How to count frequency of characters in a string?

Question

I need to write some kind of loop that can count the frequency of each letter in a string.
For example: "aasjjikkk" would count 2 'a', 1 's', 2 'j', 1 'i', 3 'k'. Ultimately id like these to end up in a map with the character as the key and the count as the value. Any good idea how to do this?

score 36 · Accepted Answer · edited Nov 17 '22 at 14:14

36

You can use a java Map and map a char to an int. You can then iterate over the characters in the string and check if they have been added to the map, if they have, you can then increment its value.

For example:

Map<Character, Integer> map = new HashMap<Character, Integer>();
String s = "aasjjikkk";
for (int i = 0; i < s.length(); i++) {
    char c = s.charAt(i);
    Integer val = map.get(c);
    if (val != null) {
        map.put(c, val + 1);
    }
    else {
       map.put(c, 1);
   }
}

At the end you will have a count of all the characters you encountered and you can extract their frequencies from that.

Alternatively, you can use Bozho's solution of using a Multiset and counting the total occurences.

edited Nov 17 '22 at 14:14

assylias

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answered Jul 15 '11 at 20:19

xunil154

527
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2

oh, but you cant instatiate Map, its abstract, just fyi. – Bill Jul 15 '11 at 20:23
My bad, it should be HashMap, not Map. Thanks for the catch – xunil154 Jul 15 '11 at 20:24
should also be . – Marcelo Jul 15 '11 at 21:59
HashMap – Bernd Elkemann Jul 15 '11 at 22:04
Could you please explain the step: if(val != null) { ... } – Christos Michael Dec 17 '15 at 10:14
@ChristosMichael if the value is null i.e. there is no entry for that character it should be increased to 1, since it is the first occurrence. – User Apr 29 '16 at 16:13

score 21 · Answer 2 · answered Jul 21 '18 at 12:59

Using the stream API as of JDK-8:

Map<Character, Long> frequency =
            str.chars()
               .mapToObj(c -> (char)c)
               .collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));

or if you want the keys as Integers:

Map<Character, Integer> frequency =
            str.chars()
               .mapToObj(c -> (char)c)
               .collect(Collectors.groupingBy(Function.identity(), Collectors.summingInt(c -> 1)));

Another variant:

Map<Character, Integer> frequency = 
            str.chars()
               .mapToObj(c -> (char)c)
               .collect(Collectors.toMap(Function.identity(), c -> 1, Math::addExact));

Sreekanth · Answer 3 · 2017-11-30T13:51:42.607

19

A concise way to do this is:

Map<Character,Integer> frequencies = new HashMap<>();
for (char ch : input.toCharArray()) 
   frequencies.put(ch, frequencies.getOrDefault(ch, 0) + 1);

We use a for-each to loop through every character. The frequencies.getOrDefault() gets value if key is present or returns(as default) its second argument.

edited Nov 30 '17 at 13:51

answered Jun 16 '17 at 01:48

Sreekanth

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6

score 14 · Answer 4 · edited May 20 '13 at 05:29

14

You can use a Multiset (from guava). It will give you the count for each object. For example:

Multiset<Character> chars = HashMultiset.create();
for (int i = 0; i < string.length(); i++) {
    chars.add(string.charAt(i));
}

Then for each character you can call chars.count('a') and it returns the number of occurrences

edited May 20 '13 at 05:29

Jeff

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answered Jul 15 '11 at 20:10

Bozho

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I need something from the standard libraries honestly. – Bill Jul 15 '11 at 20:18

score 4 · Answer 5 · edited Jul 21 '14 at 07:06

Here is another solution, dodgy as it may be.

public char getNumChar(String s) {
    char[] c = s.toCharArray();
    String alphabet = "abcdefghijklmnopqrstuvwxyz";
    int[] countArray = new int[26];
    for (char x : c) {
        for (int i = 0; i < alphabet.length(); i++) {
            if (alphabet.charAt(i) == x) {
                countArray[i]++;
            }
        }
    }

    java.util.HashMap<Integer, Character> countList = new java.util.HashMap<Integer, Character>();

    for (int i = 0; i < 26; i++) {
        countList.put(countArray[i], alphabet.charAt(i));
    }
    java.util.Arrays.sort(countArray);
    int max = countArray[25];
    return countList.get(max);
}

score 4 · Answer 6 · edited Jul 22 '18 at 17:48

Since there was no Java 8 solution, thought of posting one. Also, this solution is much neater, readable and concise than some of the other solutions mentioned here.

