Challenge. The actual question is below. I use a Pivot Table on the following datamodel:In my Pivot Table I see the spent time on a project. This works fine:People from Team “Malt” worked on Tasks which belong to Project AAA-11111 and spent 2.13 days. They also worked on Project AAA-11202 and spent 9.88 days. I like to compare those spent days with the estimations given for project AAA-11111. But if I add the estimation to the values I see all projects and not only those team Malt worked on:For Project AAA-11111 and AAA-11202 the entries are correct:How can I make Excel continue to filter on the SPENT entries but still show the ESTIMATE sums?I understand why excel behaves as it does. You can download the file I used in the examples from Dropbox.If you like to share solutions 🙂 you should be able to upload them into a dropbox folder I created. Thank you for your time reading and thinking about this.Oliver Read More
Suspected identity theft (pass-the-ticket) on multiple endpoints false positive
I have recently analysed Suspected identity theft (pass-the-ticket) alerts which I think are false positives. I’ve been digging into logs to figure this out, but I’m starting to think the reason was staring me in the face.
I’m no expert on this type of alert, but what I’ve understood is host B steals host A’s Kerberos ticket to access network resources. However, I believe Identity Protection has misidentified an IP address as a hostname. Looking forward to any opinions:
I have recently analysed Suspected identity theft (pass-the-ticket) alerts which I think are false positives. I’ve been digging into logs to figure this out, but I’m starting to think the reason was staring me in the face.I’m no expert on this type of alert, but what I’ve understood is host B steals host A’s Kerberos ticket to access network resources. However, I believe Identity Protection has misidentified an IP address as a hostname. Looking forward to any opinions: Read More
How to save entire MATLAB workspace as a JSON file?
Hi everyone, I need to export my MATLAB workspace as a JSON file.
From what I can see, the standard save command wouldn’t be useful here.
And the jsonencode help doesn’t tell you how to save the entire workspace as a JSON file:
Create JSON-formatted text from structured MATLAB data – MATLAB jsonencode – MathWorks United Kingdom
Any help on how to achieve this would be much appreciatedHi everyone, I need to export my MATLAB workspace as a JSON file.
From what I can see, the standard save command wouldn’t be useful here.
And the jsonencode help doesn’t tell you how to save the entire workspace as a JSON file:
Create JSON-formatted text from structured MATLAB data – MATLAB jsonencode – MathWorks United Kingdom
Any help on how to achieve this would be much appreciated Hi everyone, I need to export my MATLAB workspace as a JSON file.
From what I can see, the standard save command wouldn’t be useful here.
And the jsonencode help doesn’t tell you how to save the entire workspace as a JSON file:
Create JSON-formatted text from structured MATLAB data – MATLAB jsonencode – MathWorks United Kingdom
Any help on how to achieve this would be much appreciated json MATLAB Answers — New Questions
电脑蓝屏如何自行分析？
我的工作涉及到重要性，因此无法让别人处理（哪怕提供一个转储文件），我已经在微软社区提问过了“电脑蓝屏如何处理？ – Microsoft Community”那边让我来这边询问
请教教我如何自行分析电脑蓝屏或者给我提供一个分析电脑蓝屏的教程（我需要一个详细的教程）
我的工作涉及到重要性，因此无法让别人处理（哪怕提供一个转储文件），我已经在微软社区提问过了“电脑蓝屏如何处理？ – Microsoft Community”那边让我来这边询问 请教教我如何自行分析电脑蓝屏或者给我提供一个分析电脑蓝屏的教程（我需要一个详细的教程） Read More
Accessing Old High School Google Account After Switch to Outlook?
I need help accessing my old high school Google account. During high school, I used a Google account provided by the school to store all my work on Google Drive. I had always been logged into that account on my PC until last month when I accidentally logged out by clearing my browser history.
Today, I tried to log back into that account to revisit my old projects and memories, but my high school now uses Outlook instead of Google. When I attempt to log in, I’m redirected to Outlook, where my Google username isn’t recognized.
Is there any way for me to access my old files on Google Drive?
I need help accessing my old high school Google account. During high school, I used a Google account provided by the school to store all my work on Google Drive. I had always been logged into that account on my PC until last month when I accidentally logged out by clearing my browser history. Today, I tried to log back into that account to revisit my old projects and memories, but my high school now uses Outlook instead of Google. When I attempt to log in, I’m redirected to Outlook, where my Google username isn’t recognized. Is there any way for me to access my old files on Google Drive? Read More
i am trying to run biped robot and it is giving following error: Error using rlRepresentation (line 70) rlRepresentation will be removed in a future release. Unable to automat
i am trying to run biped robot and it is giving following error:
Error using rlRepresentation (line 70) rlRepresentation will be removed in a future release. Unable to automatically convert rlRepresentation to new representation object. Use the new representation objects rlValueRepresentation, rlQValueRepresentation, rlDeterministicActorRepresentation, or rlStochasticActorRepresentation instead.i am trying to run biped robot and it is giving following error:
Error using rlRepresentation (line 70) rlRepresentation will be removed in a future release. Unable to automatically convert rlRepresentation to new representation object. Use the new representation objects rlValueRepresentation, rlQValueRepresentation, rlDeterministicActorRepresentation, or rlStochasticActorRepresentation instead. i am trying to run biped robot and it is giving following error:
Error using rlRepresentation (line 70) rlRepresentation will be removed in a future release. Unable to automatically convert rlRepresentation to new representation object. Use the new representation objects rlValueRepresentation, rlQValueRepresentation, rlDeterministicActorRepresentation, or rlStochasticActorRepresentation instead. rl represetation MATLAB Answers — New Questions
Is the state variable x a column vector or a row vector?
