Month: September 2024
Can somebody solve this question. I’ve been trying for 2 hours straight. Didn’t get it yet
Find the volume of the solid generated by revolving about the x-axis the region bounded by the curve y = sqrt(x) , the x-axis, and the line y= x-2.Find the volume of the solid generated by revolving about the x-axis the region bounded by the curve y = sqrt(x) , the x-axis, and the line y= x-2. Find the volume of the solid generated by revolving about the x-axis the region bounded by the curve y = sqrt(x) , the x-axis, and the line y= x-2. volume of solid of revolution, area between curves, axis of revolution MATLAB Answers — New Questions
How can we couple MATLAB and Aspen Exchanger Design and Rating?
Dear MATLAB users,
it would be of great interest to couple MATLAB and Aspen EDR, so that at different conditions, different heat exchangers could be designed. Does anybody have experience with that? Thank you all in advance!Dear MATLAB users,
it would be of great interest to couple MATLAB and Aspen EDR, so that at different conditions, different heat exchangers could be designed. Does anybody have experience with that? Thank you all in advance! Dear MATLAB users,
it would be of great interest to couple MATLAB and Aspen EDR, so that at different conditions, different heat exchangers could be designed. Does anybody have experience with that? Thank you all in advance! aspen edr, heat exchanger design MATLAB Answers — New Questions
I can’t get engine instance when I run my java program by window service
My software system is developed using Java. One function of the program is to send variables to the MATLAB window. When I run my Java program in the development environment or command-line window, it works well and sends variables correctly to the MATLAB window. But when I start my system with a window operating system service or scheduled task, variables cannot be sent to the MATLAB window. I carefully checked and found that when running my software program using the window service, I cannot retrieve the engine instance currently running in the MATLAB window.My software system is developed using Java. One function of the program is to send variables to the MATLAB window. When I run my Java program in the development environment or command-line window, it works well and sends variables correctly to the MATLAB window. But when I start my system with a window operating system service or scheduled task, variables cannot be sent to the MATLAB window. I carefully checked and found that when running my software program using the window service, I cannot retrieve the engine instance currently running in the MATLAB window. My software system is developed using Java. One function of the program is to send variables to the MATLAB window. When I run my Java program in the development environment or command-line window, it works well and sends variables correctly to the MATLAB window. But when I start my system with a window operating system service or scheduled task, variables cannot be sent to the MATLAB window. I carefully checked and found that when running my software program using the window service, I cannot retrieve the engine instance currently running in the MATLAB window. engine, window service MATLAB Answers — New Questions
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随着云工程师和架构师开始采用容器化，确保安全通信变得至关重要。数据传输和访问控制是需要考虑的安全关键方面。SSH 隧道是一种可以帮助实现应用程序或解决方案不同组件之间安全通信的技术。SSH 隧道在现有 SSH 连接上创建加密通道，允许在本地计算机（SSH 客户端）和远程服务器（SSH 服务器）之间进行安全的数据传输。
在本文中，我们将展示如何在云中运行的容器之间建立 SSH 隧道，这些容器需要通过云虚拟机中托管的 SSH 服务器与下游资源进行通信。
SSH 隧道
在深入实现之前，让我们快速回顾一下 SSH 隧道。SSH 隧道也称为 SSH 端口转发，通过在现有 SSH 连接上创建加密隧道，允许两个端点之间进行安全通信。它使数据能够通过中间通道在本地计算机（SSH 客户端）和远程服务器（SSH 服务器）之间安全地传输。以下是可以使用 SSH 隧道的不同场景的概述：
1. 确保远程访问内部服务的安全： 出于安全原因，组织拥有内部服务（例如数据库、内部 Web 应用程序），这些服务不会暴露在公共互联网上。使用 SSH 隧道，员工可以从远程位置安全地连接到这些内部服务，而无需将服务暴露在互联网上。
2. 绕过防火墙限制： 开发人员需要访问企业防火墙后面的特定资源，但防火墙限制直接访问。通过设置 SSH 隧道，开发人员可以安全地通过防火墙转发流量，从而允许他们访问受限制的资源。
3. 保护传输中的敏感数据： 应用程序需要在不同组件或服务之间发送敏感数据，存在数据被拦截的风险。SSH 隧道可用于在数据在组件之间传输时对其进行加密，确保数据在传输过程中保持安全。
4. 安全访问远程数据库： 开发人员需要访问远程数据库服务器以进行维护或开发，但由于安全策略不允许直接访问。开发人员可以设置 SSH 隧道以安全地连接到远程数据库服务器，而无需将其暴露在公共互联网上。
5. 安全地使用不安全的协议： 应用程序使用不安全的协议（例如 FTP、HTTP）在不同服务之间进行通信。通过将不安全的协议包装在 SSH 隧道中，可以确保通信安全，从而保护数据不被拦截。
6. 远程调试： 开发人员需要调试在远程服务器上运行的应用程序，但对调试端口的直接访问受到限制。 SSH 隧道可用于将调试端口从远程服务器转发到本地计算机，从而使开发人员可以安全地调试应用程序。
7. 保护物联网设备通信： 物联网设备需要与中央服务器通信，但通信容易被拦截或篡改。通过在物联网设备和中央服务器之间建立 SSH 隧道，可以加密和保护通信，保护传输中的数据。
8. 安全文件传输： 文件需要在不同系统或位置之间安全地传输。SSH 隧道可用于通过加密隧道安全地传输文件，确保数据的机密性和完整性。
9. 访问远程服务： 用户需要安全地访问远程服务器上托管的服务或资源。通过设置 SSH 隧道，用户可以安全地访问远程服务，就像它们在本地运行一样，从而保护传输中的数据。
10. 保护网络流量： 通过不受信任的网络访问网站或网络应用程序时，需要保护网络流量。SSH 隧道可用于创建与远程服务器的安全连接，加密网络流量并保护其免遭窃听或拦截。
设想
对于本文，我们将实现以下场景：
架构组件
myInfraVNet：部署下游资源的虚拟网络。nginxVM：myInfraVNet 中运行 Nginx（Web 服务器或反向代理）的虚拟机。它被分配了一个私有 IP 地址，因此无法从互联网直接访问。nginxVM/NSG：与 nginxVM 关联的网络安全组，控制入站和出站流量。myAppVNet：部署容器应用程序的虚拟网络。容器应用环境：此环境托管两个容器化应用程序：mycontainerapp：一个简单的容器化 Python 应用程序，它从在 VM 上运行的 NGINX 服务器获取内容并将该内容与其他内容一起呈现。sshclientcontainerapp：另一个容器化应用程序，用于建立到其他资源的安全 SSH 隧道。容器注册表：存储可部署到容器应用的容器镜像。VNet Peering：允许 myAppVNet 和 myInfraVNet 中的资源相互通信。它本质上是两个 VNet 之间的桥梁，可实现低延迟、高带宽的互连。SSH 隧道：myAppVNet 中的 sshclientcontainerapp 建立到 myInfraVNet 中的 nginxVM 的 SSH 隧道，以实现容器化应用程序和 VM 之间的安全通信。网络安全组 (NSG)：nginxVM/NSG 确保只有允许的流量才能到达 nginxVM。正确配置此 NSG 以允许来自 sshclientcontainerapp 的 SSH 流量并限制不必要的访问至关重要。
编写场景脚本
基于上述场景，我们现在将编写架构实现脚本。该脚本将创建必要的资源、配置 SSH 隧道并部署容器化应用程序。
先决条件
运行脚本之前，请确保您满足以下先决条件：
