Add context value into web.xml

K

KineticUncertainty

Guest
KineticUncertainty Asks: Add context value into web.xml
I have an application where i defined the location to the external log4j2.properties. I have defined the Log4jServletContextListener in the web.xml and if i provide a path to the file it works. I am trying to figure out how to use a place holder and get the value of the context into it. Can any one help?

Here is my tomcat context entry

Code:
<?xml version="1.0" encoding="UTF-8"?>
<Context>
<Environment name="demo/log4j2.location" type="java.lang.String" value="C:/conf/log4j2.xml"/>
</Context>

I have a bean setup that's grabbing the context entry. This is working fine.

Code:
@Bean("Logging")
    public String getConfig(
            @Value("${demo/log4j2.location}") String log4jConfigLocation
    ){
        validate(log4jConfigLocation, not(equalTo("${demo/log4j2.location}")));
        return log4jConfigLocation;
    }

I can also get the context value this way if its easier to use.

Code:
@Component
public class Log4jLoader {
  
    String location;
  
    public  Log4jLoader(@Value("${construct-demo/log4j2.location}") String location){
        this.location = location;
        loadLog4jConfig();
    }

    private void loadLog4jConfig() {

in my web.xml

Code:
<context-param>
    <param-name>isLog4jAutoInitializationDisabled</param-name>
    <param-value>true</param-value>
</context-param>
<listener>
    <listener-class>org.apache.logging.log4j.web.Log4jServletContextListener</listener-class>
</listener>

<context-param>
    <param-name>log4jContextName</param-name>
    <param-value>demo</param-value>
</context-param>

<context-param>
    <param-name>log4jConfiguration</param-name>
    <param-value>file:$(location)</param-value>
</context-param>

THIS WORKS
<!--<context-param>-->
<!--<param-name>log4jConfiguration</param-name>-->
<!--<param-value>file:///C:/conf/log4j2.xml</param-value>-->
<!--</context-param>-->

is there a way to get that value into the file:$(location) ?

I have been stuck trying to figure this out for days and exhausting my googling

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[Solved] Extract Data from Corrupted TerraSync file, Pathfinder can't correct & won't open

  • Kevin Nolan
  • Geography
  • Replies: 0
Kevin Nolan Asks: Extract Data from Corrupted TerraSync file, Pathfinder can't correct & won't open
I had an employee recording points for a project with a Trimble Geo XT 7 series receiver attached to a Zephyr 2 external antenna. The receiver went to sleep between collecting episodes. Upon attempting to re-open, the file came back as corrupted and TerraSync was unable to fix and open the file. I have never had an unrecoverable corruption and didn't expect this to be a permanent problem. We stared a new file and finished collecting data.

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[Solved] How to snap lines from the same layer together?

  • Geowhat
  • Geography
  • Replies: 0
Geowhat Asks: How to snap lines from the same layer together?
I did a Network Analysis and this is the outcome. How can I snap lines from the same layer together?

enter image description here

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[Solved] Publish a Query Layer as a Feature Access Service - ArcGIS Server

a1234 Asks: Publish a Query Layer as a Feature Access Service - ArcGIS Server
I am connecting to a SQL Server 2016 database, and adding data into ArcMap 10.3.1 as a Query Layer.

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enter image description here

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Doubt about Electromagnetic Attraction b/w Electromagnet-Iron vs Electromagnet-Electromagnet

  • J.Doe
  • Physics
  • Replies: 0
J.Doe Asks: Doubt about Electromagnetic Attraction b/w Electromagnet-Iron vs Electromagnet-Electromagnet
https://www.supermagnete.de/eng/faq...the-combination-magnet-magnet-and-magnet-iron

At full contact, the attractive force between a raw magnet and an iron plate is the same as the attractive force between two raw magnets. However, with increasing distance, the attraction diminishes faster than the attraction between two raw magnets (see graph)

I am confused by this:


  1. Consider two identical Electromagnets touching each other in one case and a single electromagnet touching an iron plate in second case. Won't the attraction be more in the first case as both magnets are exerting a pull on each other where as in second case the magnet is pulling on a passive iron plate (which has temporarily induced magnetism)?


  2. For two cylindrical electromagnets with poles at distance $x$ repelling each other (they are relatively close by i.e. $10*R > x$ where $R$=radius of pole) how does the force exerted on each electromagnet change with increase in current ($I$)? Can someone help with this formula?

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Closing a Graph

sam wolfe Asks: Closing a Graph
Consider the following graph g

enter image description here

I want to "close" it by adding the following edges

enter image description here

Alternatively, I would also be happy with something like

enter image description here

where I've added four "corner" vertices.

Now, this question is, in a way, a follow up to this answer, where I initially have a picture on which I draw a graph, as follows

enter image description here

Is it possible to turn the picture into a rectangle-shaped graph and then join them in somehow? This would lead to something like the second example.

Alternatively, I also thought about simply uniting 1-degree vertices that are close to each other, creating the first example graph.

In the end, I want a graph that defines a mesh, thus the need of "closing" the graph in this fashion. Ideally, I want KVertexConnectedGraphQ[g] = True for the closed graph.

Any ideas?

