Hashmap implementation to count the occurrences of each character

Question

The below code is to count the occurence of each character and it should print the count. But with the code I have tried I get only a 1 I don't know the changes I should make.

import java.io.BufferedReader;
import java.io.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.HashMap;
import java.util.Map;

class Count_CharMap {
public static void main(String[] args) {
    try
    {
        FileInputStream file = new FileInputStream("D:\\trial.txt");
        DataInputStream dis = new DataInputStream(file);
        BufferedReader br = new BufferedReader(new InputStreamReader(dis));
        String Contents="";
        String str="";

        while ((Contents = br.readLine()) != null) {
            str+=Contents;
        }

        char[]char_array =str.toCharArray();
        int count = 0;
        char ch = char_array[count];
        Map<Character,Integer> charCounter=new HashMap<Character,Integer>();
        for(int i=0;i<str.length();i++)
        {
            if(charCounter.containsKey(char_array[i]))
            {
                charCounter.put(ch, charCounter.get(ch)+1);
            } 
            else
            {
                charCounter.put(ch, 1);
            }
       }

       for(Character key:charCounter.keySet())
       {
           System.out.println(key+""+charCounter.get(key));
       }
    } 
    catch(IOException e1){
        System.out.println(e1);
    }
    }
}

Actual output should be like If i have abcdabc in my trial.txt it should print a 2 b 2c 2 d 1.

Indent your code properly, it'll help you get help. – Andreas Wong Dec 06 '10 at 05:58 — Andreas Wong, Dec 06 '10 at 05:58

score 14 · Accepted Answer · answered Dec 06 '10 at 06:01

14

You're leaving char ch set as the same character through each execution of the loop.

It should be:

ch = char_array[i]; 
if(charCounter.containsKey(ch)){
     charCounter.put(ch, charCounter.get(ch)+1);
}
else
{
    charCounter.put(ch, 1);
}

Inside the for loop.

answered Dec 06 '10 at 06:01

Paul

2,729
20
26

hay thanks a lot. Silly i m :-) The code perfectly prints all the characters and occurences. I want to print them in alphabetical order. How to do that? – Sumithra Dec 06 '10 at 06:09
2

Use a treemap instead of a hashmap and it should do it. If not you'll have to sort it using a comparator, google it. http://download.oracle.com/javase/1.4.2/docs/api/java/util/TreeMap.html – Paul Dec 06 '10 at 06:10

Vitalii Fedorenko · Answer 2 · 2014-11-29T19:11:32.047

8

Java 8 streams:

Map<String, Long> map = 
    Arrays.stream(string.split("")).
    collect(Collectors.groupingBy(c -> c, Collectors.counting()));

Guava HashMultiset:

Multiset<Character> set = HashMultiset.create(Chars.asList("bbc".toCharArray()));
assertEquals(2, set.count('b'));

edited Nov 29 '14 at 19:11

answered Jul 28 '13 at 02:01

Vitalii Fedorenko

110,878
29
149
111

user1581734 · Answer 3 · 2013-05-21T04:54:23.140

Hai All The below code is to count the occurrence of each character and it should print the count. may be helps you..Thanks for seeing

package com.corejava;

import java.util.Map;
import java.util.TreeMap;

public class Test {
    public static void main(String[] args) {

        String str = "ramakoteswararao";

        char[] char_array = str.toCharArray();

        System.out.println("The Given String is : " + str);

    Map<Character, Integer> charCounter = new TreeMap<Character, Integer>();

        for (char i : char_array) {

    charCounter.put(i,charCounter.get(i) == null ? 1 : charCounter.get(i) + 1);

        }

    for (Character key : charCounter.keySet()) {
  System.out.println("occurrence of '" + key + "' is  "+ charCounter.get(key));
        }

    }

}

Hi, welcome to SO. When you post a solution in code, it may be helpful to include a short summary of the line of thought. — Marc Claesen, May 20 '13 at 13:29

score 1 · Answer 4 · answered Nov 07 '12 at 15:12

1

import java.util.HashMap;
import java.util.Map;
...
Map<String, Integer> freq = new HashMap<String, Integer>();
...
int count = freq.containsKey(word) ? freq.get(word) : 0;
freq.put(word, count + 1);

answered Nov 07 '12 at 15:12

atiruz

2,782
27
36

score 0 · Answer 5 · answered Apr 09 '15 at 13:21

0

inside the for loop

ch = char_array[i]; charCounter.put(charCounter.contains(ch)?charCounter.get(ch)+1:1);