安装在本地计算机上的 Azure CLI。Docker 安装在您的本地机器上。有效的 Azure 订阅。对 Azure 容器应用、Azure 容器注册表和 Azure 虚拟网络有基本的了解。
参数
首先定义脚本中要使用的参数。这些参数包括资源组名称、位置、虚拟网络名称、子网名称、VM 名称、VM 映像、VM 大小、SSH 密钥、管理员用户名、管理员密码、容器应用环境名称、容器注册表名称、容器应用映像名称、SSH 客户端容器映像名称、SSH 端口和 NGINX 端口。将生成一个随机字符串并将其附加到资源组名称、容器应用环境名称和容器注册表名称，以确保唯一性。
random=$(echo $RANDOM | tr ‘[0-9]’ ‘[a-z]’)
echo “Random:” $random
export RESOURCE_GROUP=rg-ssh-$(echo $random)
echo “RESOURCE_GROUP:” $RESOURCE_GROUP
export LOCATION=”australiaeast”
export INFRA_VNET_NAME=”myInfraVNet”
export APP_VNET_NAME=”myAppVNet”
export INFRA_SUBNET_NAME=”mySubnet”
export APP_SUBNET_NAME=”acaSubnet”
export VM_NAME=”nginxVM”
export VM_IMAGE=”Ubuntu2204″
export VM_SIZE=”Standard_DS1_v2″
export VM_KEY=mykey$(echo $random)
export ADMIN_USERNAME=”azureuser”
export ADMIN_PASSWORD=”Password123$” # Replace with your actual password
export CONTAINER_APPS_ENV=sshacae$(echo $random)
export REGISTRY_NAME=sshacr$(echo $random)
export REGISTRY_SKU=”Basic”
export CONTAINER_APP_IMAGE=”mycontainerapp:latest”
export SSH_CLIENT_CONTAINER_IMAGE=”sshclientcontainer:latest”
export CONTAINER_APP_NAME=”mycontainerapp”
export SSH_CLIENT_CONTAINER_APP_NAME=”sshclientcontainerapp”
export SSH_PORT=22
export NGINX_PORT=80
随着云工程师和架构师开始采用容器化，确保安全通信变得至关重要。数据传输和访问控制是需要考虑的安全关键方面。SSH 隧道是一种可以帮助实现应用程序或解决方案不同组件之间安全通信的技术。SSH 隧道在现有 SSH 连接上创建加密通道，允许在本地计算机（SSH 客户端）和远程服务器（SSH 服务器）之间进行安全的数据传输。在本文中，我们将展示如何在云中运行的容器之间建立 SSH 隧道，这些容器需要通过云虚拟机中托管的 SSH 服务器与下游资源进行通信。 SSH 隧道在深入实现之前，让我们快速回顾一下 SSH 隧道。SSH 隧道也称为 SSH 端口转发，通过在现有 SSH 连接上创建加密隧道，允许两个端点之间进行安全通信。它使数据能够通过中间通道在本地计算机（SSH 客户端）和远程服务器（SSH 服务器）之间安全地传输。以下是可以使用 SSH 隧道的不同场景的概述： 1. 确保远程访问内部服务的安全： 出于安全原因，组织拥有内部服务（例如数据库、内部 Web 应用程序），这些服务不会暴露在公共互联网上。使用 SSH 隧道，员工可以从远程位置安全地连接到这些内部服务，而无需将服务暴露在互联网上。2. 绕过防火墙限制： 开发人员需要访问企业防火墙后面的特定资源，但防火墙限制直接访问。通过设置 SSH 隧道，开发人员可以安全地通过防火墙转发流量，从而允许他们访问受限制的资源。3. 保护传输中的敏感数据： 应用程序需要在不同组件或服务之间发送敏感数据，存在数据被拦截的风险。SSH 隧道可用于在数据在组件之间传输时对其进行加密，确保数据在传输过程中保持安全。4. 安全访问远程数据库： 开发人员需要访问远程数据库服务器以进行维护或开发，但由于安全策略不允许直接访问。开发人员可以设置 SSH 隧道以安全地连接到远程数据库服务器，而无需将其暴露在公共互联网上。5. 安全地使用不安全的协议： 应用程序使用不安全的协议（例如 FTP、HTTP）在不同服务之间进行通信。通过将不安全的协议包装在 SSH 隧道中，可以确保通信安全，从而保护数据不被拦截。6. 远程调试： 开发人员需要调试在远程服务器上运行的应用程序，但对调试端口的直接访问受到限制。 SSH 隧道可用于将调试端口从远程服务器转发到本地计算机，从而使开发人员可以安全地调试应用程序。7. 保护物联网设备通信： 物联网设备需要与中央服务器通信，但通信容易被拦截或篡改。通过在物联网设备和中央服务器之间建立 SSH 隧道，可以加密和保护通信，保护传输中的数据。8. 安全文件传输： 文件需要在不同系统或位置之间安全地传输。SSH 隧道可用于通过加密隧道安全地传输文件，确保数据的机密性和完整性。9. 访问远程服务： 用户需要安全地访问远程服务器上托管的服务或资源。通过设置 SSH 隧道，用户可以安全地访问远程服务，就像它们在本地运行一样，从而保护传输中的数据。10. 保护网络流量： 通过不受信任的网络访问网站或网络应用程序时，需要保护网络流量。SSH 隧道可用于创建与远程服务器的安全连接，加密网络流量并保护其免遭窃听或拦截。 设想对于本文，我们将实现以下场景： 架构组件myInfraVNet：部署下游资源的虚拟网络。nginxVM：myInfraVNet 中运行 Nginx（Web 服务器或反向代理）的虚拟机。它被分配了一个私有 IP 地址，因此无法从互联网直接访问。nginxVM/NSG：与 nginxVM 关联的网络安全组，控制入站和出站流量。myAppVNet：部署容器应用程序的虚拟网络。容器应用环境：此环境托管两个容器化应用程序：mycontainerapp：一个简单的容器化 Python 应用程序，它从在 VM 上运行的 NGINX 服务器获取内容并将该内容与其他内容一起呈现。sshclientcontainerapp：另一个容器化应用程序，用于建立到其他资源的安全 SSH 隧道。容器注册表：存储可部署到容器应用的容器镜像。VNet Peering：允许 myAppVNet 和 myInfraVNet 中的资源相互通信。它本质上是两个 VNet 之间的桥梁，可实现低延迟、高带宽的互连。SSH 隧道：myAppVNet 中的 sshclientcontainerapp 建立到 myInfraVNet 中的 nginxVM 的 SSH 隧道，以实现容器化应用程序和 VM 之间的安全通信。网络安全组 (NSG)：nginxVM/NSG 确保只有允许的流量才能到达 nginxVM。正确配置此 NSG 以允许来自 sshclientcontainerapp 的 SSH 流量并限制不必要的访问至关重要。编写场景脚本基于上述场景，我们现在将编写架构实现脚本。该脚本将创建必要的资源、配置 SSH 隧道并部署容器化应用程序。 先决条件运行脚本之前，请确保您满足以下先决条件：安装在本地计算机上的 Azure CLI。Docker 安装在您的本地机器上。有效的 Azure 订阅。对 Azure 容器应用、Azure 容器注册表和 Azure 虚拟网络有基本的了解。参数首先定义脚本中要使用的参数。这些参数包括资源组名称、位置、虚拟网络名称、子网名称、VM 名称、VM 映像、VM 大小、SSH 密钥、管理员用户名、管理员密码、容器应用环境名称、容器注册表名称、容器应用映像名称、SSH 客户端容器映像名称、SSH 端口和 NGINX 端口。将生成一个随机字符串并将其附加到资源组名称、容器应用环境名称和容器注册表名称，以确保唯一性。 random=$(echo $RANDOM | tr ‘[0-9]’ ‘[a-z]’)
echo “Random:” $random
export RESOURCE_GROUP=rg-ssh-$(echo $random)
echo “RESOURCE_GROUP:” $RESOURCE_GROUP
export LOCATION=”australiaeast”
export INFRA_VNET_NAME=”myInfraVNet”
export APP_VNET_NAME=”myAppVNet”
export INFRA_SUBNET_NAME=”mySubnet”
export APP_SUBNET_NAME=”acaSubnet”
export VM_NAME=”nginxVM”
export VM_IMAGE=”Ubuntu2204″
export VM_SIZE=”Standard_DS1_v2″
export VM_KEY=mykey$(echo $random)
export ADMIN_USERNAME=”azureuser”
export ADMIN_PASSWORD=”Password123$” # Replace with your actual password
export CONTAINER_APPS_ENV=sshacae$(echo $random)
export REGISTRY_NAME=sshacr$(echo $random)
export REGISTRY_SKU=”Basic”
export CONTAINER_APP_IMAGE=”mycontainerapp:latest”
export SSH_CLIENT_CONTAINER_IMAGE=”sshclientcontainer:latest”
export CONTAINER_APP_NAME=”mycontainerapp”
export SSH_CLIENT_CONTAINER_APP_NAME=”sshclientcontainerapp”
export SSH_PORT=22
export NGINX_PORT=80 Read More
How to get a and b using fit(x,y,power1)
I have XY data and fit it with y=ax^b. I use f = fit(x,y,power1) but don’t know how to get "a" and "b" to continue working with them. f(1) gives a, [a,b] = fit(x,y,power1) doesn’t help also.I have XY data and fit it with y=ax^b. I use f = fit(x,y,power1) but don’t know how to get "a" and "b" to continue working with them. f(1) gives a, [a,b] = fit(x,y,power1) doesn’t help also. I have XY data and fit it with y=ax^b. I use f = fit(x,y,power1) but don’t know how to get "a" and "b" to continue working with them. f(1) gives a, [a,b] = fit(x,y,power1) doesn’t help also. curve fitting toolbox MATLAB Answers — New Questions
SharePoint Library Shows 0 Items but Counts Files Count Is more in Site Content
I’m facing an odd issue in my SharePoint online document library. I had a folder where I added a few items and then deleted them. Now, the library shows 0 items, but the site content shows a count of 3.