In a state space equation such as xdot=Ax+Bu, if the system has only two state variables, x1 and x2, shouldn’t the x matrix be just a 2*1 matrix? Suppose I control the system by means of an LQR linear controller and get the corresponding results for the state variables x1 and x2, at this point I want to represent them by means of an image. At this point I just want to know the representation of x1 in the image, is it correct to use the plot(t,x(:,1)); statement. If the plotting is correct at this point, but x(:,1) is queried in the help as referring to the elements of all rows in the first column of the matrix, this suggests that the value of x1 is stored in the first column of the x matrix, but as stated at the beginning the x matrix should be a 2 row, 1 column matrix.
About why the plot(t,x(:,1)); statement is used to get the image of x1 is something I can’t understand at the moment.
Hope someone can help me with this query, thanks a lot!!!!
Translated with www.DeepL.com/Translator (free version)
clc;clear;close all;
%% 定义参数
g=10;
d=1;
%% 定义矩阵
A=[0 1;g/d 0];
B=[0;1];
C = [1, 0];
D = 0;
%% 建立状态空间方程表达式
sys = ss(A,B,C,D);
%% 定义初始状态
z0=[pi/20;0];
%% 定义权重系数，求K
q1=[100 0;0 1];
r1=1;
[K1, z, l] = lqr (sys, q1, r1);
q2=[1 0;0 100];
r2=1;
[K2, z, l] = lqr (sys, q2, r2);
q3=[1 0;0 1];
r3=100;
[K3, z, l] = lqr (sys, q3, r3);
%% 定义闭环系统
sys_cl1=ss(A-B*K1,[0;0],C,D);
sys_cl2=ss(A-B*K2,[0;0],C,D);
sys_cl3=ss(A-B*K3,[0;0],C,D);
%% 仿真
%% 对初始条件的响应
t=0:0.01:5;
[y1,t,z1]=initial(sys_cl1,z0,t);
[y2,t,z2]=initial(sys_cl2,z0,t);
[y3,t,z3]=initial(sys_cl3,z0,t);
%% 绘图
%% 状态变量1
subplot(3,1,1);
plot(t,z1(:,1));
hold on;
plot(t,z2(:,1));
hold on;
plot(t,z3(:,1));
legend(‘z1_1′,’z1_2′,’z1_3’);
grid on;
hold off;
%% 状态变量2
subplot(3,1,2);
plot(t,z1(:,2));
hold on;
plot(t,z2(:,2));
hold on;
plot(t,z3(:,2));
legend(‘z2_1′,’z2_2′,’z2_3’);
grid on;
hold off;
%% 系统输入
subplot(3,1,3);
plot(t,-K1*z1′);
hold on;
plot(t,-K2*z3′);
hold on;
plot(t,-K3*z3′);
legend(‘u_1′,’u_2′,’u_3’);
grid on;
hold off;In a state space equation such as xdot=Ax+Bu, if the system has only two state variables, x1 and x2, shouldn’t the x matrix be just a 2*1 matrix? Suppose I control the system by means of an LQR linear controller and get the corresponding results for the state variables x1 and x2, at this point I want to represent them by means of an image. At this point I just want to know the representation of x1 in the image, is it correct to use the plot(t,x(:,1)); statement. If the plotting is correct at this point, but x(:,1) is queried in the help as referring to the elements of all rows in the first column of the matrix, this suggests that the value of x1 is stored in the first column of the x matrix, but as stated at the beginning the x matrix should be a 2 row, 1 column matrix.
About why the plot(t,x(:,1)); statement is used to get the image of x1 is something I can’t understand at the moment.
Hope someone can help me with this query, thanks a lot!!!!
Translated with www.DeepL.com/Translator (free version)
clc;clear;close all;
%% 定义参数
g=10;
d=1;
%% 定义矩阵
A=[0 1;g/d 0];
B=[0;1];
C = [1, 0];
D = 0;
%% 建立状态空间方程表达式
sys = ss(A,B,C,D);
%% 定义初始状态
z0=[pi/20;0];
%% 定义权重系数，求K
q1=[100 0;0 1];
r1=1;
[K1, z, l] = lqr (sys, q1, r1);
q2=[1 0;0 100];
r2=1;
[K2, z, l] = lqr (sys, q2, r2);
q3=[1 0;0 1];
r3=100;
[K3, z, l] = lqr (sys, q3, r3);
%% 定义闭环系统
sys_cl1=ss(A-B*K1,[0;0],C,D);
sys_cl2=ss(A-B*K2,[0;0],C,D);
sys_cl3=ss(A-B*K3,[0;0],C,D);
%% 仿真
%% 对初始条件的响应
t=0:0.01:5;
[y1,t,z1]=initial(sys_cl1,z0,t);
[y2,t,z2]=initial(sys_cl2,z0,t);
[y3,t,z3]=initial(sys_cl3,z0,t);
%% 绘图
%% 状态变量1
subplot(3,1,1);
plot(t,z1(:,1));
hold on;
plot(t,z2(:,1));
hold on;
plot(t,z3(:,1));
legend(‘z1_1′,’z1_2′,’z1_3’);
grid on;
hold off;
%% 状态变量2
subplot(3,1,2);
plot(t,z1(:,2));
hold on;
plot(t,z2(:,2));
hold on;
plot(t,z3(:,2));
legend(‘z2_1′,’z2_2′,’z2_3’);
grid on;
hold off;
%% 系统输入
subplot(3,1,3);
plot(t,-K1*z1′);
hold on;
plot(t,-K2*z3′);
hold on;
plot(t,-K3*z3′);
legend(‘u_1′,’u_2′,’u_3’);
grid on;
hold off; In a state space equation such as xdot=Ax+Bu, if the system has only two state variables, x1 and x2, shouldn’t the x matrix be just a 2*1 matrix? Suppose I control the system by means of an LQR linear controller and get the corresponding results for the state variables x1 and x2, at this point I want to represent them by means of an image. At this point I just want to know the representation of x1 in the image, is it correct to use the plot(t,x(:,1)); statement. If the plotting is correct at this point, but x(:,1) is queried in the help as referring to the elements of all rows in the first column of the matrix, this suggests that the value of x1 is stored in the first column of the x matrix, but as stated at the beginning the x matrix should be a 2 row, 1 column matrix.
