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Jul 31st, 2008

occurence of a word in a text.

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i was trying to write a code for counting the number of ocurence of a particular word from a sentence.it gives me a wrong output it starts checking for the same alphabets in the sentence rather than the word.

Java Syntax (Toggle Plain Text)
import java.lang.String;
 
   public class StringCount2
 {
   public static void main(String args[])
   {
    String searchFor = "is";
    String base = "This is object oriented programming language  ";
    int len = searchFor.length();
    //System.out.println( "length of object =" +len);
    int result = 0;
    if (len > 0) 
    { 
     int start = base.indexOf(searchFor); 
    // System.out.println("index of object =" + start);
     while (start != -1) 
    {
     result++;
     start = base.indexOf(searchFor, start+len);
 
    }
    }
     System.out.println(result);
    }
 }
import java.lang.String;

   public class StringCount2
 {
   public static void main(String args[])
   {
    String searchFor = "is";
    String base = "This is object oriented programming language  ";
    int len = searchFor.length();
    //System.out.println( "length of object =" +len);
    int result = 0;
    if (len > 0) 
    { 
     int start = base.indexOf(searchFor); 
    // System.out.println("index of object =" + start);
     while (start != -1) 
    {
     result++;
     start = base.indexOf(searchFor, start+len);
     
    }
    }
     System.out.println(result);
    }
 }

can any one help me with this.

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newtechie

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Jul 31st, 2008

Re: occurence of a word in a text.

If you need to search for a particular word, use the Regex API --

http://java.sun.com/javase/6/docs/ap...x/Pattern.html

-- though, be warned. It is not fond of the light-hearted occurrence-searcher.

Alex Edwards

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Posting Shark

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Jul 31st, 2008

Re: occurence of a word in a text.

Condition on incrementing the result must do the trick.
Count should increment only if the word you are searching for in the String(Sentence) does not have any letters of Alphabet before or after it.

Try out this..

Java Syntax (Toggle Plain Text)
 public class StringCount2
 {
   public static void main(String args[])
   {
        String searchFor = "is";
        String base = "This is object oriented programming language.is is my favourite.";
        int len = searchFor.length();
        int result = 0;
        if (len > 0) 
        { 
             int start = base.indexOf(searchFor);
             while (start != -1)
             {
                int a=(int)(base.charAt(start-1));//Gives the ASCII equivalent of the character before the search String.
                int b=(int)(base.charAt(start+searchFor.length()));//Gives the ASCII equivalent of the character after the search String.
                //Check if the Character before or after (or both) the search String(*searchFor) is an Alphabet or not.
                if(((a>=65 && a<=90) || (a>=97 && a<=122)) || ((b>=65 && b<=90) || (b>=97 && b>=122)))
                {
                    //Not to be incremented as string searchFor is a part of a base.
                }
                else
                    result++;//Update the result.
                start = base.indexOf(searchFor, (start+len));
             }
         }
        System.out.println(result);
    }
 }
 public class StringCount2
 {
   public static void main(String args[])
   {
        String searchFor = "is";
        String base = "This is object oriented programming language.is is my favourite.";
        int len = searchFor.length();
        int result = 0;
        if (len > 0) 
        { 
             int start = base.indexOf(searchFor);
             while (start != -1)
             {
                int a=(int)(base.charAt(start-1));//Gives the ASCII equivalent of the character before the search String.
                int b=(int)(base.charAt(start+searchFor.length()));//Gives the ASCII equivalent of the character after the search String.
                //Check if the Character before or after (or both) the search String(*searchFor) is an Alphabet or not.
                if(((a>=65 && a<=90) || (a>=97 && a<=122)) || ((b>=65 && b<=90) || (b>=97 && b>=122)))
                {
                    //Not to be incremented as string searchFor is a part of a base.
                }
                else
                    result++;//Update the result.
                start = base.indexOf(searchFor, (start+len));
             }
         }
        System.out.println(result);
    }
 }

tyagi

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Jul 31st, 2008

Re: occurence of a word in a text.

thanks tyagi it works.

newtechie

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Jul 31st, 2008

Re: occurence of a word in a text.

hi thyagi, i could not understand the if part

Quote ...

if(((a>=65 && a<=90) || (a>=97 && a<=122)) || ((b>=65 && b<=90) || (b>=97 && b>=122)))

how you got those values for and b
pl. clarify

newtechie

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Aug 1st, 2008

Re: occurence of a word in a text.

Look at an ASCII value table. You will get your answer.

tuse

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Junior Poster

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Aug 1st, 2008

Re: occurence of a word in a text.

Yes.The if part is to check whether the ASCII equivalent of the character adjacent the required word in the string on either sides is a letter of the Alphabet or not. Note that it is 'OR' so as to account for non increment in cases like 'this' in the example you have mentioned, which implies that the ASCII equivalent of the character adjacent the required word in the string falls in the range of that of [a-z] or [A-Z] and hence the variable 'result' is not to be incremented.

tyagi

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No one has posted to this discussion for at least three months. Please let old threads die and do not reply to them unless you feel you have something new and valuable to contribute that absolutely must be added to make the discussion complete. Otherwise, please start a new thread in this forum instead.

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