Could this be a syncing issue? It’s been over 24 hours since I created a new library. I even checked the Recycle Bin (1st & 2nd stage) and cleared my browser cache. I’m the site collection admin, so permissions shouldn’t be a problem.
FYI, When I get data from the API, it’s still returning those 2 files, and the paths are valid and these are
not some hidden folders
Has anyone else experienced this? Any insights would be greatly appreciated!
I’m facing an odd issue in my SharePoint online document library. I had a folder where I added a few items and then deleted them. Now, the library shows 0 items, but the site content shows a count of 3. Could this be a syncing issue? It’s been over 24 hours since I created a new library. I even checked the Recycle Bin (1st & 2nd stage) and cleared my browser cache. I’m the site collection admin, so permissions shouldn’t be a problem. FYI, When I get data from the API, it’s still returning those 2 files, and the paths are valid and these arenot some hidden folders Has anyone else experienced this? Any insights would be greatly appreciated! Read More
When I run Matlab on Ubuntu 20.04.1 and load Matlab project, Matlab crashes
WHen I launch Matlab with terminal command, it shows some failure. I don’t know if they are relevant to the issue?
MATLAB is selecting SOFTWARE OPENGL rendering.
MESA-LOADER: failed to open iris: /usr/lib/dri/iris_dri.so: cannot open shared object file: No such file or directory (search paths /usr/lib/x86_64-linux-gnu/dri:$${ORIGIN}/dri:/usr/lib/dri, suffix _dri)
failed to load driver: iris
MESA-LOADER: failed to open kms_swrast: /usr/lib/dri/kms_swrast_dri.so: cannot open shared object file: No such file or directory (search paths /usr/lib/x86_64-linux-gnu/dri:$${ORIGIN}/dri:/usr/lib/dri, suffix _dri)
failed to load driver: kms_swrast
MESA-LOADER: failed to open swrast: /usr/lib/dri/swrast_dri.so: cannot open shared object file: No such file or directory (search paths /usr/lib/x86_64-linux-gnu/dri:$${ORIGIN}/dri:/usr/lib/dri, suffix _dri)
failed to load swrast driverWHen I launch Matlab with terminal command, it shows some failure. I don’t know if they are relevant to the issue?
MATLAB is selecting SOFTWARE OPENGL rendering.
MESA-LOADER: failed to open iris: /usr/lib/dri/iris_dri.so: cannot open shared object file: No such file or directory (search paths /usr/lib/x86_64-linux-gnu/dri:$${ORIGIN}/dri:/usr/lib/dri, suffix _dri)
failed to load driver: iris
MESA-LOADER: failed to open kms_swrast: /usr/lib/dri/kms_swrast_dri.so: cannot open shared object file: No such file or directory (search paths /usr/lib/x86_64-linux-gnu/dri:$${ORIGIN}/dri:/usr/lib/dri, suffix _dri)
failed to load driver: kms_swrast
MESA-LOADER: failed to open swrast: /usr/lib/dri/swrast_dri.so: cannot open shared object file: No such file or directory (search paths /usr/lib/x86_64-linux-gnu/dri:$${ORIGIN}/dri:/usr/lib/dri, suffix _dri)
failed to load swrast driver WHen I launch Matlab with terminal command, it shows some failure. I don’t know if they are relevant to the issue?
MATLAB is selecting SOFTWARE OPENGL rendering.
MESA-LOADER: failed to open iris: /usr/lib/dri/iris_dri.so: cannot open shared object file: No such file or directory (search paths /usr/lib/x86_64-linux-gnu/dri:$${ORIGIN}/dri:/usr/lib/dri, suffix _dri)
failed to load driver: iris
MESA-LOADER: failed to open kms_swrast: /usr/lib/dri/kms_swrast_dri.so: cannot open shared object file: No such file or directory (search paths /usr/lib/x86_64-linux-gnu/dri:$${ORIGIN}/dri:/usr/lib/dri, suffix _dri)
failed to load driver: kms_swrast
MESA-LOADER: failed to open swrast: /usr/lib/dri/swrast_dri.so: cannot open shared object file: No such file or directory (search paths /usr/lib/x86_64-linux-gnu/dri:$${ORIGIN}/dri:/usr/lib/dri, suffix _dri)
failed to load swrast driver #matlab, #ubuntu MATLAB Answers — New Questions
Dot indexing is not supported for variables of this type.
I get this error while running the code.
Dot indexing is not supported for variables of this type.
Error in Fun_Objective (line 23)
Conv = Aspen.Tree.FindNode("DataResults SummaryRun-StatusOutputPER_ERROR").Value; %Convergence
I am getting an error suddenly after certain interations ( sometimes 1/5/20 iterations) that dot indexing not supported in the statement above. When I go and check the aspen file it just shows required input complete, but does not run. At the start I have reinit the code and I do run the engine, but it still shows required input complete, and the simulation does not run. Does anyone face a similar problem like this?I get this error while running the code.
Dot indexing is not supported for variables of this type.
Error in Fun_Objective (line 23)
Conv = Aspen.Tree.FindNode("DataResults SummaryRun-StatusOutputPER_ERROR").Value; %Convergence
I am getting an error suddenly after certain interations ( sometimes 1/5/20 iterations) that dot indexing not supported in the statement above. When I go and check the aspen file it just shows required input complete, but does not run. At the start I have reinit the code and I do run the engine, but it still shows required input complete, and the simulation does not run. Does anyone face a similar problem like this? I get this error while running the code.
Dot indexing is not supported for variables of this type.
Error in Fun_Objective (line 23)
Conv = Aspen.Tree.FindNode("DataResults SummaryRun-StatusOutputPER_ERROR").Value; %Convergence