About why the plot(t,x(:,1)); statement is used to get the image of x1 is something I can’t understand at the moment.
Hope someone can help me with this query, thanks a lot!!!!
Translated with www.DeepL.com/Translator (free version)
clc;clear;close all;
%% 定义参数
g=10;
d=1;
%% 定义矩阵
A=[0 1;g/d 0];
B=[0;1];
C = [1, 0];
D = 0;
%% 建立状态空间方程表达式
sys = ss(A,B,C,D);
%% 定义初始状态
z0=[pi/20;0];
%% 定义权重系数，求K
q1=[100 0;0 1];
r1=1;
[K1, z, l] = lqr (sys, q1, r1);
q2=[1 0;0 100];
r2=1;
[K2, z, l] = lqr (sys, q2, r2);
q3=[1 0;0 1];
r3=100;
[K3, z, l] = lqr (sys, q3, r3);
%% 定义闭环系统
sys_cl1=ss(A-B*K1,[0;0],C,D);
sys_cl2=ss(A-B*K2,[0;0],C,D);
sys_cl3=ss(A-B*K3,[0;0],C,D);
%% 仿真
%% 对初始条件的响应
t=0:0.01:5;
[y1,t,z1]=initial(sys_cl1,z0,t);
[y2,t,z2]=initial(sys_cl2,z0,t);
[y3,t,z3]=initial(sys_cl3,z0,t);
%% 绘图
%% 状态变量1
subplot(3,1,1);
plot(t,z1(:,1));
hold on;
plot(t,z2(:,1));
hold on;
plot(t,z3(:,1));
legend(‘z1_1′,’z1_2′,’z1_3’);
grid on;
hold off;
%% 状态变量2
subplot(3,1,2);
plot(t,z1(:,2));
hold on;
plot(t,z2(:,2));
hold on;
plot(t,z3(:,2));
legend(‘z2_1′,’z2_2′,’z2_3’);
grid on;
hold off;
%% 系统输入
subplot(3,1,3);
plot(t,-K1*z1′);
hold on;
plot(t,-K2*z3′);
hold on;
plot(t,-K3*z3′);
legend(‘u_1′,’u_2′,’u_3’);
grid on;
hold off; state space, vector MATLAB Answers — New Questions
Simulink error: Size of data type ‘CAN_MESSAGE’ has not been set
Hello everyone, I am working on a model using CAN bus to receive data.But when building the model it pops out an ERROR:"Size of data type ‘CAN_MESSAGE’ has not been set" and I can’t find solution anywhere.
My environment: Matlab2018b, Visual Studio 2015, Freescale s32ds v1.0. I am coding for a VCU which contains a MPC5744 as CPU (perhaps)
And I have asked the guy from the VCU company, he said the model seems to be fine, the problem may be about my environment or something else…
Sincere appreciation for any advice!!Hello everyone, I am working on a model using CAN bus to receive data.But when building the model it pops out an ERROR:"Size of data type ‘CAN_MESSAGE’ has not been set" and I can’t find solution anywhere.
My environment: Matlab2018b, Visual Studio 2015, Freescale s32ds v1.0. I am coding for a VCU which contains a MPC5744 as CPU (perhaps)
And I have asked the guy from the VCU company, he said the model seems to be fine, the problem may be about my environment or something else…
Sincere appreciation for any advice!! Hello everyone, I am working on a model using CAN bus to receive data.But when building the model it pops out an ERROR:"Size of data type ‘CAN_MESSAGE’ has not been set" and I can’t find solution anywhere.
My environment: Matlab2018b, Visual Studio 2015, Freescale s32ds v1.0. I am coding for a VCU which contains a MPC5744 as CPU (perhaps)
And I have asked the guy from the VCU company, he said the model seems to be fine, the problem may be about my environment or something else…
Sincere appreciation for any advice!! can communication, simulink vehicle network toolbox MATLAB Answers — New Questions
Plot intersection between two surfaces and a plane.
I would like to plot the xy intersection of two surfaces coming from
surf command and a plane from the patch command.
I tried with the contourf, but it seems to be not possible.
Can this be done?I would like to plot the xy intersection of two surfaces coming from
surf command and a plane from the patch command.
I tried with the contourf, but it seems to be not possible.
Can this be done? I would like to plot the xy intersection of two surfaces coming from
surf command and a plane from the patch command.
I tried with the contourf, but it seems to be not possible.
