in my homework I need to extract the server name from the url
at the same time, I need to take into account that there may not be a slash after the server name
I'm not allowed to use a loop
At the same time, I am once again trying to redo a remark from my teacher:
"Now substring can be done twice (if it goes into if). You need to make sure that only one substring is made for any variant of the function execution"
how can this be fixed? I've tried everything
public class Url {
public static String getServerName(String url) {
int index1 = url.indexOf("://") + 3;
String serverName = url.substring(index1);
int index2 = serverName.indexOf("/");
if (index2 >= 0) {
return url.substring(index1, index1 + index2);
}
return serverName;
}
public static void main(String[] args) {
String url = "https://SomeServerName";
System.out.println(getServerName(url));
}
}
edit: The ArrayList wasn't needed to reproduce the "error". Sorry for this delay, but know it should be much clearer.
Why is:
c2.number.equals(c3.number) = false
I really expected a true here. There must be something wrong with my equals method?
Why on earth do I need to write more text...
package com.example.mypackage;
import java.util.ArrayList;
import java.util.Scanner;
class Contact {
public String name;
public String number;
public Contact(String name, String number) {
this.name = name;
this.number = number;
}
public void print(){
System.out.println(name+number);
}
#Override
public boolean equals(Object obj) {
if (this == obj) {
return true;
} else if (obj == null) {
return false;
} else if (obj instanceof Contact) {
Contact contact = (Contact) obj;
if ((contact.name == this.name && contact.number == this.number)) {
return true;
}
}
return false;
}
}
public class Main {
private static Scanner scanner = new Scanner(System.in);
public static void main(String[] args) {
Contact c1 = new Contact("ben", "1");
c1.print();
Contact c2 = new Contact("ben", "1");
c2.print();
System.out.println("name : ");
String name=scanner.nextLine();
System.out.println("number");
String number=scanner.nextLine();
Contact c3=new Contact(name, number);
c3.print();
System.out.println("c1.equals(c2) = "+c1.equals(c2));
System.out.println("c3 instanceof Contact = "+(c3 instanceof Contact));
System.out.println("c2.name.equals(c3.name) = "+c2.name.equals(c3.name));
System.out.println("c2.number.equals(c3.number) = "+c2.number.equals(c3.number));
System.out.println("c2.number.equals(c3.number) = "+c3.equals(c2));
}
}
Output is:
ben1
ben1
name :
ben
number
1
ben1
c1.equals(c2) = true
c3 instanceof Contact = true
c2.name.equals(c3.name) = true
c2.number.equals(c3.number) = true
c2.number.equals(c3.number) = false
Process finished with exit code 0
Why is:
c2.number.equals(c3.number) = false
I really expected a true here. There must be something wrong with my equals method?
Why on earth do I need to write more text...
Why is:
c2.number.equals(c3.number) = false
I really expected a true here. There must be something wrong with my equals method?
Why on earth do I need to write more text...
Why is:
c2.number.equals(c3.number) = false
I really expected a true here. There must be something wrong with my equals method?
Why on earth do I need to write more text...
Ah finally I got it. The error is in the equals method.
I must use "equals()" instead of "==" there. For some reason this comparison does work with c1 and c2 but not with c3.
-1 is returned if it wasn't found in the list.
Did you forget to myList.add() it?
The only add I see is when you added c1.
You need to myList.add(c3) after you get the input, or it won't be in the list to find an index of.
I'm trying to evaluate user input, but the while statement seems to go into an infinite loop without asking for input.
import javax.swing.JOptionPane;
public class StringMethodsTwo {
public static void main(String[] args)
{
String sFullName = " ";
String prompt = "Please enter your full name:";
while(sFullName.startsWith(" "));
{
sFullName = getInput(prompt);
if(sFullName.length() < 2)
{
prompt = "Please enter your full name, \"<first> <middle> <last>\":";
}
}
}
public static String getInput(String prompt)
{
return JOptionPane.showInputDialog(null, prompt);
}
}
You are doing the loop
while(sFullName.startsWith(" ")); // while(true);
{
Delete the ";"
Lose the semicolon at the end of your "while" line.
I want to get the highest available string value in java how can i achieve this.