String string = "aasjjikkk";

Map<Character, Long> characterFrequency = string.chars()  // creates an IntStream
    .mapToObj(c -> (char) c) // converts the IntStream to Stream<Character>
    .collect(Collectors.groupingBy(c -> c, Collectors.counting())); // creates a
                                                                    // Map<Character, Long> 
                                                                    // where the Long is
                                                                    // the frequency

score 3 · Answer 7 · answered Jul 15 '11 at 20:12

Well, two ways come to mind and it depends on your preference:

Sort the array by characters. Then, counting each character becomes trivial. But you will have to make a copy of the array first.
Create another integer array of size 26 (say freq) and str is the array of characters.

for(int i = 0; i < str.length; i ++)

freq[str[i] - 'a'] ++; //Assuming all characters are in lower case

So the number of 'a' 's will be stored at freq[0] and the number of 'z' 's will be at freq[25]

Donald Raab · Answer 8 · 2019-11-10T01:58:02.600

You can use a CharAdapter and a CharBag from Eclipse Collections and avoid boxing to Character and Integer.

CharBag bag = Strings.asChars("aasjjikkk").toBag();

Assert.assertEquals(2, bag.occurrencesOf('a'));
Assert.assertEquals(1, bag.occurrencesOf('s'));
Assert.assertEquals(2, bag.occurrencesOf('j'));
Assert.assertEquals(1, bag.occurrencesOf('i'));
Assert.assertEquals(3, bag.occurrencesOf('k'));

Note: I am a committer for Eclipse Collections.

score 2 · Answer 9 · answered Apr 29 '19 at 18:35

2

String s = "aaaabbbbcccddddd";
Map<Character, Integer> map = new HashMap<>();

Using one line in Java8

s.chars().forEach(e->map.put((char)e, map.getOrDefault((char)e, 0) + 1));

answered Apr 29 '19 at 18:35

Ankit Sharma

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Echoinacup · Answer 10 · 2020-03-19T12:00:23.527

There is one more option and it looks quite nice. Since java 8 there is new method merge java doc

  public static void main(String[] args) {
    String s = "aaabbbcca";

    Map<Character, Integer> freqMap = new HashMap<>();
    for (int i = 0; i < s.length(); i++) {
        Character c = s.charAt(i);
        freqMap.merge(c, 1, (a, b) -> a + b);
    }
    freqMap.forEach((k, v) -> System.out.println(k + " and " + v));
}

Or even cleaner with ForEach

       for (Character c : s.toCharArray()) {
            freqMapSecond.merge(c, 1, Integer::sum);
           }

score 2 · Answer 11 · answered Jul 15 '11 at 20:34

Here is a solution:

Define your own Pair:

public class Pair
{
    private char letter;
    private int count;
    public Pair(char letter, int count)
    {
        this.letter = letter;
        this.count= count;
    }
    public char getLetter(){return key;}
    public int getCount(){return count;}
}

Then you could do:

public static Pair countCharFreq(String s)
{
    String temp = s;
    java.util.List<Pair> list = new java.util.ArrayList<Pair>();
    while(temp.length() != 0)
    {
        list.add(new Pair(temp.charAt(0), countOccurrences(temp, temp.charAt(0))));
        temp.replaceAll("[" + temp.charAt(0) +"]","");
    }
}

public static int countOccurrences(String s, char c)
{
    int count = 0;
    for(int i = 0; i < s.length(); i++)
    {
        if(s.charAt(i) == c) count++;
    }
    return count;
}