I am getting an error suddenly after certain interations ( sometimes 1/5/20 iterations) that dot indexing not supported in the statement above. When I go and check the aspen file it just shows required input complete, but does not run. At the start I have reinit the code and I do run the engine, but it still shows required input complete, and the simulation does not run. Does anyone face a similar problem like this? matlab, aspen, tac, ga MATLAB Answers — New Questions
果博东方客服电话【微4766168】
介绍
在为 ETL（提取、转换、加载）管道构建强大的基础时，Azure 数据工厂或 Azure Synapse Analytics、Azure Batch 和 Azure 存储这三者不可或缺。这些工具可实现跨不同数据源的高效数据移动、转换和处理，从而帮助我们实现战略目标。
本文档提供了有关如何使用 SAMI 对 Azure Batch 和使用 Synapse SAMI 对 Azure Storage 进行身份验证的全面指南。这可以实现用户驱动的存储连接，从而促进数据提取。此外，它还允许使用自定义活动（例如高性能计算 (HPC)）来处理提取的数据。
这些功能的关键推动因素是 Synapse Pipeline。作为主要的协调器，Synapse Pipeline 擅长以安全的方式集成各种 Azure 资源。其功能可以扩展到 Azure 数据工厂 (ADF)，从而提供更广泛的数据管理和转换范围。
通过本指南，您将深入了解如何利用这些强大的 Azure 服务来优化您的数据处理工作流程。
服务概述
在此过程中，我们将使用不同的服务，下面您可以找到有关每种服务的详细信息。
Azure Synapse 分析/数据工厂
Azure Synapse Analytics 是一项企业分析服务，可加快跨数据仓库和大数据系统的洞察速度。 Azure Synapse 汇集了企业数据仓库中使用的最佳 SQL 技术、用于大数据的 Spark 技术、用于日志和时间序列分析的数据资源管理器、用于数据集成和 ETL/ELT 的管道，以及与其他 Azure 服务（如 Power BI、CosmosDB 和 AzureML）的深度集成。文档：什么是 Azure Synapse Analytics？ – Azure Synapse Analytics | Microsoft LearnAzure 数据工厂简介 – Azure 数据工厂 | Microsoft Learn
Azure 批处理
Azure Batch 是一种功能强大的平台服务，旨在在云中运行大规模并行和高性能计算 (HPC) 应用程序。文档：Azure Batch 在云中运行大型并行作业 – Azure Batch | Microsoft Learn
Azure 存储
Azure 存储为各种数据类型提供可扩展且安全的存储服务，包括Azure Blob 存储、 Azure 表存储和 Azure 队列存储等服务 。文档：Azure 存储简介 – Azure 上的云存储 | Microsoft Learn
托管标识
Azure 托管标识是 Azure Active Directory 的一项功能，可自动管理应用程序在连接到支持 Azure AD 身份验证的资源时使用的凭据。它们消除了开发人员管理机密、凭据、证书和密钥的需要。托管标识有两种类型：系统分配：与您的应用程序相关。用户分配：可分配给你的应用的独立 Azure 资源文档： Azure 资源的托管标识 – Azure 资源的托管标识 | Microsoft Learn
设想
运行 ADF/Synapse 管道，提取位于存储帐户中的脚本，并使用用户分配的托管标识 (UAMI) 将其执行到批处理节点中以对存储进行身份验证，并使用系统分配的托管标识 (SAMI) 对批处理进行身份验证。
先决条件
ADF/Synapse 工作区文档：快速入门：创建 Synapse 工作区 – Azure Synapse Analytics | Microsoft LearnUA 管理身份文档：管理用户分配的托管标识 – Azure 资源的托管标识 | Microsoft Learn博客文档：https://techcommunity.microsoft.com/t5/azure-data-factory-blog/support-for-user-assigned-managed-ide…存储帐户Storage Account文档：创建存储帐户 – Azure 存储 | Microsoft Learn
程序概述
在此过程中，我们将逐步完成以下操作：
创建 UAMI 凭证为存储和批处理帐户创建链接服务将 UAMI 和 SAMI 添加到存储和批处理账户创建、配置和执行 ADF/Synapse 管道为了避免冗余，在所有练习和示例中，我们将把 ADF（门户、工作区、管道、作业、链接服务）称为 Synapse。调试
程序
创建 UAMI 凭证
1. 在您的 Synapse Portal 中，转到管理 -> 凭据 -> 新建并填写详细信息，然后单击创建。
为存储和批处理创建链接服务连接
2. 在 Synapse 门户中，转到“管理 – 链接服务”->“新建”->“Azure Blob 存储”->“继续”并填写表格
a. 身份验证类型：UAMI
b. Azure 订阅：选择您的订阅
c. 存储帐户名称：选择要使用的脚本所在的存储帐户
d. 凭证：选择在步骤 #1 中创建的
e. 点击创建
3. 在 Azure 门户中，转到您的批处理帐户 -> 密钥，并复制下一步要使用的批处理帐户名称和帐户端点，还要复制此示例要使用的池名称。
4. 在 Synapse 门户中，转到“管理”->“链接服务”->“新建”->“Azure Batch”->“继续”并填写信息
a. 身份验证方法：SAMI（复制托管身份名称，稍后会用到）
b. 账户名称、批处理 URL 和池名称：在此处粘贴从步骤 3 复制的值
c.存储链接服务名称：选择步骤#2中创建的名称
5. 发布所有更改
将 UAMI RBAC 角色添加到存储帐户
6. 在 Azure 门户中，转到您的存储帐户 -> 访问控制 (IAM)
a. 单击“添加选项”，然后单击“添加角色分配”并搜索“存储 Blob 数据贡献者”，然后单击“下一步”。
b. 选择托管身份并选择您的 UAMI，单击选择，然后单击下一步、下一步和审核 + 分配。
介绍在为 ETL（提取、转换、加载）管道构建强大的基础时，Azure 数据工厂或 Azure Synapse Analytics、Azure Batch 和 Azure 存储这三者不可或缺。这些工具可实现跨不同数据源的高效数据移动、转换和处理，从而帮助我们实现战略目标。 本文档提供了有关如何使用 SAMI 对 Azure Batch 和使用 Synapse SAMI 对 Azure Storage 进行身份验证的全面指南。这可以实现用户驱动的存储连接，从而促进数据提取。此外，它还允许使用自定义活动（例如高性能计算 (HPC)）来处理提取的数据。 这些功能的关键推动因素是 Synapse Pipeline。作为主要的协调器，Synapse Pipeline 擅长以安全的方式集成各种 Azure 资源。其功能可以扩展到 Azure 数据工厂 (ADF)，从而提供更广泛的数据管理和转换范围。 通过本指南，您将深入了解如何利用这些强大的 Azure 服务来优化您的数据处理工作流程。 服务概述在此过程中，我们将使用不同的服务，下面您可以找到有关每种服务的详细信息。 Azure Synapse 分析/数据工厂Azure Synapse Analytics 是一项企业分析服务，可加快跨数据仓库和大数据系统的洞察速度。 Azure Synapse 汇集了企业数据仓库中使用的最佳 SQL 技术、用于大数据的 Spark 技术、用于日志和时间序列分析的数据资源管理器、用于数据集成和 ETL/ELT 的管道，以及与其他 Azure 服务（如 Power BI、CosmosDB 和 AzureML）的深度集成。文档：什么是 Azure Synapse Analytics？ – Azure Synapse Analytics | Microsoft LearnAzure 数据工厂简介 – Azure 数据工厂 | Microsoft Learn Azure 批处理Azure Batch 是一种功能强大的平台服务，旨在在云中运行大规模并行和高性能计算 (HPC) 应用程序。文档：Azure Batch 在云中运行大型并行作业 – Azure Batch | Microsoft Learn Azure 存储Azure 存储为各种数据类型提供可扩展且安全的存储服务，包括Azure Blob 存储、 Azure 表存储和 Azure 队列存储等服务 。文档：Azure 存储简介 – Azure 上的云存储 | Microsoft Learn托管标识Azure 托管标识是 Azure Active Directory 的一项功能，可自动管理应用程序在连接到支持 Azure AD 身份验证的资源时使用的凭据。它们消除了开发人员管理机密、凭据、证书和密钥的需要。托管标识有两种类型：系统分配：与您的应用程序相关。用户分配：可分配给你的应用的独立 Azure 资源文档： Azure 资源的托管标识 – Azure 资源的托管标识 | Microsoft Learn 设想运行 ADF/Synapse 管道，提取位于存储帐户中的脚本，并使用用户分配的托管标识 (UAMI) 将其执行到批处理节点中以对存储进行身份验证，并使用系统分配的托管标识 (SAMI) 对批处理进行身份验证。 先决条件ADF/Synapse 工作区文档：快速入门：创建 Synapse 工作区 – Azure Synapse Analytics | Microsoft LearnUA 管理身份文档：管理用户分配的托管标识 – Azure 资源的托管标识 | Microsoft Learn博客文档：https://techcommunity.microsoft.com/t5/azure-data-factory-blog/support-for-user-assigned-managed-ide…存储帐户Storage Account文档：创建存储帐户 – Azure 存储 | Microsoft Learn 程序概述在此过程中，我们将逐步完成以下操作： 创建 UAMI 凭证为存储和批处理帐户创建链接服务将 UAMI 和 SAMI 添加到存储和批处理账户创建、配置和执行 ADF/Synapse 管道为了避免冗余，在所有练习和示例中，我们将把 ADF（门户、工作区、管道、作业、链接服务）称为 Synapse。调试 程序创建 UAMI 凭证1. 在您的 Synapse Portal 中，转到管理 -> 凭据 -> 新建并填写详细信息，然后单击创建。 为存储和批处理创建链接服务连接2. 在 Synapse 门户中，转到“管理 – 链接服务”->“新建”->“Azure Blob 存储”->“继续”并填写表格a. 身份验证类型：UAMIb. Azure 订阅：选择您的订阅c. 存储帐户名称：选择要使用的脚本所在的存储帐户d. 凭证：选择在步骤 #1 中创建的e. 点击创建 3. 在 Azure 门户中，转到您的批处理帐户 -> 密钥，并复制下一步要使用的批处理帐户名称和帐户端点，还要复制此示例要使用的池名称。 4. 在 Synapse 门户中，转到“管理”->“链接服务”->“新建”->“Azure Batch”->“继续”并填写信息a. 身份验证方法：SAMI（复制托管身份名称，稍后会用到）b. 账户名称、批处理 URL 和池名称：在此处粘贴从步骤 3 复制的值c.存储链接服务名称：选择步骤#2中创建的名称 5. 发布所有更改 将 UAMI RBAC 角色添加到存储帐户6. 在 Azure 门户中，转到您的存储帐户 -> 访问控制 (IAM)a. 单击“添加选项”，然后单击“添加角色分配”并搜索“存储 Blob 数据贡献者”，然后单击“下一步”。 b. 选择托管身份并选择您的 UAMI，单击选择，然后单击下一步、下一步和审核 + 分配。 Read More
Summariing Data
Hi,
I am trying to summarise data from a number of tabs, the data having the same format as reflected below.