Edit 1: Note that my goal is to be able to do this for general graphs. For example, considering the graph

enter image description here

I want to get something like

enter image description here

I guess using something like ConvexHull (which doesn't correspond to what is drawn) in some matter could help my goal, but at this point I'm entirely sure how.

Edit 2: In order to be more practical consider the graph given by

Code:
g = Graph[{1 \[UndirectedEdge] 10, 2 \[UndirectedEdge] 9, 
   3 \[UndirectedEdge] 9, 4 \[UndirectedEdge] 12, 
   5 \[UndirectedEdge] 8, 6 \[UndirectedEdge] 13, 
   7 \[UndirectedEdge] 14, 8 \[UndirectedEdge] 11, 
   8 \[UndirectedEdge] 17, 9 \[UndirectedEdge] 20, 
   10 \[UndirectedEdge] 11, 10 \[UndirectedEdge] 21, 
   11 \[UndirectedEdge] 25, 12 \[UndirectedEdge] 18, 
   12 \[UndirectedEdge] 19, 13 \[UndirectedEdge] 18, 
   13 \[UndirectedEdge] 27, 14 \[UndirectedEdge] 15, 
   14 \[UndirectedEdge] 19, 16 \[UndirectedEdge] 17, 
   17 \[UndirectedEdge] 23, 18 \[UndirectedEdge] 24, 
   19 \[UndirectedEdge] 22, 20 \[UndirectedEdge] 21, 
   20 \[UndirectedEdge] 27, 21 \[UndirectedEdge] 34, 
   22 \[UndirectedEdge] 26, 22 \[UndirectedEdge] 29, 
   23 \[UndirectedEdge] 31, 23 \[UndirectedEdge] 33, 
   24 \[UndirectedEdge] 29, 24 \[UndirectedEdge] 30, 
   25 \[UndirectedEdge] 31, 25 \[UndirectedEdge] 34, 
   27 \[UndirectedEdge] 28, 28 \[UndirectedEdge] 32, 
   28 \[UndirectedEdge] 38, 29 \[UndirectedEdge] 37, 
   30 \[UndirectedEdge] 32, 30 \[UndirectedEdge] 35, 
   31 \[UndirectedEdge] 39, 32 \[UndirectedEdge] 41, 
   34 \[UndirectedEdge] 36, 35 \[UndirectedEdge] 42, 
   35 \[UndirectedEdge] 44, 36 \[UndirectedEdge] 45, 
   36 \[UndirectedEdge] 54, 37 \[UndirectedEdge] 40, 
   37 \[UndirectedEdge] 44, 38 \[UndirectedEdge] 45, 
   38 \[UndirectedEdge] 48, 39 \[UndirectedEdge] 47, 
   39 \[UndirectedEdge] 53, 41 \[UndirectedEdge] 42, 
   41 \[UndirectedEdge] 48, 42 \[UndirectedEdge] 56, 
   43 \[UndirectedEdge] 46, 44 \[UndirectedEdge] 51, 
   45 \[UndirectedEdge] 55, 46 \[UndirectedEdge] 47, 
   46 \[UndirectedEdge] 52, 47 \[UndirectedEdge] 49, 
   48 \[UndirectedEdge] 50},
  VertexCoordinates -> {{102.5`, 175.5`}, {84.5`, 152.5`}, {108.5`, 
     175.5`}, {133.5`, 153.5`}, {152.5`, 175.5`}, {244.5`, 
     175.5`}, {254.5`, 148.5`}, {43.5`, 174.5`}, {43.5`, 
     170.5`}, {196.5`, 174.5`}, {202.5`, 147.5`}, {297.5`, 
     174.5`}, {309.5`, 147.5`}, {63.5`, 148.5`}, {10.5`, 
     141.5`}, {143.5`, 117.5`}, {119.5`, 109.5`}, {67.5`, 
     94.5`}, {236.5`, 131.5`}, {293.5`, 127.5`}, {180.5`, 
     89.5`}, {312.5`, 146.5`}, {4.5`, 143.5`}, {18.5`, 
     97.5`}, {253.5`, 95.5`}, {301.5`, 98.5`}, {110.5`, 
     75.5`}, {313.5`, 93.5`}, {286.5`, 83.5`}, {52.5`, 80.5`}, {4.5`, 
     76.5`}, {236.5`, 82.5`}, {181.5`, 86.5`}, {187.5`, 
     80.5`}, {168.5`, 31.5`}, {297.5`, 37.5`}, {244.5`, 
     49.5`}, {59.5`, 29.5`}, {216.5`, 27.5`}, {125.5`, 
     38.5`}, {225.5`, 26.5`}, {280.5`, 23.5`}, {152.5`, 
     20.5`}, {110.5`, 3.5`}, {313.5`, 29.5`}, {199.5`, 7.5`}, {32.5`, 
     7.5`}, {85.5`, 3.5`}, {236.5`, 3.5`}, {4.5`, 25.5`}, {10.5`, 
     16.5`}, {281.5`, 4.5`}, {155.5`, 3.5`}, {4.5`, 3.5`}, {34.5`, 
     4.5`}, {199.5`, 4.5`}},
  VertexSize -> 3 {1, 1}, VertexStyle -> Red, 
  EdgeStyle -> Directive[Black]]

which yields the first graph g. Then, the code

Code:
hm = ConvexHullMesh[
      Transpose[
Select[{GraphEmbedding[g], VertexDegree[g]} // 
      Transpose, #[[2]] == 1 &]][[1]]]
gb = Graph[hm["Edges"], VertexCoordinates -> MeshCoordinates[hm], 
  VertexSize -> 3 {1, 1}, VertexStyle -> Red, 
  EdgeStyle -> Directive[Black]]

yields

enter image description here

Now, how do I merge both graphs? I tried GraphUnion, but I would need the correct VertexCoordinates. Could it be simply an ordering problem? Any suggestion?