RishiKesh Pathak · Answer 12 · 2016-03-29T05:01:16.997

package com.rishi.zava;

import java.util.HashMap;
import java.util.Map;
import java.util.Map.Entry;

public class ZipString {
    public static void main(String arg[]) {
        String input = "aaaajjjgggtttssvvkkllaaiiikk";
        int len = input.length();
        Map<Character, Integer> zip = new HashMap<Character, Integer>();
        for (int j = 0; len > j; j++) {
            int count = 0;
            for (int i = 0; len > i; i++) {
                if (input.charAt(j) == input.charAt(i)) {
                    count++;
                }
            }
            zip.put(input.charAt(j), count);
        }
        StringBuffer myValue = new StringBuffer();
        String myMapKeyValue = "";
        for (Entry<Character, Integer> entry : zip.entrySet()) {
            myMapKeyValue = Character.toString(entry.getKey()).concat(
                    Integer.toString(entry.getValue()));
            myValue.append(myMapKeyValue);
        }
        System.out.println(myValue);
    }
}

Input = aaaajjjgggtttssvvkkllaaiiikk

Output = a6s2t3v2g3i3j3k4l2

score 0 · Answer 13 · answered Feb 28 '14 at 06:16

This is similar to xunil154's answer, except that a string is made a char array and a linked hashmap is used to maintain the insertion order of the characters.

String text = "aasjjikkk";
char[] charArray = text.toCharArray();
Map<Character, Integer> freqList = new LinkedHashMap<Character, Integer>();

        for(char key : charArray) {
            if(freqList.containsKey(key)) {
               freqList.put(key, freqList.get(key) + 1);
            } else
                freqList.put(key, 1);
        }

score 0 · Answer 14 · edited Jul 29 '20 at 21:29

import java.util.HashMap;
import java.util.Iterator;
import java.util.Map;
import java.util.Map.Entry;
import java.util.Scanner;


public class FrequenceyOfCharacters {

    public static void main(String[] args) {
        System.out.println("Please enter the string to count each character frequencey: ");
        Scanner sc=new  Scanner(System.in);
        String s =sc.nextLine();
        String input = s.replaceAll("\\s",""); // To remove space.
        frequenceyCount(input);
        
        
    }

    private static void frequenceyCount(String input) {
        
        Map<Character,Integer> hashCount=new HashMap<>();
        Character c;
        for(int i=0; i<input.length();i++)
        {
           c =input.charAt(i);
           if(hashCount.get(c)!=null){
               hashCount.put(c, hashCount.get(c)+1);
           }else{
               hashCount.put(c, 1);
           }
        }
        Iterator it = hashCount.entrySet().iterator();
        System.out.println("char : frequency");
        while (it.hasNext()) {
            Map.Entry pairs = (Map.Entry)it.next();
            System.out.println(pairs.getKey() + " : " + pairs.getValue());
            it.remove(); 
        }
        
    }

}

score 0 · Answer 15 · edited Dec 25 '15 at 07:48

import java.util.*;
class Charfrequency
{
 public static void main(String a[]){

        Scanner sc=new Scanner(System.in);
        System.out.println("Enter Your String :");
        String s1=sc.nextLine();
        int count,j=1;
        char var='a';
        char ch[]=s1.toCharArray();
        while(j<=26)
        {
           count=0;
                for(int i=0; i<s1.length(); i++)
                {
                    if(ch[i]==var || ch[i]==var-32)
                    {
                        count++;
                    }
                }
                if(count>0){
                System.out.println("Frequency of "+var+" is "+count);
                }
                var++;
                j++;
        }
 }
}

score 0 · Answer 16 · answered Apr 03 '16 at 23:07

The shorted possible code using a HashMap. (With no forceful line saves)

private static Map<Character, Integer> findCharacterFrequency(String str) {

        Map<Character, Integer> map = new HashMap<>();

        for (char ch : str.toCharArray()) {
            /* Using getOrDefault(), since Java1.8 */
            map.put(ch, map.getOrDefault(ch, 0) + 1);
        }
        return map;
    }

naveen prasanna · Answer 17 · 2017-11-21T07:35:18.950

Please try the given code below, hope it will helpful to you,

import java.util.Scanner;

class String55 {
    public static int frequency(String s1,String s2)
    {

        int count=0;
        char ch[]=s1.toCharArray();
        char ch1[]=s2.toCharArray();
        for (int i=0;i<ch.length-1; i++)
        {



                int k=i;

                int j1=i+1;
                int j=0;
                int j11=j;
                int j2=j+1;
                {
                    while(k<ch.length && j11<ch1.length && ch[k]==ch1[j11])
                    {
                    k++;
                    j11++;