Can this be done? plane, intersection, xyplot, double surface MATLAB Answers — New Questions
PPO | RL | Training for 10000 rounds still doesn’t get effective training, and the reward curve is very flat
With reinforcement learning using PPO algorithm, the Episode reward will oscillate in a wide range all the time, and the average reward curve will change very little.
My question is whether the Episode reward value for such oscillations is due to the large difference in the initial random environment settings, the large difference in the rewards obtained per episode, or the unreasonable parameter settings of the training.
The initial environment generates random locations through distance and angle variables
angle=-4*pi/8+sign(2*rand(1)-1)*rand(1)*pi/8;
dist=10000+rand(1)*4000;
The current agent settings and training settings are as follows：
actorOpts = rlOptimizerOptions(LearnRate=1e-4,GradientThreshold=1);
criticOpts = rlOptimizerOptions(LearnRate=1e-4,GradientThreshold=1);
agentOpts = rlPPOAgentOptions(…
ActorOptimizerOptions=actorOpts,…
CriticOptimizerOptions=criticOpts,…
ExperienceHorizon=2500,…
ClipFactor=0.1,…
EntropyLossWeight=0.02,…
MiniBatchSize=256,…
NumEpoch=9,…
AdvantageEstimateMethod="gae",…
GAEFactor=0.95,…
SampleTime=0.01,…
DiscountFactor=0.99);
trainOpts = rlTrainingOptions(…
‘MaxEpisodes’,100000, …
‘MaxStepsPerEpisode’,2500, …
‘Verbose’,false, …
‘StopTrainingCriteria’,"AverageReward",…
‘StopTrainingValue’,1000,…
‘ScoreAveragingWindowLength’,1000);
trainOpts.UseParallel = true;
trainOpts.ParallelizationOptions.StepsUntilDataIsSent = 2^11;With reinforcement learning using PPO algorithm, the Episode reward will oscillate in a wide range all the time, and the average reward curve will change very little.
My question is whether the Episode reward value for such oscillations is due to the large difference in the initial random environment settings, the large difference in the rewards obtained per episode, or the unreasonable parameter settings of the training.
The initial environment generates random locations through distance and angle variables
angle=-4*pi/8+sign(2*rand(1)-1)*rand(1)*pi/8;
dist=10000+rand(1)*4000;
The current agent settings and training settings are as follows：
actorOpts = rlOptimizerOptions(LearnRate=1e-4,GradientThreshold=1);
criticOpts = rlOptimizerOptions(LearnRate=1e-4,GradientThreshold=1);
agentOpts = rlPPOAgentOptions(…
ActorOptimizerOptions=actorOpts,…
CriticOptimizerOptions=criticOpts,…
ExperienceHorizon=2500,…
ClipFactor=0.1,…
EntropyLossWeight=0.02,…
MiniBatchSize=256,…
NumEpoch=9,…
AdvantageEstimateMethod="gae",…
GAEFactor=0.95,…
SampleTime=0.01,…
DiscountFactor=0.99);
trainOpts = rlTrainingOptions(…
‘MaxEpisodes’,100000, …
‘MaxStepsPerEpisode’,2500, …
‘Verbose’,false, …
‘StopTrainingCriteria’,"AverageReward",…
‘StopTrainingValue’,1000,…
‘ScoreAveragingWindowLength’,1000);
trainOpts.UseParallel = true;
trainOpts.ParallelizationOptions.StepsUntilDataIsSent = 2^11; With reinforcement learning using PPO algorithm, the Episode reward will oscillate in a wide range all the time, and the average reward curve will change very little.
My question is whether the Episode reward value for such oscillations is due to the large difference in the initial random environment settings, the large difference in the rewards obtained per episode, or the unreasonable parameter settings of the training.
The initial environment generates random locations through distance and angle variables
angle=-4*pi/8+sign(2*rand(1)-1)*rand(1)*pi/8;
dist=10000+rand(1)*4000;
The current agent settings and training settings are as follows：
actorOpts = rlOptimizerOptions(LearnRate=1e-4,GradientThreshold=1);
criticOpts = rlOptimizerOptions(LearnRate=1e-4,GradientThreshold=1);
agentOpts = rlPPOAgentOptions(…
ActorOptimizerOptions=actorOpts,…
CriticOptimizerOptions=criticOpts,…
ExperienceHorizon=2500,…
ClipFactor=0.1,…
EntropyLossWeight=0.02,…
MiniBatchSize=256,…
NumEpoch=9,…
AdvantageEstimateMethod="gae",…
GAEFactor=0.95,…
SampleTime=0.01,…
DiscountFactor=0.99);
trainOpts = rlTrainingOptions(…
‘MaxEpisodes’,100000, …
‘MaxStepsPerEpisode’,2500, …
‘Verbose’,false, …
‘StopTrainingCriteria’,"AverageReward",…
‘StopTrainingValue’,1000,…
‘ScoreAveragingWindowLength’,1000);
trainOpts.UseParallel = true;
trainOpts.ParallelizationOptions.StepsUntilDataIsSent = 2^11; ppo, reinforcement learning, reward curve MATLAB Answers — New Questions
Issues with Converting Binary Array to Decimal Value in MATLAB DLL for C++ Integration
Hello everyone,
I want to convert a binary array to a decimal value. When using MATLAB functions like bi2de and bit2int and then compiling them into a DLL using the library compiler, calling them from C++ results in errors indicating that the library could not be initialized. Is there an alternative MATLAB function that can be compiled into a DLL and called from C++ without issues?