Example: hello jameswangfron
I want to get the highest string "jameswangfron"
String Text = request.getParameter("hello jameswangfron");
Please code example.
public class HelloWorld{
public static void main(String []args){
String text = "hello jameswangfron";
String[] textArray = text.split(" ");
String biggestString = "";
for(int i=0; i<textArray.length; i++){
if(i==0) {
textArray[i].length();
biggestString = textArray[i];
} else {
if(textArray[i].length()>textArray[i-1].length()){
biggestString = textArray[i];
}
}
}
System.out.println("Biggest String : "+biggestString);
}
}
And it shows the output as
Biggest String : jameswangfron
Maybe this will be easyer to understand
public class HelloWorld {
public static void main(String[] args) {
System.out.println(StringManipulator.getMaxLengthString("hello jameswangfron", " "));
}
}
class StringManipulator{
public static String getMaxLengthString(String data, String separator){
String[] stringArray = data.split(separator);
String toReturn = "";
int maxLengthSoFar = 0;
for (String string : stringArray) {
if(string.length()>maxLengthSoFar){
maxLengthSoFar = string.length();
toReturn = string;
}
}
return toReturn;
}
}
But there is a catch. If you pay attention to split method from class String, you will find out that the spliter is actually a regex. For your code, i see that you want to separate the words (which means blank space). if you want an entire text to search, you have to pass a regex.
Here's a tip. If you want your words to be separated by " ", ".", "," (you get the ideea) then you should replace the " " from getMaxLengthString method with the following
"[^a-zA-Z0-9]"
If you want digits to split up words, simply put
"[^a-zA-Z]"
This tells us that we use the separators as anything that is NOT a lower case letter or upper case letter. (the ^ character means you don't want the characters you listed in your brackets [])
Here is another way of doing this
"[^\\w]"
\w it actually means word characters. so if you negate this (with ^) you should be fine
I'm using the Stanford Named Entity Recognizer http://nlp.stanford.edu/software/CRF-NER.shtml and it's working fine. This is
List<List<CoreLabel>> out = classifier.classify(text);
for (List<CoreLabel> sentence : out) {
for (CoreLabel word : sentence) {
if (!StringUtils.equals(word.get(AnswerAnnotation.class), "O")) {
namedEntities.add(word.word().trim());
}
}
}
However the problem I'm finding is identifying names and surnames. If the recognizer encounters "Joe Smith", it is returning "Joe" and "Smith" separately. I'd really like it to return "Joe Smith" as one term.
Could this be achieved through the recognizer maybe through a configuration? I didn't find anything in the javadoc till now.
Thanks!
This is because your inner for loop is iterating over individual tokens (words) and adding them separately. You need to change things to add whole names at once.
One way is to replace the inner for loop with a regular for loop with a while loop inside it which takes adjacent non-O things of the same class and adds them as a single entity.*
Another way would be to use the CRFClassifier method call:
List<Triple<String,Integer,Integer>> classifyToCharacterOffsets(String sentences)
which will give you whole entities, which you can extract the String form of by using substring on the original input.
*The models that we distribute use a simple raw IO label scheme, where things are labeled PERSON or LOCATION, and the appropriate thing to do is simply to coalesce adjacent tokens with the same label. Many NER systems use more complex labels such as IOB labels, where codes like B-PERS indicates where a person entity starts. The CRFClassifier class and feature factories support such labels, but they're not used in the models we currently distribute (as of 2012).
The counterpart of the classifyToCharacterOffsets method is that (AFAIK) you can't access the label of the entities.
As proposed by Christopher, here is an example of a loop which assembles "adjacent non-O things". This example also counts the number of occurrences.