I have tried to use Powery Query to summarise in a table format Employee name, type, shift date, Department, Role, Rate and Net Hours but I could not determine how to bring employee name and type into the table? Shift date needs to run down the page as reflected above. If anyoen can help with how to summarise the data this would be great.
Thanks. Matt.
Hi,I am trying to summarise data from a number of tabs, the data having the same format as reflected below. I have tried to use Powery Query to summarise in a table format Employee name, type, shift date, Department, Role, Rate and Net Hours but I could not determine how to bring employee name and type into the table? Shift date needs to run down the page as reflected above. If anyoen can help with how to summarise the data this would be great.Thanks. Matt. Read More
EXCEL CLOSE WHEN DONT FIND IMAGE
Hi, i have a file that load multiple image by name on a column.
But when don’t find the image file… the file close him self.
The file and the folder with the images are in the same place.
There a way to stop that? I try use On error resume next, but the error continues…
Windows
excel 365
64
Hi, i have a file that load multiple image by name on a column.But when don’t find the image file… the file close him self.The file and the folder with the images are in the same place.There a way to stop that? I try use On error resume next, but the error continues…Windowsexcel 36564 Read More
need data about tach signal?
greetings !!
I want to know about tach signal(like signal from tachometer) in MATLAB and its generation in matlab like attached image(Might be aperiodic).. And i want to add some Gaussian noise to that signal to find frequency b/w peeks.
I tried using "pulse train" but not getting as i like..greetings !!
I want to know about tach signal(like signal from tachometer) in MATLAB and its generation in matlab like attached image(Might be aperiodic).. And i want to add some Gaussian noise to that signal to find frequency b/w peeks.
I tried using "pulse train" but not getting as i like.. greetings !!
I want to know about tach signal(like signal from tachometer) in MATLAB and its generation in matlab like attached image(Might be aperiodic).. And i want to add some Gaussian noise to that signal to find frequency b/w peeks.
I tried using "pulse train" but not getting as i like.. signal generation, pulse generation, tach signal MATLAB Answers — New Questions
Latest Dhan Referral Code 2024 is SHAR68 – Best Signup Rewards
Dhan App Referral Code 2024
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Earn Referral Rewards, for lifetime: When your invited friends trade on the Dhan App, you will earn a lifetime share of 20% of the brokerage revenue generated by them.Receive daily credit for referral: Rewards are accrued every day, similar to daily dividends.
Dhan Trading App Refer & Earn Conclusion.
Considering all the offers, and features, The Dhan Referral program is an excellent refer and earn program. Through this program, you have the great opportunity to earn lifelong incentives in the form of 20% brokerage shares on each referral. Therefore, use the Dhan referral Code — SHAR68 right now to create your free Demat account.
What is Dhan Referral Code 2024? Apply Dhan Referral Code – SHAR68 when signing up for a Free Demat account & avail 0 AMC charges, best technology-enabled app and a web interface for traders and investors without any hidden charges. Dhan App Referral Code 2024ExchangeDhan AppDhan Referral CodeSHAR68Sign Up RewardsChanges time to timePer Referral20% brokerage CommissionsWhy Choose Dhan?Best trading platform in India Dhan App Referral Code is SHAR68Just like Angel One Referral Code , Dhan App also has a Referral program. Using Dhan App Referral Code — SHAR68, you will get many benefits like Free Demat Account, Rs.0 charges on Equity Delivery, 50% Off on Brokerage, Flat Rs.20 Brokerage plan & more.So 👉Open FREE Dhan Account in 2 Minutes and get started in your Investment Journey. Read More here about Dhan Referral Code .Benefits of Applying Dhan Referral CodeGet Lifetime Free Demat Account in 2023, 2024, 2025 & Coming Years.Zero Annual Maintenance ChargesFlat 50% Off on Brokerage for Women (Now Removed)Zero Brokerage Charges on Equity Delivery Trade, IPO, ETFFlat Rs.20 Brokerage plan for F&O, Intraday Trading After 1 Month PeriodAdvantages of Dhan Referral ProgramIf you tell your relatives, and friends about the Dhan Trading app and they download & applied your referral code — SHAR68 or link that you provide, you will get Flat 20% of the brokerage fees paid by those people. These earnings will be for a lifetime. So it’s an excellent opportunity to participate in Dhan App Refer and Earn program and make most of it. Dhan Referral Money is like real money you will get in your bank account.Why to Choose Dhan AppTrading IQ and ChartIQ: Dhan have some real time charts which is phenomenal and also have upgraded and interactive features, you will be able to trade more quickly and effectively. So for Trader Dhan is a good to go trading App.Dhan Trading App has been rapidly gaining market share because of its user-friendly and practical feature set. With all the features Trading is very easy and quick on Dhan App. So for Daily Trader it’s a great choice.Easy access to more than a hundred different technical indicators which will help you make money trading stocks, currencies, futures, and options.The Dhan Brokerage Calculator is also a very good feature that can assist you in determining the amount of brokerage fees that are associated with your transactions on the NSE, BSE, and MCX exchanges.Participate in IPO, Equity Investment, ETF investment at Rs.0 Brokerage charges. Additionally, you can apply Dhan Stocks App referral code — SHAR68.In the Future, you can also Invest in Direct Mutual Funds via Dhan App & Dhan Website.What is the latest Dhan Referral Code & Referral Link?Dhan App Referral Code 2024, 2025 is same — SHAR68. Basically Dhan App Referral Code — SHAR68 is a permanent code that doesn’t change with year. So Even in Year 2026, 2027, 2028, 2029, 2030 this code will continue to work. Apply Dhan Referral Code Today & Avail Benifits.Dhan Refer and Earn Offer Terms and ConditionsEarn Referral Rewards, for lifetime: When your invited friends trade on the Dhan App, you will earn a lifetime share of 20% of the brokerage revenue generated by them.Receive daily credit for referral: Rewards are accrued every day, similar to daily dividends.Dhan Trading App Refer & Earn Conclusion.Considering all the offers, and features, The Dhan Referral program is an excellent refer and earn program. Through this program, you have the great opportunity to earn lifelong incentives in the form of 20% brokerage shares on each referral. Therefore, use the Dhan referral Code — SHAR68 right now to create your free Demat account. Read More
Full screen figure with xticklabels multiline problem
Hi,
Short decription of what I need:
full screean figure of some bars
xticklabels multiline
What is wrong?
the text of xtickslabels is not fully displaying (down part is cut, not depends on the xtickangle)
no space between the text of xticklabels and the axis
figure(‘units’,’normalized’,’outerposition’,[0 0 1 1]);
set(gcf,’Visible’, ‘off’);
bar(1,CO2_o(1),’FaceColor’,’#572280′);
hold on
bar(2,CO2_o(2),’FaceColor’,’#953FBA’)
hold on
bar(3,CO2_o(3),’FaceColor’,’#A279CB’);
hold on
bar(4,CO2_o(4),’FaceColor’,’#D7B3FA’)
hold on
bar(5,CO2_o(5),’FaceColor’,’#DA5900′)
hold on
bar(6,CO2_o(6),’FaceColor’,’#FF7C01′);
hold on
bar(7,CO2_o(7),’FaceColor’,’#FFAB02′)
hold on
bar(8,CO2_o(9),’FaceColor’,’#EDD263′)
hold on
bar(9,CO2_o(9),’FaceColor’,’#AB0909′)
hold on
bar(10,CO2_o(10),’FaceColor’,’#C04655′);
hold on
bar(11,CO2_o(11),’FaceColor’,’#F17A89′)
hold on
bar(12,CO2_o(12),’FaceColor’,’#F6AFAF’);
hold on
bar(13,CO2_o(13),’FaceColor’,’#2865CA’);
hold on
bar(14,CO2_o(14),’FaceColor’,’#B1CCF9′)
hold off
title(‘EMISJA CO_{2} – JAZDA OSZCZĘDNA’,’fontSize’,18)
ylabel([‘$[frac{kg}{1000 hspace{1mm}paskm}]$’], ‘Interpreter’, ‘latex’,’fontweight’,’bold’,’fontSize’,14)
h=[];
ax=gca;
ax.XTickLabel = ”;
myLabels={‘Alstom EMU250 ED250’, ‘Alstom EMU250 ED250’,…
‘Alstom EMU250 ED250’, ‘Alstom EMU250 ED250’,…
‘Pesa Elf’, ‘Pesa Elf’,…
‘Pesa Elf’, ‘Pesa Elf’,…
‘ADtranz 113E EU11’, ‘ADtranz 113E EU11’,…
‘ADtranz 113E EU11’, ‘ADtranz 113E EU11’,…
‘Pafawag 104E EP09’, ‘Pafawag 104E EP09’;
‘Koniecpol-Opole’, ‘ Koniecpol-Opole’, ‘Opole-Koniecpol’ ,’Opole-Koniecpol’,…
‘Koniecpol-Opole’, ‘ Koniecpol-Opole’, ‘Opole-Koniecpol’ ,’Opole-Koniecpol’,…
‘Koniecpol-Opole’, ‘ Koniecpol-Opole’, ‘Opole-Koniecpol’ ,’Opole-Koniecpol’,…
‘Koniecpol-Opole’, ‘Opole-Koniecpol’;
”,’REKU’,”,’REKU’,…
”,’REKU’,”,’REKU’,…
”,’REKU’,”,’REKU’,…
”,”};
for i = 1:length(myLabels)
h(i)= text(sprintf(‘%sn%sn%s’, myLabels{:,i}), …
‘horizontalalignment’, ‘right’, ‘verticalalignment’, ‘middle’);
set(h,’Rotation’,60,’fontsize’,10);
endHi,
Short decription of what I need:
full screean figure of some bars
xticklabels multiline
What is wrong?