                    }


                int l=k+j1;
                int m=j11+j2;
                if( l== m)
                {
                    count=1;
                    count++;
                }





        }
        }
        return count;


    }
    public static void main (String[] args) {
        Scanner sc=new Scanner(System.in);
        System.out.println("enter the pattern");
        String s1=sc.next();
            System.out.println("enter the String");
            String s2=sc.next();
            int res=frequency(s1, s2);
            System.out.println("FREQUENCY==" +res);

    }
}

SAMPLE OUTPUT: enter the pattern man enter the String dhimanman FREQUENCY==2

Thank-you.Happy coding.

score 0 · Answer 18 · answered Dec 10 '18 at 23:46

package com.dipu.string;

import java.util.HashMap;
import java.util.Map;

public class RepetativeCharInString {
    public static void main(String[] args) {
        String data = "aaabbbcccdddffffrss";
        char[] charArray = data.toCharArray();
        Map<Character, Integer> map = new HashMap<>();
        for (char c : charArray) {
            if (map.containsKey(c)) {
                map.put(c, map.get(c) + 1);
            } else {
                map.put(c, 1);
            }
        }
        System.out.println(map);

    }
}

Please remember to always add an explanation of your code. – mypetlion Dec 11 '18 at 00:06 — mypetlion, Dec 11 '18 at 00:06

MinakshiJ · Answer 19 · 2019-02-08T18:31:32.870

*import java.util.ArrayList;
import java.util.Collections;

public class Freq {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        String temp="zsaaqaaaaaaaabbbbbcc";
    List<String> temp1= new ArrayList<String> ();
    ArrayList<Integer>freq=new ArrayList<Integer>();
    for(int i=0;i<temp.length()-1;i++)
    {       
        temp1.add(Character.toString(temp.charAt(i)));      
    }
    Set<String> uniqset=new HashSet<String>(temp1);
    for(String s:uniqset)
    {
        freq.add(Collections.frequency(temp1, s));
        System.out.println(s+" -->>"+Collections.frequency(temp1, s));
    }
    }

}
           ------Output-------
       a -->>10
       b -->>5
       c -->>1
       q -->>1
       s -->>1
       z -->>1

Use collections frequency method to count frequency of char*

score 0 · Answer 20 · answered May 06 '19 at 14:54

0

Uffh. Don't you think this is the simplest solution?

    char inputChar = '|';
    int freq = "|fd|fdfd|f dfd|fd".replaceAll("[^" + inputChar +"]", "").length();
    System.out.println("freq " + freq);

answered May 06 '19 at 14:54

Chandrayya G K

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5
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Baidyanath · Answer 21 · 2019-10-26T09:26:13.847

0

We can use frequency method of Collections class for this. Split the string into string array. Use HashSet to remove duplicates and check frequency of each object in HashSet using frequency method of Collections

void usingCollections(){

  String input = "cuttack";

  String [] stringArray = input.split("");

  Set<String> s = new HashSet(Arrays.asList(stringArray));

  for(String abc : s){

    System.out.println (abc + ":"+Collections.frequency(Arrays.asList(stringArray),abc));

  }
}

edited Oct 26 '19 at 09:26

answered Oct 19 '19 at 13:11

Baidyanath

11
2

1

While this code may solve the question, [including an explanation](//meta.stackexchange.com/q/114762) of how and why this solves the problem would really help to improve the quality of your post, and probably result in more up-votes. Remember that you are answering the question for readers in the future, not just the person asking now. Please [edit] your answer to add explanations and give an indication of what limitations and assumptions apply. – Dharman Oct 19 '19 at 22:15