public HashMap<String, HashMap<String, Integer>> extractEntities(String text){
HashMap<String, HashMap<String, Integer>> entities =
new HashMap<String, HashMap<String, Integer>>();
for (List<CoreLabel> lcl : classifier.classify(text)) {
Iterator<CoreLabel> iterator = lcl.iterator();
if (!iterator.hasNext())
continue;
CoreLabel cl = iterator.next();
while (iterator.hasNext()) {
String answer =
cl.getString(CoreAnnotations.AnswerAnnotation.class);
if (answer.equals("O")) {
cl = iterator.next();
continue;
}
if (!entities.containsKey(answer))
entities.put(answer, new HashMap<String, Integer>());
String value = cl.getString(CoreAnnotations.ValueAnnotation.class);
while (iterator.hasNext()) {
cl = iterator.next();
if (answer.equals(
cl.getString(CoreAnnotations.AnswerAnnotation.class)))
value = value + " " +
cl.getString(CoreAnnotations.ValueAnnotation.class);
else {
if (!entities.get(answer).containsKey(value))
entities.get(answer).put(value, 0);
entities.get(answer).put(value,
entities.get(answer).get(value) + 1);
break;
}
}
if (!iterator.hasNext())
break;
}
}
return entities;
}
I had the same problem, so I looked it up, too. The method proposed by Christopher Manning is efficient, but the delicate point is to know how to decide which kind of separator is appropriate. One could say only a space should be allowed, e.g. "John Zorn" >> one entity. However, I may find the form "J.Zorn", so I should also allow certain punctuation marks. But what about "Jack, James and Joe" ? I might get 2 entities instead of 3 ("Jack James" and "Joe").
By digging a bit in the Stanford NER classes, I actually found a proper implementation of this idea. They use it to export entities under the form of single String objects. For instance, in the method PlainTextDocumentReaderAndWriter.printAnswersTokenizedInlineXML, we have:
private void printAnswersInlineXML(List<IN> doc, PrintWriter out) {
final String background = flags.backgroundSymbol;
String prevTag = background;
for (Iterator<IN> wordIter = doc.iterator(); wordIter.hasNext();) {
IN wi = wordIter.next();
String tag = StringUtils.getNotNullString(wi.get(AnswerAnnotation.class));
String before = StringUtils.getNotNullString(wi.get(BeforeAnnotation.class));
String current = StringUtils.getNotNullString(wi.get(CoreAnnotations.OriginalTextAnnotation.class));
if (!tag.equals(prevTag)) {
if (!prevTag.equals(background) && !tag.equals(background)) {
out.print("</");
out.print(prevTag);
out.print('>');
out.print(before);
out.print('<');
out.print(tag);
out.print('>');
} else if (!prevTag.equals(background)) {
out.print("</");
out.print(prevTag);
out.print('>');
out.print(before);
} else if (!tag.equals(background)) {
out.print(before);
out.print('<');
out.print(tag);
out.print('>');
}
} else {
out.print(before);
}
out.print(current);
String afterWS = StringUtils.getNotNullString(wi.get(AfterAnnotation.class));
if (!tag.equals(background) && !wordIter.hasNext()) {
out.print("</");
out.print(tag);
out.print('>');
prevTag = background;
} else {
prevTag = tag;
}
out.print(afterWS);
}
}
They iterate over each word, checking if it has the same class (answer) than the previous, as explained before. For this, they take advantage of the fact expressions considered as not being entities are flagged using the so-called backgroundSymbol (class "O"). They also use the property BeforeAnnotation, which represents the string separating the current word from the previous one. This last point allows solving the problem I initially raised, regarding the choice of an appropriate separator.
Code for the above:
<List> result = classifier.classifyToCharacterOffsets(text);
for (Triple<String, Integer, Integer> triple : result)
{
System.out.println(triple.first + " : " + text.substring(triple.second, triple.third));
}
List<List<CoreLabel>> out = classifier.classify(text);
for (List<CoreLabel> sentence : out) {
String s = "";
String prevLabel = null;
for (CoreLabel word : sentence) {
if(prevLabel == null || prevLabel.equals(word.get(CoreAnnotations.AnswerAnnotation.class)) ) {
s = s + " " + word;
prevLabel = word.get(CoreAnnotations.AnswerAnnotation.class);
}
else {
if(!prevLabel.equals("O"))
System.out.println(s.trim() + '/' + prevLabel + ' ');
s = " " + word;
prevLabel = word.get(CoreAnnotations.AnswerAnnotation.class);
}
}
if(!prevLabel.equals("O"))
System.out.println(s + '/' + prevLabel + ' ');
}
I just wrote a small logic and it's working fine. what I did is group words with same label if they are adjacent.
