Word association counting

Question

I am new to java. I need to count word associations with each other in a sentence. For example, for the sentence, "Dog is a Dog and Cat is a Cat", the final association count will be- The first row: Dog-Dog(0), Dog-is(2), Dog-a(2) Dog-and(1), Dog-Cat(2)

and so on.

It is kind of developing an association matrix. Any suggestion on how that can be developed?

Interesting! Can you elaborate what is the use of this and also why is the count 2 for "dog-is". See if this process helps: http://it.toolbox.com/blogs/enterprise-solutions/building-an-association-matrix-15499 — Aravind Yarram, Dec 19 '10 at 00:39
@Pangea: Well, in the sentence "Dog" is present with 2 "is", that is why Dog-is pair will get the value 2. Making the matrix with a table is easy, but during implementation, I am lost. — Rushdi Shams, Dec 19 '10 at 06:50
I am sorry but I see the "dog is" occurring only once right. "Dog is a Dog and Cat is a Cat" — Aravind Yarram, Dec 19 '10 at 11:59

score 3 · Answer 1 · answered Dec 19 '10 at 00:43

Thanks Roman. I can split the words from the sentences-

String sentence=null;
    String target="Dog is a Dog and Cat is a Cat";
    int index = 0;
    Locale currentLocale = new Locale ("en","US");
    BreakIterator wordIterator = BreakIterator.getWordInstance(currentLocale);
    //Creating the sentence iterator
    BreakIterator bi = BreakIterator.getSentenceInstance();
    bi.setText(target);

    while (bi.next() != BreakIterator.DONE) {

        sentence = target.substring(index, bi.current());
        System.out.println(sentence);
        wordIterator.setText(sentence);
        int start = wordIterator.first();
        int end = wordIterator.next();

        while (end!=BreakIterator.DONE){

            String word = sentence.substring(start,end);
             if (Character.isLetterOrDigit(word.charAt(0))) {

                System.out.println(word);

             }//if (Character.isLetterOrDigit(word.charAt(0)))

             start = end;
             end = wordIterator.next();
        }//while (end!=BreakIterator.DONE)
        index = bi.current();
    }  //  while (bi.next() != BreakIterator.DONE)

But did not get your other two points. Thanks.

it's overkill, IMHO. `target.split("\\s")` should be enough and it can replace all this overcomplicated code. — Roman, Dec 19 '10 at 01:01

Roman · Answer 2 · 2010-12-19T00:59:09.450

Split the sentence into separate words.
Generate pairs.
Merge the same pairs.

It's as simple as:

String[] words = sentence.split("\\s"); //first step
List<List<String>> pairs = 
    new ArrayList<List<String>>((int)(((words.length) / 2.0) * (words.length - 1)));
for (int i = 0; i < words.length - 1; i++) {
    for (int j = i + 1; j < words.length; j++) {
         List<String> pair = Arrays.asList(words[i], words[j]);
         Collections.sort(pair);
         pairs.add(pair);
    }
} //second step
Map<List<String>, Integer> pair2count = new LinkedHashMap<List<String>, Integer>();
for (List<String> pair : pairs) {
    if (pair2count.containsKey(pair)) {
        pair2count.put(pair, pair2count.get(pair) + 1);
    } else {
        pair2count.put(pair, 1);
    }
} //third step

//output
System.out.println(pair2count);

Word association counting

2 Answers2

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