score 0 · Answer 22 · answered Dec 13 '19 at 18:50

This is more Effective way to count frequency of characters in a string

public class demo {
    public static void main(String[] args) {
        String s = "babdcwertyuiuygf";
        Map<Character, Integer> map = new TreeMap<>();
        s.chars().forEach(e->map.put((char)e, map.getOrDefault((char)e, 0) + 1));
        StringBuffer myValue = new StringBuffer();
        String myMapKeyValue = "";
        for (Map.Entry<Character, Integer> entry : map.entrySet()) {
            myMapKeyValue = Character.toString(entry.getKey()).concat(
                Integer.toString(entry.getValue()));
            myValue.append(myMapKeyValue);
        }
        System.out.println(myValue);
    }
}

score 0 · Answer 23 · answered Mar 17 '20 at 04:50

0

Another way using map merge method

   Map<Character, Integer> map = new HashMap<>();
   String s = "aasjjikkk";
    for (int i = 0; i < s.length(); i++) {
        map.merge(s.charAt(i), 1, (l, r) -> l + r);

answered Mar 17 '20 at 04:50

Gajraj Tanwar

81
2

score 0 · Answer 24 · answered Jul 15 '11 at 20:12

If this does not need to be super-fast just create an array of integers, one integer for each letter (only alphabetic so 2*26 integers? or any binary data possible?). go through the string one char at a time, get the index of the responsible integer (e.g. if you only have alphabetic chars you can have 'A' be at index 0 and get that index by subtracting any 'A' to 'Z' by 'A' just as an example of how you can get reasonably fast indices) and increment the value in that index.

There are various micro-optimizations to make this faster (if necessary).

Mike Caputo · Answer 25 · 2011-07-15T20:36:05.163

0

You can use a Hashtable with each character as the key and the total count becomes the value.

Hashtable<Character,Integer> table = new Hashtable<Character,Integer>();
String str = "aasjjikkk";
for( c in str ) {
    if( table.get(c) == null )
        table.put(c,1);
    else
        table.put(c,table.get(c) + 1);
}

for( elem in table ) {
    println "elem:" + elem;
}

edited Jul 15 '11 at 20:36

answered Jul 15 '11 at 20:22

Mike Caputo

1,156
17
33

score -1 · Answer 26 · answered Jul 08 '14 at 12:18

import java.io.FileInputStream;
import java.util.HashSet;
import java.util.Iterator;
public class CountFrequencyOfCharater {
public static void main(String args[]) throws Exception
{
    HashSet hs=new HashSet();
    String str="hey how are you?";
    char arr[]=new char[str.length()];
    for(int i=0;i<str.length();i++)
    {
        arr[i]=str.charAt(i);
    }
    for(int j=0;j<str.length();j++)
    {
        int c=0;
        for(int k=0;k<str.length();k++)
        {
            if(arr[j]==arr[k])
            c++;
        }
        hs.add(arr[j]+"="+c+",");
    }
        Iterator it=hs.iterator();
        while(it.hasNext())
        {
             System.out.print(it.next());
        }
  }
}

Please add some description or explanation as to why this works instead of just offering code. This will make it more meaningful for people in the future looking at it. — Aelphaeis, Jul 08 '14 at 12:37

Suraj Verma · Answer 27 · 2020-07-31T09:15:54.453

-2

#From C language

 #include<stdio.h>`
 #include <string.h>`
  int main()
{
    char s[1000];  
    int  i,j,k,count=0,n;
    printf("Enter  the string : ");
    gets(s);
    for(j=0;s[j];j++);
    n=j; 
    printf(" frequency count character in string:\n");
    for(i=0;i<n;i++)  
    {
        count=1;
        if(s[i])
        {
        
          for(j=i+1;j<n;j++)  
          {   
            
            if(s[i]==s[j])
            {
                 count++;
                 s[j]='\0';
            }
          }  
          printf(" '%c' = %d \n",s[i],count);
       }
    } 
    return 0;
}

edited Jul 31 '20 at 09:15

answered Jul 27 '19 at 12:14

Suraj Verma

463
6
8

While this code may answer the question, providing additional context regarding why and/or how this code answers the question improves its long-term value. – adiga Jul 27 '19 at 13:06
just curious, why would you answer a question tagged for java with a different programming language? – luckyguy73 Jul 26 '20 at 21:38

How to count frequency of characters in a string?

27 Answers27

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