I have also tried using a combination of bin2dec(num2str()), which loads the DLL correctly after compilation, but the execution results are incorrect. The result is displayed as a huge decimal number. What could be the reason for this?Hello everyone,
I want to convert a binary array to a decimal value. When using MATLAB functions like bi2de and bit2int and then compiling them into a DLL using the library compiler, calling them from C++ results in errors indicating that the library could not be initialized. Is there an alternative MATLAB function that can be compiled into a DLL and called from C++ without issues?
I have also tried using a combination of bin2dec(num2str()), which loads the DLL correctly after compilation, but the execution results are incorrect. The result is displayed as a huge decimal number. What could be the reason for this? Hello everyone,
I want to convert a binary array to a decimal value. When using MATLAB functions like bi2de and bit2int and then compiling them into a DLL using the library compiler, calling them from C++ results in errors indicating that the library could not be initialized. Is there an alternative MATLAB function that can be compiled into a DLL and called from C++ without issues?
I have also tried using a combination of bin2dec(num2str()), which loads the DLL correctly after compilation, but the execution results are incorrect. The result is displayed as a huge decimal number. What could be the reason for this? binary array, decimal value, matlab compiler, dll MATLAB Answers — New Questions
Calculate the volume of shape descibed by 3d points.
Given a set of 3d data points, without using the Image Processing Toolbox, what is the best method for working put the volume of the object described by the points?Given a set of 3d data points, without using the Image Processing Toolbox, what is the best method for working put the volume of the object described by the points? Given a set of 3d data points, without using the Image Processing Toolbox, what is the best method for working put the volume of the object described by the points? convex hull, convhull, trisurf MATLAB Answers — New Questions
PcaWallpaperAppDetect error
Hi just upgraded to 24h2 and receiving this error. it says error in pcaSvc.dll.
How do i fix it. Although it is not causing any problems but still i want to know what causing it.
.
Hi just upgraded to 24h2 and receiving this error. it says error in pcaSvc.dll. How do i fix it. Although it is not causing any problems but still i want to know what causing it.. Read More
DevOps in the era of Generative AI: Foundations of LLMOps
Spotlight on AI in your DevOps Lifecycle
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DevOps in the era of Generative AI: Foundations of LLMOps
With the advent of generative AI, the development life cycle of intelligent applications has undergone a significant change. This shift from classical ML to LLMs-based solutions leads to implications not only on how we build applications but also in how we test, evaluate, deploy, and monitor them. The introduction of LLMOps is an important development that requires understanding the foundations of this new approach to DevOps.
The session “DevOps in the era of Generative AI: Foundations of LLMOps” will explore the basics of LLMOps, providing examples of tools and practices available in the Azure ecosystem. This talk will be held on June 12th, 2024, from 4:00 PM to 5:00 PM (UTC).
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Continuous Integration and Continuous Delivery (CI/CD) for AI
The session “Continuous Integration and Continuous Delivery (CI/CD) for AI” will focus on MLOps for machine learning and AI projects. This talk will cover how to set up CI/CD and collaborate with others using GitHub. It will also discuss version control, automated testing, and deployment strategies.
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Monitoring, Logging, and AI Model Performance
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Microsoft Tech Community – Latest Blogs –Read More
finding 4 neighbor of pixel in grayscale image.
i am trying to find the 4 neighbor of pixel x0,y0 in grayscale image but i getting empty x,y . her is the code i am using
function [x,y]= neighbours(image,x0,y0,thres)
image=double(image);
interest = image(x0,y0);
pixel =zeros(size(image));
pixel(x0,y0)=1;
x=[];
y=[];
up= image(conv2(pixel,[0 1 0; 0 0 0; 0 0 0],’same’)>0);
if abs(up- interest)<=thres
if isempty(up)~=1
x(1)=x0;
y(1)=y0-1;
end
end
down= image(conv2(pixel,[0 0 0; 0 0 0; 0 1 0],’same’)>0);
if abs(down- interest)<=thres
if isempty(down)~=1
x(2)=x0;
y(2)=y0+1;
end
end
left= image(conv2(pixel,[0 0 0; 1 0 0; 0 0 0],’same’)>0);
if abs(left- interest)<=thres
if isempty(left)~=1
x(3)=x0-1;
y(3)=y0;
end
end
right= image(conv2(pixel,[0 0 0; 0 0 1; 0 0 0],’same’)>0);