Make use of the classifiers already provided to you. I believe this is what you are looking for:
private static String combineNERSequence(String text) {
String serializedClassifier = "edu/stanford/nlp/models/ner/english.all.3class.distsim.crf.ser.gz";
AbstractSequenceClassifier<CoreLabel> classifier = null;
try {
classifier = CRFClassifier
.getClassifier(serializedClassifier);
} catch (ClassCastException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (ClassNotFoundException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
System.out.println(classifier.classifyWithInlineXML(text));
// FOR TSV FORMAT //
//System.out.print(classifier.classifyToString(text, "tsv", false));
return classifier.classifyWithInlineXML(text);
}
Here is my full code, I use Stanford core NLP and write algorithm to concatenate Multi Term names.
import edu.stanford.nlp.ling.CoreAnnotations;
import edu.stanford.nlp.ling.CoreLabel;
import edu.stanford.nlp.pipeline.Annotation;
import edu.stanford.nlp.pipeline.StanfordCoreNLP;
import edu.stanford.nlp.util.CoreMap;
import org.apache.log4j.Logger;
import java.util.ArrayList;
import java.util.List;
import java.util.Properties;
/**
* Created by Chanuka on 8/28/14 AD.
*/
public class FindNameEntityTypeExecutor {
private static Logger logger = Logger.getLogger(FindNameEntityTypeExecutor.class);
private StanfordCoreNLP pipeline;
public FindNameEntityTypeExecutor() {
logger.info("Initializing Annotator pipeline ...");
Properties props = new Properties();
props.setProperty("annotators", "tokenize, ssplit, pos, lemma, ner");
pipeline = new StanfordCoreNLP(props);
logger.info("Annotator pipeline initialized");
}
List<String> findNameEntityType(String text, String entity) {
logger.info("Finding entity type matches in the " + text + " for entity type, " + entity);
// create an empty Annotation just with the given text
Annotation document = new Annotation(text);
// run all Annotators on this text
pipeline.annotate(document);
List<CoreMap> sentences = document.get(CoreAnnotations.SentencesAnnotation.class);
List<String> matches = new ArrayList<String>();
for (CoreMap sentence : sentences) {
int previousCount = 0;
int count = 0;
// traversing the words in the current sentence
// a CoreLabel is a CoreMap with additional token-specific methods
for (CoreLabel token : sentence.get(CoreAnnotations.TokensAnnotation.class)) {
String word = token.get(CoreAnnotations.TextAnnotation.class);
int previousWordIndex;
if (entity.equals(token.get(CoreAnnotations.NamedEntityTagAnnotation.class))) {
count++;
if (previousCount != 0 && (previousCount + 1) == count) {
previousWordIndex = matches.size() - 1;
String previousWord = matches.get(previousWordIndex);
matches.remove(previousWordIndex);
previousWord = previousWord.concat(" " + word);
matches.add(previousWordIndex, previousWord);
} else {
matches.add(word);
}
previousCount = count;
}
else
{
count=0;
previousCount=0;
}
}
}
return matches;
}
}
Another approach to deal with multi words entities.
This code combines multiple tokens together if they have the same annotation and go in a row.
Restriction:
If the same token has two different annotations, the last one will be saved.
private Document getEntities(String fullText) {
Document entitiesList = new Document();
NERClassifierCombiner nerCombClassifier = loadNERClassifiers();
if (nerCombClassifier != null) {
List<List<CoreLabel>> results = nerCombClassifier.classify(fullText);
for (List<CoreLabel> coreLabels : results) {
String prevLabel = null;
String prevToken = null;
for (CoreLabel coreLabel : coreLabels) {
String word = coreLabel.word();
String annotation = coreLabel.get(CoreAnnotations.AnswerAnnotation.class);
if (!"O".equals(annotation)) {
if (prevLabel == null) {
prevLabel = annotation;
prevToken = word;
} else {
if (prevLabel.equals(annotation)) {
prevToken += " " + word;
} else {
prevLabel = annotation;
prevToken = word;
}
}
} else {
if (prevLabel != null) {
entitiesList.put(prevToken, prevLabel);
prevLabel = null;
}
}
}
}
}
return entitiesList;
}
Imports:
Document: org.bson.Document;
NERClassifierCombiner: edu.stanford.nlp.ie.NERClassifierCombiner;