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Showing an inequality in vector space.

Hamdiken Asks: Showing an inequality in vector space.
Let $E$ be a normed vector space. We define a linear form $T:E\rightarrow \mathbb R$. Denote $H=Ker(T)$.

  • Show that if $T$ is continuous, then $H$ is a closed set.

$H$ is the continuous inverse of a closed set ($\{0\}$), hence closed.

  • Suppose that $H$ is closed. Let $a\in E$ such that $T(a)=1$. Show that the set $a+H$ is closed and does not contain $0$.

My guess is that since the set $\{a\}$ is bounded, and since $H$ is closed in a vector space, which means it's also bounded, gives the result that $a+H$ is closed. As for the $0$, we have for every $y\in a+H$, $T(y)=T(a+x)=T(a)+T(x)=1$. However $T(0)=0\neq1$, so $0\notin a+H$.

  • Deduce the existence of $r>0$ such that $B(0,r)\cap(a+H)=\emptyset$.

Since $0\notin a+H$, there exists a neighbourhood $V_0$ of $0$ such that $V_0\cap(a+H)=\emptyset$, hence the existence of an open ball $B(0,r)$ such that $B(0,r)\cap(a+H)=\emptyset$.

  • Show that for every $x\in B(0,r)$, we have $$|Tx|\leq 1$$

This is where I stopped. I thought of the orthogonal projection but didn't understand how to proceed.

Any help and correction of the answers above will be highly appreciated.

SolveForum.com may not be responsible for the answers or solutions given to any question asked by the users. All Answers or responses are user generated answers and we do not have proof of its validity or correctness. Please vote for the answer that helped you in order to help others find out which is the most helpful answer. Questions labeled as solved may be solved or may not be solved depending on the type of question and the date posted for some posts may be scheduled to be deleted periodically. Do not hesitate to share your thoughts here to help others.

Monotone likelihood ratio for logistic distribution $f(x;\theta) = e^{-x - \theta} (1 + e^{-x - \theta})^{-2}$

Balkys Asks: Monotone likelihood ratio for logistic distribution $f(x;\theta) = e^{-x - \theta} (1 + e^{-x - \theta})^{-2}$
This is a question from Problems 9.4 of An Introduction to Probability and Statistics by Rohatgi.

Let $X$ have logistic distribution with the PDF $$f(x;\theta) = e^{-x - \theta} (1 + e^{-x - \theta})^{-2}, x \in \mathbb{R}.$$ Does ${f}$ have a monotone likelihood ratio?

I have tried to look at this related question, but the theory was beyond what I have learned.

My attempt: Let $\theta_1, \theta_0 \in \mathbb{R}$. Let $\theta_1 > \theta_0.$

Consider the likelihood ratio, $T_{NP}(\mathbf{x}) = \dfrac{f_n(\mathbf{x};\theta_1)}{f_n(\mathbf{x};\theta_0)} = \dfrac{ e^{-n \bar{x} - n \theta_1} \prod (1+e^{-x_i - \theta_1})^{-2} }{ e^{ -n \bar{x} - n \theta_0} \prod (1+e^{-x_i - \theta_0})^{-2} }.$

After simplifying more, I get $T_{NP}(\mathbf{x}) = e^{n(\theta_0 - \theta_1)} \prod \left(\dfrac{1+e^{-x_i - \theta_0}}{1+e^{-x_i - \theta_1}}\right)^2.$

I found that $\dfrac{d}{dx} \dfrac{1+e^{-x - \theta_0}}{1+e^{-x - \theta_1}}$ was negative for all $x$.

I tried to relate this to what I had done so far. I used that $\log$ is monotone increasing. Taking the log of the likelihood ratio, I get that the log-likelihood ratio is decreasing in $\sum \log \left(\dfrac{1+e^{-x_i - \theta_0}}{1+e^{-x_i - \theta_1}}\right).$ However, I am not sure how to get the required test statistic that is independent of the parameter $\theta$.

I think that a relevant result is that for a distribution from the exponential family, $f(x;\theta) = c(\theta)h(x)\exp(\pi(\theta)T(x))$, the likelihood ratio is monotone in $T(x)$.

I tried to use this by writing $f(x;\theta) = \exp{(-x-\theta)} \exp{(-2\log(1+e^{-x-\theta}))}.$ But I cannot identify $T(x)$ from this form.

Could someone please help me? Thank you very much.

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