if abs(right- interest)<=thres
if isempty(right)~=1
x(4)=x0+1;
y(4)=y0;
end
end
endi am trying to find the 4 neighbor of pixel x0,y0 in grayscale image but i getting empty x,y . her is the code i am using
function [x,y]= neighbours(image,x0,y0,thres)
image=double(image);
interest = image(x0,y0);
pixel =zeros(size(image));
pixel(x0,y0)=1;
x=[];
y=[];
up= image(conv2(pixel,[0 1 0; 0 0 0; 0 0 0],’same’)>0);
if abs(up- interest)<=thres
if isempty(up)~=1
x(1)=x0;
y(1)=y0-1;
end
end
down= image(conv2(pixel,[0 0 0; 0 0 0; 0 1 0],’same’)>0);
if abs(down- interest)<=thres
if isempty(down)~=1
x(2)=x0;
y(2)=y0+1;
end
end
left= image(conv2(pixel,[0 0 0; 1 0 0; 0 0 0],’same’)>0);
if abs(left- interest)<=thres
if isempty(left)~=1
x(3)=x0-1;
y(3)=y0;
end
end
right= image(conv2(pixel,[0 0 0; 0 0 1; 0 0 0],’same’)>0);
if abs(right- interest)<=thres
if isempty(right)~=1
x(4)=x0+1;
y(4)=y0;
end
end
end i am trying to find the 4 neighbor of pixel x0,y0 in grayscale image but i getting empty x,y . her is the code i am using
function [x,y]= neighbours(image,x0,y0,thres)
image=double(image);
interest = image(x0,y0);
pixel =zeros(size(image));
pixel(x0,y0)=1;
x=[];
y=[];
up= image(conv2(pixel,[0 1 0; 0 0 0; 0 0 0],’same’)>0);
if abs(up- interest)<=thres
if isempty(up)~=1
x(1)=x0;
y(1)=y0-1;
end
end
down= image(conv2(pixel,[0 0 0; 0 0 0; 0 1 0],’same’)>0);
if abs(down- interest)<=thres
if isempty(down)~=1
x(2)=x0;
y(2)=y0+1;
end
end
left= image(conv2(pixel,[0 0 0; 1 0 0; 0 0 0],’same’)>0);
if abs(left- interest)<=thres
if isempty(left)~=1
x(3)=x0-1;
y(3)=y0;
end
end
right= image(conv2(pixel,[0 0 0; 0 0 1; 0 0 0],’same’)>0);
if abs(right- interest)<=thres
if isempty(right)~=1
x(4)=x0+1;
y(4)=y0;
end
end
end image segmentation, image processing, image analysis, matlab, digital image processing MATLAB Answers — New Questions
i have made a code which finds a specific value ‘no’ from the excel sheet but it does the job one by one. i want to automate it so it gives all values with ‘no’ at once.
this is the code i’m using.
data = readtable(‘FINALconvergencecheck_ksst.xlsx’);
data.Properties.VariableNames(1) = "an6cd";
data.Properties.VariableNames(3) = "an6cl";
data.Properties.VariableNames(5) = "an6cm";
data.Properties.VariableNames(7) = "ap0cd";
data.Properties.VariableNames(9) = "ap0cl";
data.Properties.VariableNames(11) = "ap0cm";
data.Properties.VariableNames(13) = "ap3cd";
data.Properties.VariableNames(15) = "ap3cl";
data.Properties.VariableNames(17) = "ap3cm";
data.Properties.VariableNames(19) = "ap7cd";
data.Properties.VariableNames(21) = "ap7cl";
data.Properties.VariableNames(23) = "ap7cm";
data.Properties.VariableNames(25) = "ap10cd";
data.Properties.VariableNames(27) = "ap10cl";
data.Properties.VariableNames(29) = "ap10cm";
data.Properties.VariableNames(31) = "ap12cd";
data.Properties.VariableNames(33) = "ap12cl";
data.Properties.VariableNames(35) = "ap12cm";
data.Properties.VariableNames(37) = "ap14cd";
data.Properties.VariableNames(39) = "ap14cl";
data.Properties.VariableNames(41) = "ap14cm";
data.Properties.VariableNames(43) = "ap15cd";
data.Properties.VariableNames(45) = "ap15cl";
data.Properties.VariableNames(47) = "ap15cm";
data.Properties.VariableNames(49) = "ap17cd";
data.Properties.VariableNames(51) = "ap17cl";
data.Properties.VariableNames(53) = "ap17cm";
data.Properties.VariableNames(2) = "an6cd_r";
data.Properties.VariableNames(4) = "an6cl_r";
data.Properties.VariableNames(6) = "an6cm_r";
data.Properties.VariableNames(8) = "ap0cd_r";
data.Properties.VariableNames(10) = "ap0cl_r";
data.Properties.VariableNames(12) = "ap0cm_r";
data.Properties.VariableNames(14) = "ap3cd_r";
data.Properties.VariableNames(16) = "ap3cl_r";
data.Properties.VariableNames(18) = "ap3cm_r";
data.Properties.VariableNames(20) = "ap7cd_r";
data.Properties.VariableNames(22) = "ap7cl_r";
data.Properties.VariableNames(24) = "ap7cm_r";
data.Properties.VariableNames(26) = "ap10cd_r";
data.Properties.VariableNames(28) = "ap10cl_r";
data.Properties.VariableNames(30) = "ap10cm_r";
data.Properties.VariableNames(32) = "ap12cd_r";
data.Properties.VariableNames(34) = "ap12cl_r";
data.Properties.VariableNames(36) = "ap12cm_r";
data.Properties.VariableNames(38) = "ap14cd_r";
data.Properties.VariableNames(40) = "ap14cl_r";
data.Properties.VariableNames(42) = "ap14cm_r";
data.Properties.VariableNames(44) = "ap15cd_r";
data.Properties.VariableNames(46) = "ap15cl_r";
data.Properties.VariableNames(48) = "ap15cm_r";
data.Properties.VariableNames(50) = "ap17cd_r";
data.Properties.VariableNames(52) = "ap17cl_r";
data.Properties.VariableNames(54) = "ap17cm_r";
display = find(data.ap10cl_r=="no");
k=display;
k1=string(k);
switch k1
case "yes"
disp(‘convereged’)
case "no"
disp(‘not convereged’)
end
here, i have to change the coulum again and again. i want to automate it in a way that it gives all the values where ‘no’ is at once.