answered Apr 09 '15 at 13:21

shyam karwa

213
1
3
8

score 0 · Answer 6 · edited Jun 12 '17 at 12:42

0

import java.util.TreeMap;

public class OccuranceDemo {
    public static void main(String[] args) {
        TreeMap<String , Integer> mp=new TreeMap();
        String s="rain rain go away";
        String[] arr = s.split(" ");
        int length=arr.length;
        for(int i=0;i<length;i++)
       {
         String h = arr[i];
        mp.put(h, mp.get(h)==null?1:mp.get(h)+1);
    }
    System.out.println(mp.get("go"));
  }
}

edited Jun 12 '17 at 12:42

HDJEMAI

9,436
46
67
93

answered Jun 12 '17 at 12:22

Teju J Gowda

1

If we need we get any other key in the above example we have taken go – Teju J Gowda Jun 12 '17 at 12:23
Hi, welcome to SO. A good first try, but your answer doesn't actually meet a number of requirements: (1) It counts the words, not the characters (2) It does not print the words and the counts at the end - doesn't work for arbitrary strings (3) It does not Hashmap, as asked (4) It does not actually fix the asker's code, but offers another solution. Please read the question carefully in the future before asking. See this guide: https://stackoverflow.com/help/how-to-answer – kenny_k Jun 12 '17 at 12:45

score 0 · Answer 7 · answered Sep 14 '17 at 18:31

    String str=new String("aabbbcddddee");
    char[] ch=str.toCharArray();
    HashMap<Character,Integer> hm=new HashMap<Character,Integer>();
    for(char ch1:ch)
    {

        if(hm.containsKey(ch1))
        {
            hm.put(ch1,hm.get(ch1)+1);
        }
        else
        {
            hm.put(ch1,1);
        }
    }

    Set s1=hm.entrySet();
    Iterator itr=s1.iterator();

    while(itr.hasNext())
    {
        Map.Entry m1=(Map.Entry)itr.next();
        System.out.println(m1);
    }

score 0 · Answer 8 · edited Jan 29 '18 at 09:50

import java.util.*;

public class Test {
    public static void main(String[] args) {

        String str = "STACKOVERFLOW";

        char[] char_array = str.toCharArray();

        System.out.println("The Given String is : " + str);

    Map<Character, Integer> charCounter = new TreeMap<Character, Integer>();

        for (char i : char_array) {

    charCounter.put(i,charCounter.get(i) == null ? 1 : charCounter.get(i) + 1);

        }

    for (Character key : charCounter.keySet()) {
  System.out.println("occurrence of '" + key + "' is  "+ charCounter.get(key));
        }

    }

}

Kailas Kakade · Answer 9 · 2018-08-23T15:55:15.930

public void mapPractices() {

    Map<Character, Integer> map = new HashMap<>();
    String dataString = "!@#$%^&*()__)(*&^%$#@!@#$%^&*(*&^%$#@!@#$%^&*()(*&^%$#@!@#$%^&*()(*&^%$#@!@#$%^&*()(*&^%$#";
    for (int i = 0; i < dataString.length(); i++) {

        char charAt = dataString.charAt(i);

        if (map.containsKey(charAt)) {
            int val = map.get(charAt);
            map.put(charAt, val+1);

        } else {

            map.put(charAt, +1);
        }

    }
    System.out.println("Characters ant the string=" + map);

}

score 0 · Answer 10 · answered Nov 26 '20 at 16:34

a lambda one-liner
After the bug in the old-school-solution is fixed, here is an alternative solution in lambda that does the same thing:

Map<Character,Long> counts =
"an example string to work with".codePoints().boxed().collect(
    groupingBy( t -> (char)(int)t, counting() ) );

gets: { =5, a=2, e=2, g=1, h=1, i=2, k=1, l=1, m=1, n=2, o=2, p=1, r=2, s=1, t=3, w=2, x=1}
You can get the number of a specific character eg. 't' by:
counts.get( 't' )

(I also write a lambda solution out of morbid curiosity to find out how slow my solution is, preferably from the guy with the 10-line solution.)

score 0 · Answer 11 · answered Sep 01 '23 at 15:24

Paul's answer contains exact fix, but instead of lengthy if-else checks, you may use getOrDefault method available since Java 8 to achieve the same results:

Map<Character, Integer> charCounter = new HashMap<>();
for (char c : str.toCharArray()) {
  charCounter.put(c, charCounter.getOrDefault(c, 0) + 1);
}

In this code snippet, we iterate over each character in the input string and update the count of that character in the map. If the character is not already present in the map, it is added with a default value of 0 using getOrDefault.

Hashmap implementation to count the occurrences of each character

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