the text of xtickslabels is not fully displaying (down part is cut, not depends on the xtickangle)
no space between the text of xticklabels and the axis
figure(‘units’,’normalized’,’outerposition’,[0 0 1 1]);
set(gcf,’Visible’, ‘off’);
bar(1,CO2_o(1),’FaceColor’,’#572280′);
hold on
bar(2,CO2_o(2),’FaceColor’,’#953FBA’)
hold on
bar(3,CO2_o(3),’FaceColor’,’#A279CB’);
hold on
bar(4,CO2_o(4),’FaceColor’,’#D7B3FA’)
hold on
bar(5,CO2_o(5),’FaceColor’,’#DA5900′)
hold on
bar(6,CO2_o(6),’FaceColor’,’#FF7C01′);
hold on
bar(7,CO2_o(7),’FaceColor’,’#FFAB02′)
hold on
bar(8,CO2_o(9),’FaceColor’,’#EDD263′)
hold on
bar(9,CO2_o(9),’FaceColor’,’#AB0909′)
hold on
bar(10,CO2_o(10),’FaceColor’,’#C04655′);
hold on
bar(11,CO2_o(11),’FaceColor’,’#F17A89′)
hold on
bar(12,CO2_o(12),’FaceColor’,’#F6AFAF’);
hold on
bar(13,CO2_o(13),’FaceColor’,’#2865CA’);
hold on
bar(14,CO2_o(14),’FaceColor’,’#B1CCF9′)
hold off
title(‘EMISJA CO_{2} – JAZDA OSZCZĘDNA’,’fontSize’,18)
ylabel([‘$[frac{kg}{1000 hspace{1mm}paskm}]$’], ‘Interpreter’, ‘latex’,’fontweight’,’bold’,’fontSize’,14)
h=[];
ax=gca;
ax.XTickLabel = ”;
myLabels={‘Alstom EMU250 ED250’, ‘Alstom EMU250 ED250’,…
‘Alstom EMU250 ED250’, ‘Alstom EMU250 ED250’,…
‘Pesa Elf’, ‘Pesa Elf’,…
‘Pesa Elf’, ‘Pesa Elf’,…
‘ADtranz 113E EU11’, ‘ADtranz 113E EU11’,…
‘ADtranz 113E EU11’, ‘ADtranz 113E EU11’,…
‘Pafawag 104E EP09’, ‘Pafawag 104E EP09’;
‘Koniecpol-Opole’, ‘ Koniecpol-Opole’, ‘Opole-Koniecpol’ ,’Opole-Koniecpol’,…
‘Koniecpol-Opole’, ‘ Koniecpol-Opole’, ‘Opole-Koniecpol’ ,’Opole-Koniecpol’,…
‘Koniecpol-Opole’, ‘ Koniecpol-Opole’, ‘Opole-Koniecpol’ ,’Opole-Koniecpol’,…
‘Koniecpol-Opole’, ‘Opole-Koniecpol’;
”,’REKU’,”,’REKU’,…
”,’REKU’,”,’REKU’,…
”,’REKU’,”,’REKU’,…
”,”};
for i = 1:length(myLabels)
h(i)= text(sprintf(‘%sn%sn%s’, myLabels{:,i}), …
‘horizontalalignment’, ‘right’, ‘verticalalignment’, ‘middle’);
set(h,’Rotation’,60,’fontsize’,10);
end Hi,
Short decription of what I need:
full screean figure of some bars
xticklabels multiline
What is wrong?
the text of xtickslabels is not fully displaying (down part is cut, not depends on the xtickangle)
no space between the text of xticklabels and the axis
figure(‘units’,’normalized’,’outerposition’,[0 0 1 1]);
set(gcf,’Visible’, ‘off’);
bar(1,CO2_o(1),’FaceColor’,’#572280′);
hold on
bar(2,CO2_o(2),’FaceColor’,’#953FBA’)
hold on
bar(3,CO2_o(3),’FaceColor’,’#A279CB’);
hold on
bar(4,CO2_o(4),’FaceColor’,’#D7B3FA’)
hold on
bar(5,CO2_o(5),’FaceColor’,’#DA5900′)
hold on
bar(6,CO2_o(6),’FaceColor’,’#FF7C01′);
hold on
bar(7,CO2_o(7),’FaceColor’,’#FFAB02′)
hold on
bar(8,CO2_o(9),’FaceColor’,’#EDD263′)
hold on
bar(9,CO2_o(9),’FaceColor’,’#AB0909′)
hold on
bar(10,CO2_o(10),’FaceColor’,’#C04655′);
hold on
bar(11,CO2_o(11),’FaceColor’,’#F17A89′)
hold on
bar(12,CO2_o(12),’FaceColor’,’#F6AFAF’);
hold on
bar(13,CO2_o(13),’FaceColor’,’#2865CA’);
hold on
bar(14,CO2_o(14),’FaceColor’,’#B1CCF9′)
hold off
title(‘EMISJA CO_{2} – JAZDA OSZCZĘDNA’,’fontSize’,18)
ylabel([‘$[frac{kg}{1000 hspace{1mm}paskm}]$’], ‘Interpreter’, ‘latex’,’fontweight’,’bold’,’fontSize’,14)
h=[];
ax=gca;
ax.XTickLabel = ”;
myLabels={‘Alstom EMU250 ED250’, ‘Alstom EMU250 ED250’,…
‘Alstom EMU250 ED250’, ‘Alstom EMU250 ED250’,…
‘Pesa Elf’, ‘Pesa Elf’,…
‘Pesa Elf’, ‘Pesa Elf’,…
‘ADtranz 113E EU11’, ‘ADtranz 113E EU11’,…
‘ADtranz 113E EU11’, ‘ADtranz 113E EU11’,…
‘Pafawag 104E EP09’, ‘Pafawag 104E EP09’;
‘Koniecpol-Opole’, ‘ Koniecpol-Opole’, ‘Opole-Koniecpol’ ,’Opole-Koniecpol’,…
‘Koniecpol-Opole’, ‘ Koniecpol-Opole’, ‘Opole-Koniecpol’ ,’Opole-Koniecpol’,…
‘Koniecpol-Opole’, ‘ Koniecpol-Opole’, ‘Opole-Koniecpol’ ,’Opole-Koniecpol’,…
‘Koniecpol-Opole’, ‘Opole-Koniecpol’;
”,’REKU’,”,’REKU’,…
”,’REKU’,”,’REKU’,…
”,’REKU’,”,’REKU’,…
”,”};
for i = 1:length(myLabels)
h(i)= text(sprintf(‘%sn%sn%s’, myLabels{:,i}), …
‘horizontalalignment’, ‘right’, ‘verticalalignment’, ‘middle’);
set(h,’Rotation’,60,’fontsize’,10);
end xticklabels, multiline, chart, bar, fullscreen, screensize MATLAB Answers — New Questions
Few users do not show interactive user sign in logs, only non-interactive
we have few users that are not showing Windows Sign in data in Azure Sign in logs as interactive. It even happened to me. Suddenly. I did not show any data for one day in as sign-in interactive tab. We re using Windows Hello for Business. Usually for values interactive and non interactive values show up for Windows sign in. Any ideas why this is happening randomly? Found this link, not sure if this is the reason why: https://learn.microsoft.com/en-us/entra/identity-platform/configurable-token-lifetimes
we have few users that are not showing Windows Sign in data in Azure Sign in logs as interactive. It even happened to me. Suddenly. I did not show any data for one day in as sign-in interactive tab. We re using Windows Hello for Business. Usually for values interactive and non interactive values show up for Windows sign in. Any ideas why this is happening randomly? Found this link, not sure if this is the reason why: https://learn.microsoft.com/en-us/entra/identity-platform/configurable-token-lifetimes Read More
how to draw a peak line
Hi, i’ve data and i want to draw a line that corresponds to the peaks (where there is a zero the previous value >0 must be pasted) (See black line painted)
Is there a function or graph type that does this to me?Hi, i’ve data and i want to draw a line that corresponds to the peaks (where there is a zero the previous value >0 must be pasted) (See black line painted)
Is there a function or graph type that does this to me? Hi, i’ve data and i want to draw a line that corresponds to the peaks (where there is a zero the previous value >0 must be pasted) (See black line painted)
Is there a function or graph type that does this to me? how to draw a peak line MATLAB Answers — New Questions
I continue to get this error and can’t figure out how to fix it, Unable to perform assignment because the size of the left side is 1-by-4 and the size of the right side is 1-b
% Given temperature-depth data
z = [0, -2.3, -4.6, -6.9, -9.2, -11.5, -13.8, -16.1]; % Depth in meters
T = [22.8, 22.7, 22.5, 20.6, 13.9, 11.7, 11.2, 11.1]; % Temperature in Celsius
% Step 1: Plot the data points
figure;