kindly help.this is the code i’m using.
data = readtable(‘FINALconvergencecheck_ksst.xlsx’);
data.Properties.VariableNames(1) = "an6cd";
data.Properties.VariableNames(3) = "an6cl";
data.Properties.VariableNames(5) = "an6cm";
data.Properties.VariableNames(7) = "ap0cd";
data.Properties.VariableNames(9) = "ap0cl";
data.Properties.VariableNames(11) = "ap0cm";
data.Properties.VariableNames(13) = "ap3cd";
data.Properties.VariableNames(15) = "ap3cl";
data.Properties.VariableNames(17) = "ap3cm";
data.Properties.VariableNames(19) = "ap7cd";
data.Properties.VariableNames(21) = "ap7cl";
data.Properties.VariableNames(23) = "ap7cm";
data.Properties.VariableNames(25) = "ap10cd";
data.Properties.VariableNames(27) = "ap10cl";
data.Properties.VariableNames(29) = "ap10cm";
data.Properties.VariableNames(31) = "ap12cd";
data.Properties.VariableNames(33) = "ap12cl";
data.Properties.VariableNames(35) = "ap12cm";
data.Properties.VariableNames(37) = "ap14cd";
data.Properties.VariableNames(39) = "ap14cl";
data.Properties.VariableNames(41) = "ap14cm";
data.Properties.VariableNames(43) = "ap15cd";
data.Properties.VariableNames(45) = "ap15cl";
data.Properties.VariableNames(47) = "ap15cm";
data.Properties.VariableNames(49) = "ap17cd";
data.Properties.VariableNames(51) = "ap17cl";
data.Properties.VariableNames(53) = "ap17cm";
data.Properties.VariableNames(2) = "an6cd_r";
data.Properties.VariableNames(4) = "an6cl_r";
data.Properties.VariableNames(6) = "an6cm_r";
data.Properties.VariableNames(8) = "ap0cd_r";
data.Properties.VariableNames(10) = "ap0cl_r";
data.Properties.VariableNames(12) = "ap0cm_r";
data.Properties.VariableNames(14) = "ap3cd_r";
data.Properties.VariableNames(16) = "ap3cl_r";
data.Properties.VariableNames(18) = "ap3cm_r";
data.Properties.VariableNames(20) = "ap7cd_r";
data.Properties.VariableNames(22) = "ap7cl_r";
data.Properties.VariableNames(24) = "ap7cm_r";
data.Properties.VariableNames(26) = "ap10cd_r";
data.Properties.VariableNames(28) = "ap10cl_r";
data.Properties.VariableNames(30) = "ap10cm_r";
data.Properties.VariableNames(32) = "ap12cd_r";
data.Properties.VariableNames(34) = "ap12cl_r";
data.Properties.VariableNames(36) = "ap12cm_r";
data.Properties.VariableNames(38) = "ap14cd_r";
data.Properties.VariableNames(40) = "ap14cl_r";
data.Properties.VariableNames(42) = "ap14cm_r";
data.Properties.VariableNames(44) = "ap15cd_r";
data.Properties.VariableNames(46) = "ap15cl_r";
data.Properties.VariableNames(48) = "ap15cm_r";
data.Properties.VariableNames(50) = "ap17cd_r";
data.Properties.VariableNames(52) = "ap17cl_r";
data.Properties.VariableNames(54) = "ap17cm_r";
display = find(data.ap10cl_r=="no");
k=display;
k1=string(k);
switch k1
case "yes"
disp(‘convereged’)
case "no"
disp(‘not convereged’)
end
here, i have to change the coulum again and again. i want to automate it in a way that it gives all the values where ‘no’ is at once.
kindly help. this is the code i’m using.
data = readtable(‘FINALconvergencecheck_ksst.xlsx’);
data.Properties.VariableNames(1) = "an6cd";
data.Properties.VariableNames(3) = "an6cl";
data.Properties.VariableNames(5) = "an6cm";
data.Properties.VariableNames(7) = "ap0cd";
data.Properties.VariableNames(9) = "ap0cl";
data.Properties.VariableNames(11) = "ap0cm";
data.Properties.VariableNames(13) = "ap3cd";
data.Properties.VariableNames(15) = "ap3cl";
data.Properties.VariableNames(17) = "ap3cm";
data.Properties.VariableNames(19) = "ap7cd";
data.Properties.VariableNames(21) = "ap7cl";
data.Properties.VariableNames(23) = "ap7cm";
data.Properties.VariableNames(25) = "ap10cd";
data.Properties.VariableNames(27) = "ap10cl";
data.Properties.VariableNames(29) = "ap10cm";
data.Properties.VariableNames(31) = "ap12cd";
data.Properties.VariableNames(33) = "ap12cl";
data.Properties.VariableNames(35) = "ap12cm";
data.Properties.VariableNames(37) = "ap14cd";
data.Properties.VariableNames(39) = "ap14cl";
data.Properties.VariableNames(41) = "ap14cm";
data.Properties.VariableNames(43) = "ap15cd";
data.Properties.VariableNames(45) = "ap15cl";
data.Properties.VariableNames(47) = "ap15cm";
data.Properties.VariableNames(49) = "ap17cd";
data.Properties.VariableNames(51) = "ap17cl";
data.Properties.VariableNames(53) = "ap17cm";
data.Properties.VariableNames(2) = "an6cd_r";
data.Properties.VariableNames(4) = "an6cl_r";
data.Properties.VariableNames(6) = "an6cm_r";
data.Properties.VariableNames(8) = "ap0cd_r";
data.Properties.VariableNames(10) = "ap0cl_r";
data.Properties.VariableNames(12) = "ap0cm_r";
data.Properties.VariableNames(14) = "ap3cd_r";
data.Properties.VariableNames(16) = "ap3cl_r";
data.Properties.VariableNames(18) = "ap3cm_r";
data.Properties.VariableNames(20) = "ap7cd_r";
data.Properties.VariableNames(22) = "ap7cl_r";
data.Properties.VariableNames(24) = "ap7cm_r";
data.Properties.VariableNames(26) = "ap10cd_r";
data.Properties.VariableNames(28) = "ap10cl_r";
data.Properties.VariableNames(30) = "ap10cm_r";
data.Properties.VariableNames(32) = "ap12cd_r";
data.Properties.VariableNames(34) = "ap12cl_r";
data.Properties.VariableNames(36) = "ap12cm_r";
data.Properties.VariableNames(38) = "ap14cd_r";
