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Count how many letters in the input file?

Posted on 2003-11-23
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Last Modified: 2007-12-19
i tried to make a program to count lettres in  an input file and i need result like this(for example it depends on your file and the output must write the only letter in the file )
Letter  Frequency
A           3
V           4
M           2

but my reslut is
Letter  Frequency
A           2
B           2
C           2
........ els

it gives me the number of the first letter in my input file correctly and then write all the ltters whith the same number of the first letter. as i mention above i dont need all the letter i need just what is written in the file with correct frequency.
this is my code

import java.io.FileReader;
import java.io.IOException;

public class Stats2
{

      public static void main(String[] args )
      {
            if (args.length < 0)
            System.out.println("No filename specified");
            else
            try
            {

                  FileReader reader = new FileReader("text.txt");
          }
          catch (IOException ioe)
          {
                  System.err.println(" Error on  opening file text.data\n" + ioe.toString() );
                  System.exit(1);
          }
          try
          {
                  char[] capital = { 'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J','K', 'L', 'M', 'N',
                                    'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z'};

            char[] small =   { 'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n',
                                  'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z' };
                  System.out.println("Leter     Frequency");
                  FileReader reader = new FileReader("text.txt");
                  int nextChar;
                  char ch;
                  int count=0;
            for (int i = 0; i < 26; i++)
            {


      while ( (nextChar = reader.read() ) != -1 )
      {

      ch = (char) nextChar;
      if( ch== capital[i] || ch == small[i] )
      {
       count++;
                }

      }

              System.out.print("  " + capital[i]);
              System.out.println("          " + count);

             }

                  reader.close();

                  System.exit(0);
            }
            catch( IOException ioe)
            {
                  System.err.println("Input/output error");
                  System.exit(2);
            }
      }
}
0
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Question by:alqaoud
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2 Comments
 
LVL 86

Accepted Solution

by:
CEHJ earned 150 total points
ID: 9806220
You need to maintain a Map where the key is the letter and the value is the number of occurrences. Here is a version for words. Use the same principle for letters. Here you would need to create Character objects as keys




import java.util.*;
import java.io.*;

public class WordCount {
  private StreamTokenizer streamTokenizer;
  private TreeMap treeMap;
  private Reader reader;
  private Writer writer;

 class Counter {
   private int i;

   public Counter(int i) {
     this.i = i;
   }
   public void increaseCount() {
     i++;
   }
   public String toString() {
     return Integer.toString(i);
   }
 }

 public WordCount(String fileName) {
   try {
     reader = new BufferedReader(new FileReader(fileName));
   }
   catch (FileNotFoundException fnfe) {
     System.out.println(fnfe.toString());
   }
   streamTokenizer = new StreamTokenizer(reader);

    // you can use this method to customize your needs
   streamTokenizer.ordinaryChar('.');

   treeMap = new TreeMap();
   try {
     while (streamTokenizer.nextToken() != StreamTokenizer.TT_EOF) {
       String s;
       switch (streamTokenizer.ttype) {
            case StreamTokenizer.TT_EOL :
              s = new String("EOL");
              break;
            case StreamTokenizer.TT_WORD :
               s = streamTokenizer.sval;
              break;
            case StreamTokenizer.TT_NUMBER :
              s = Double.toString(streamTokenizer.nval);
              break;
            default :
              s = String.valueOf((char)streamTokenizer.ttype);
       }
       if (treeMap.containsKey(s))
         ((Counter)treeMap.get(s)).increaseCount();
       else
         treeMap.put(s, new Counter(1));
     }
    }
    catch(IOException ioe) {
      System.out.println(ioe.toString());
    }
 }
 public Set entrySet() {
   return treeMap.entrySet();
 }

 public Iterator iterator() {
   return entrySet().iterator();
 }

 public void printResult() {
   Iterator iterator = iterator();
   while (iterator.hasNext()){
     System.out.println(iterator.next().toString());
   }
 }

 public static void main(String[] args) {
   if(args.length < 1){
     System.out.println("Usage: java WordCount <file>");
     System.exit(-1);
   }
   new WordCount(args[0]).printResult();
 }
}
0
 
LVL 86

Expert Comment

by:CEHJ
ID: 9809959
8-)
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