plot(T, z, ‘r*’); % Plotting temperature vs depth with red stars
xlabel(‘Temperature (°C)’);
ylabel(‘Depth (m)’);
title(‘Temperature vs Depth’);
set(gca, ‘YDir’,’reverse’); % Inverting the y-axis
grid on;
% Step 2: Sort the data since the depth must be in increasing order
[z_sorted, sortIndex] = sort(z);
T_sorted = T(sortIndex);
% Step 3: Create a cubic spline interpolation
cs = spline(z_sorted, T_sorted);
% Step 4: Evaluate the spline and its derivatives on a finer grid
z_fine = linspace(min(z_sorted), max(z_sorted), 500);
T_spline = ppval(cs, z_fine);
% Step 5: Find the first and second derivatives of the spline
[breaks,coefs,l,k,d] = unmkpp(cs); % Extracts the pieces of the cubic spline
dcoefs = coefs; % Derivative coefficients
% Each row of dcoefs will be the coefficients of the polynomial of a piece
for j = 1:l
dcoefs(j,:) = polyder(dcoefs(j,:));
end
% Make a pp-form of derivative
csd1 = mkpp(breaks,dcoefs(:,1:k-1));
% First derivative evaluation
T_spline_deriv = ppval(csd1, z_fine);
% Find the second derivative
for j = 1:l
dcoefs(j,:) = polyder(dcoefs(j,:));
end
% Make a pp-form of second derivative
csd2 = mkpp(breaks,dcoefs(:,1:k-2));
% Second derivative evaluation
T_spline_second_deriv = ppval(csd2, z_fine);
% Step 6: Locate the thermocline by finding the depth where the second derivative
% changes sign, and the first derivative is a maximum
inflection_points = find(diff(sign(T_spline_second_deriv)) ~= 0) + 1;
[~, max_gradient_index] = max(abs(T_spline_deriv(inflection_points)));
thermocline_depth = z_fine(inflection_points(max_gradient_index));
thermocline_temperature = T_spline(inflection_points(max_gradient_index));
% Display the thermocline depth and temperature
fprintf(‘The thermocline is located at a depth of %.2f m with a temperature of %.2f°C.n’, …
thermocline_depth, thermocline_temperature);% Given temperature-depth data
z = [0, -2.3, -4.6, -6.9, -9.2, -11.5, -13.8, -16.1]; % Depth in meters
T = [22.8, 22.7, 22.5, 20.6, 13.9, 11.7, 11.2, 11.1]; % Temperature in Celsius
% Step 1: Plot the data points
figure;
plot(T, z, ‘r*’); % Plotting temperature vs depth with red stars
xlabel(‘Temperature (°C)’);
ylabel(‘Depth (m)’);
title(‘Temperature vs Depth’);
set(gca, ‘YDir’,’reverse’); % Inverting the y-axis
grid on;
% Step 2: Sort the data since the depth must be in increasing order
[z_sorted, sortIndex] = sort(z);
T_sorted = T(sortIndex);
% Step 3: Create a cubic spline interpolation
cs = spline(z_sorted, T_sorted);
% Step 4: Evaluate the spline and its derivatives on a finer grid
z_fine = linspace(min(z_sorted), max(z_sorted), 500);
T_spline = ppval(cs, z_fine);
% Step 5: Find the first and second derivatives of the spline
[breaks,coefs,l,k,d] = unmkpp(cs); % Extracts the pieces of the cubic spline
dcoefs = coefs; % Derivative coefficients
% Each row of dcoefs will be the coefficients of the polynomial of a piece
for j = 1:l
dcoefs(j,:) = polyder(dcoefs(j,:));
end
% Make a pp-form of derivative
csd1 = mkpp(breaks,dcoefs(:,1:k-1));
% First derivative evaluation
T_spline_deriv = ppval(csd1, z_fine);
% Find the second derivative
for j = 1:l
dcoefs(j,:) = polyder(dcoefs(j,:));
end
% Make a pp-form of second derivative
csd2 = mkpp(breaks,dcoefs(:,1:k-2));
% Second derivative evaluation
T_spline_second_deriv = ppval(csd2, z_fine);
% Step 6: Locate the thermocline by finding the depth where the second derivative
% changes sign, and the first derivative is a maximum
inflection_points = find(diff(sign(T_spline_second_deriv)) ~= 0) + 1;
[~, max_gradient_index] = max(abs(T_spline_deriv(inflection_points)));
thermocline_depth = z_fine(inflection_points(max_gradient_index));
thermocline_temperature = T_spline(inflection_points(max_gradient_index));
% Display the thermocline depth and temperature
fprintf(‘The thermocline is located at a depth of %.2f m with a temperature of %.2f°C.n’, …
thermocline_depth, thermocline_temperature); % Given temperature-depth data
z = [0, -2.3, -4.6, -6.9, -9.2, -11.5, -13.8, -16.1]; % Depth in meters
T = [22.8, 22.7, 22.5, 20.6, 13.9, 11.7, 11.2, 11.1]; % Temperature in Celsius
% Step 1: Plot the data points
figure;
plot(T, z, ‘r*’); % Plotting temperature vs depth with red stars
xlabel(‘Temperature (°C)’);
ylabel(‘Depth (m)’);
title(‘Temperature vs Depth’);
set(gca, ‘YDir’,’reverse’); % Inverting the y-axis
grid on;
% Step 2: Sort the data since the depth must be in increasing order
[z_sorted, sortIndex] = sort(z);
T_sorted = T(sortIndex);
% Step 3: Create a cubic spline interpolation
cs = spline(z_sorted, T_sorted);
% Step 4: Evaluate the spline and its derivatives on a finer grid
z_fine = linspace(min(z_sorted), max(z_sorted), 500);
T_spline = ppval(cs, z_fine);
% Step 5: Find the first and second derivatives of the spline
[breaks,coefs,l,k,d] = unmkpp(cs); % Extracts the pieces of the cubic spline
dcoefs = coefs; % Derivative coefficients
% Each row of dcoefs will be the coefficients of the polynomial of a piece
for j = 1:l
dcoefs(j,:) = polyder(dcoefs(j,:));
end
% Make a pp-form of derivative
csd1 = mkpp(breaks,dcoefs(:,1:k-1));
% First derivative evaluation
T_spline_deriv = ppval(csd1, z_fine);
% Find the second derivative
for j = 1:l
dcoefs(j,:) = polyder(dcoefs(j,:));
end
% Make a pp-form of second derivative
csd2 = mkpp(breaks,dcoefs(:,1:k-2));
% Second derivative evaluation
T_spline_second_deriv = ppval(csd2, z_fine);
% Step 6: Locate the thermocline by finding the depth where the second derivative
% changes sign, and the first derivative is a maximum
inflection_points = find(diff(sign(T_spline_second_deriv)) ~= 0) + 1;
[~, max_gradient_index] = max(abs(T_spline_deriv(inflection_points)));
thermocline_depth = z_fine(inflection_points(max_gradient_index));
thermocline_temperature = T_spline(inflection_points(max_gradient_index));
% Display the thermocline depth and temperature
fprintf(‘The thermocline is located at a depth of %.2f m with a temperature of %.2f°C.n’, …
thermocline_depth, thermocline_temperature); matrices MATLAB Answers — New Questions
5GB email storage instead of 15GB?
Under the free plan, my email storage should be 15GB, according to the list of plans found here: https://www.microsoft.com/en-ca/microsoft-365/outlook/outlook-personal-email-plans?ocid=cmmasasml2s .