data.Properties.VariableNames(40) = "ap14cl_r";
data.Properties.VariableNames(42) = "ap14cm_r";
data.Properties.VariableNames(44) = "ap15cd_r";
data.Properties.VariableNames(46) = "ap15cl_r";
data.Properties.VariableNames(48) = "ap15cm_r";
data.Properties.VariableNames(50) = "ap17cd_r";
data.Properties.VariableNames(52) = "ap17cl_r";
data.Properties.VariableNames(54) = "ap17cm_r";
display = find(data.ap10cl_r=="no");
k=display;
k1=string(k);
switch k1
case "yes"
disp(‘convereged’)
case "no"
disp(‘not convereged’)
end
here, i have to change the coulum again and again. i want to automate it in a way that it gives all the values where ‘no’ is at once.
kindly help. excel, loop MATLAB Answers — New Questions
Process gain in System Identification seems off by 100
Hello,
Running a simple first order test process through System Identification, to make sure I understand the results.
The .mat file is attached with data, after which I import into a iddata.
The output of my data changes from 1000 down to 0 (data does not run far enough to get all the way to 0) and the input changes from 6 to 5.
I would expect a process gain of 1000 / 1 = 1000. However, SI is giving me 10?
Time constant is correct at 0.71sec. Delay is correct at 0.
load ("data.mat", "import");
model = iddata(import.out, import.co, 0.001)Hello,
Running a simple first order test process through System Identification, to make sure I understand the results.
The .mat file is attached with data, after which I import into a iddata.
The output of my data changes from 1000 down to 0 (data does not run far enough to get all the way to 0) and the input changes from 6 to 5.
I would expect a process gain of 1000 / 1 = 1000. However, SI is giving me 10?
Time constant is correct at 0.71sec. Delay is correct at 0.
load ("data.mat", "import");
model = iddata(import.out, import.co, 0.001) Hello,
Running a simple first order test process through System Identification, to make sure I understand the results.
The .mat file is attached with data, after which I import into a iddata.
The output of my data changes from 1000 down to 0 (data does not run far enough to get all the way to 0) and the input changes from 6 to 5.
I would expect a process gain of 1000 / 1 = 1000. However, SI is giving me 10?
Time constant is correct at 0.71sec. Delay is correct at 0.
load ("data.mat", "import");
model = iddata(import.out, import.co, 0.001) process gain, fopdt, handles MATLAB Answers — New Questions
Prevent Windows Update from shutting down Matlab session
I run Matlab simulations that sometimes take weeks to finish and, when Windows restarts due to an auto update, I lose all my work so I really need a way to stop auto restarts from interrupting my Matlab simulation.
Windows has an API that enables a process that has a window to block a system shutdown indefinitely until the user manually chooses what to do next. Applications like Matlab receive shutdown notifications through the WM_QUERYENDSESSION and WM_ENDSESSION messages.
How can I use this to stop my simulations from shutting down and trashing all my work?
Sometimes my sims are straightforward and easy to start up again where I left off but often they are very complex simulations with lots of complex states using vast amounts of memory that it would be very difficult to get back to without running from the start again.I run Matlab simulations that sometimes take weeks to finish and, when Windows restarts due to an auto update, I lose all my work so I really need a way to stop auto restarts from interrupting my Matlab simulation.
Windows has an API that enables a process that has a window to block a system shutdown indefinitely until the user manually chooses what to do next. Applications like Matlab receive shutdown notifications through the WM_QUERYENDSESSION and WM_ENDSESSION messages.
How can I use this to stop my simulations from shutting down and trashing all my work?
Sometimes my sims are straightforward and easy to start up again where I left off but often they are very complex simulations with lots of complex states using vast amounts of memory that it would be very difficult to get back to without running from the start again. I run Matlab simulations that sometimes take weeks to finish and, when Windows restarts due to an auto update, I lose all my work so I really need a way to stop auto restarts from interrupting my Matlab simulation.
Windows has an API that enables a process that has a window to block a system shutdown indefinitely until the user manually chooses what to do next. Applications like Matlab receive shutdown notifications through the WM_QUERYENDSESSION and WM_ENDSESSION messages.
How can I use this to stop my simulations from shutting down and trashing all my work?
Sometimes my sims are straightforward and easy to start up again where I left off but often they are very complex simulations with lots of complex states using vast amounts of memory that it would be very difficult to get back to without running from the start again. windows, restart, shutdown, auto update MATLAB Answers — New Questions