But I’m getting a message on outlook, which says “you’re running low on storage”…it says that I’ve used up 4.2GB of my 5GB storage. When I go to “manage storage,” it shows that all of my storage used is in email attachments and zero is onedrive.
If you look at the bottom of the screenshot, you see that it says right there that I have 15GB of email storage under this plan.
What gives? What happened to 15GB? How to I rectify this?
Under the free plan, my email storage should be 15GB, according to the list of plans found here: https://www.microsoft.com/en-ca/microsoft-365/outlook/outlook-personal-email-plans?ocid=cmmasasml2s . But I’m getting a message on outlook, which says “you’re running low on storage”…it says that I’ve used up 4.2GB of my 5GB storage. When I go to “manage storage,” it shows that all of my storage used is in email attachments and zero is onedrive.If you look at the bottom of the screenshot, you see that it says right there that I have 15GB of email storage under this plan.What gives? What happened to 15GB? How to I rectify this? Read More
Gráficos y tablas
Hola Comunidad
Bajé 2 plantillas de presupuesto mensual, pero no logro cambiar los nombres en las tablas y/o gráficos.
Alguien me puede ayudar por favor.
Hola ComunidadBajé 2 plantillas de presupuesto mensual, pero no logro cambiar los nombres en las tablas y/o gráficos.Alguien me puede ayudar por favor. Read More
how do i get a proper time plot for this qudratic equation?
the script seems to be not behaving properly,as it is supposed to be ?
% Define the range for x
x = linspace(-5, 5, 10); % 10 points between -5 and 5
dx1 = 2*x.^2 – 8; % Compute the derivative (velocity field)
% Create a figure
figure;
hold on;
% Plot the real line (x-axis)
plot(x, zeros(size(x)), ‘k’); % The x-axis (real line)
% Plot the vector field as arrows
for i = 1:length(x)
if dx1(i) > 0
% Arrow pointing to the right (positive dx)
quiver(x(i), 0, 0.5, 0, ‘r’, ‘MaxHeadSize’, 0.5);
elseif dx1(i) < 0
% Arrow pointing to the left (negative dx)
quiver(x(i), 0, -0.5, 0, ‘b’, ‘MaxHeadSize’, 0.5);
end
end
% Plot equilibria points
plot([-2, 2], [0, 0], ‘bo’, ‘MarkerFaceColor’, ‘b’, ‘MarkerSize’, 8); % Equilibria at x = -2 and x = 2
% Labels and title
xlabel(‘x’);
ylabel(‘x-dot’);
title(‘1D Vector Field for x-dot = 2x^2 – 8’);
ylim([-1 1]); % Set y-axis limits to keep focus on x-axis
xlim([-4 4]); % Set x-axis limits
grid on; % Add grid
% Add text annotations for equilibrium points
text(-2, 0.2, ‘Stable Equilibrium: x = -2’, ‘HorizontalAlignment’, ‘center’);
text(2, 0.2, ‘Unstable Equilibrium: x = 2’, ‘HorizontalAlignment’, ‘center’);
hold off;
t = [-5 5];
ode_function = @(t,x) 2*x.^2-8;
x0 = [-2 2];
[time, x] = ode45(ode_function, t, x0);
figure;
plot(time, x, ‘LineWidth’,0.5);
title(‘Solution of dx/dt = 2x^2 – 8 vs Time’);
xlabel(‘Time t’);
ylabel(‘x(t)’);
xlim([-t(2) t(2)]); % Adjust x-axis limits for clarity
ylim([-3 3]); % Set y-axis limits to see behavior around -2 and 2
grid on;
the first part works perfectly fine, but the 2nd part where i have to plot wrt time, is where i am having trouble? its supposed to be going from x-2 being stable , and 2 being unstable, the plot should depict that.
but I am unable to get that done properlythe script seems to be not behaving properly,as it is supposed to be ?
% Define the range for x
x = linspace(-5, 5, 10); % 10 points between -5 and 5
dx1 = 2*x.^2 – 8; % Compute the derivative (velocity field)
% Create a figure
figure;
hold on;
% Plot the real line (x-axis)
plot(x, zeros(size(x)), ‘k’); % The x-axis (real line)
% Plot the vector field as arrows
for i = 1:length(x)
if dx1(i) > 0
% Arrow pointing to the right (positive dx)
quiver(x(i), 0, 0.5, 0, ‘r’, ‘MaxHeadSize’, 0.5);
elseif dx1(i) < 0
% Arrow pointing to the left (negative dx)
quiver(x(i), 0, -0.5, 0, ‘b’, ‘MaxHeadSize’, 0.5);
end
end
% Plot equilibria points
plot([-2, 2], [0, 0], ‘bo’, ‘MarkerFaceColor’, ‘b’, ‘MarkerSize’, 8); % Equilibria at x = -2 and x = 2
% Labels and title
xlabel(‘x’);
ylabel(‘x-dot’);
title(‘1D Vector Field for x-dot = 2x^2 – 8’);
ylim([-1 1]); % Set y-axis limits to keep focus on x-axis
xlim([-4 4]); % Set x-axis limits
grid on; % Add grid
% Add text annotations for equilibrium points
text(-2, 0.2, ‘Stable Equilibrium: x = -2’, ‘HorizontalAlignment’, ‘center’);
text(2, 0.2, ‘Unstable Equilibrium: x = 2’, ‘HorizontalAlignment’, ‘center’);
hold off;
t = [-5 5];
ode_function = @(t,x) 2*x.^2-8;
x0 = [-2 2];
[time, x] = ode45(ode_function, t, x0);
figure;
plot(time, x, ‘LineWidth’,0.5);
title(‘Solution of dx/dt = 2x^2 – 8 vs Time’);
xlabel(‘Time t’);
ylabel(‘x(t)’);
xlim([-t(2) t(2)]); % Adjust x-axis limits for clarity
ylim([-3 3]); % Set y-axis limits to see behavior around -2 and 2
grid on;
the first part works perfectly fine, but the 2nd part where i have to plot wrt time, is where i am having trouble? its supposed to be going from x-2 being stable , and 2 being unstable, the plot should depict that.
but I am unable to get that done properly the script seems to be not behaving properly,as it is supposed to be ?
% Define the range for x
x = linspace(-5, 5, 10); % 10 points between -5 and 5
dx1 = 2*x.^2 – 8; % Compute the derivative (velocity field)
% Create a figure
figure;
hold on;
% Plot the real line (x-axis)
plot(x, zeros(size(x)), ‘k’); % The x-axis (real line)
% Plot the vector field as arrows
for i = 1:length(x)
if dx1(i) > 0
% Arrow pointing to the right (positive dx)
quiver(x(i), 0, 0.5, 0, ‘r’, ‘MaxHeadSize’, 0.5);
elseif dx1(i) < 0
% Arrow pointing to the left (negative dx)
quiver(x(i), 0, -0.5, 0, ‘b’, ‘MaxHeadSize’, 0.5);
end
end
% Plot equilibria points
plot([-2, 2], [0, 0], ‘bo’, ‘MarkerFaceColor’, ‘b’, ‘MarkerSize’, 8); % Equilibria at x = -2 and x = 2
% Labels and title
xlabel(‘x’);
ylabel(‘x-dot’);
title(‘1D Vector Field for x-dot = 2x^2 – 8’);
ylim([-1 1]); % Set y-axis limits to keep focus on x-axis
xlim([-4 4]); % Set x-axis limits
grid on; % Add grid
% Add text annotations for equilibrium points
text(-2, 0.2, ‘Stable Equilibrium: x = -2’, ‘HorizontalAlignment’, ‘center’);
text(2, 0.2, ‘Unstable Equilibrium: x = 2’, ‘HorizontalAlignment’, ‘center’);
hold off;
t = [-5 5];
ode_function = @(t,x) 2*x.^2-8;
x0 = [-2 2];
[time, x] = ode45(ode_function, t, x0);
figure;
plot(time, x, ‘LineWidth’,0.5);
title(‘Solution of dx/dt = 2x^2 – 8 vs Time’);
xlabel(‘Time t’);
ylabel(‘x(t)’);
xlim([-t(2) t(2)]); % Adjust x-axis limits for clarity
ylim([-3 3]); % Set y-axis limits to see behavior around -2 and 2
grid on;
the first part works perfectly fine, but the 2nd part where i have to plot wrt time, is where i am having trouble? its supposed to be going from x-2 being stable , and 2 being unstable, the plot should depict that.
but I am unable to get that done properly code generation, ode45, equation MATLAB Answers — New Questions
