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?

Molarity

You’ll find more good Answers on this duplicate: How to convert stream of character to Map<Character, Integer> – Curagh

Answer 1

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.

Recreate

oh, but you cant instatiate Map, its abstract, just fyi. – Molarity

My bad, it should be HashMap, not Map. Thanks for the catch – Recreate

<char, int> should also be <Character, Integer>. – Mercurialism

HashMap<char... should not work right? HashMap requires its generic types to be reference types not immediates right? You could use boxed Character here, or you could take a look at my array-solution. – Intestine

Could you please explain the step: if(val != null) { ... } – Piatt

@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. – Sublimate

Answer 2

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 values 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));

Lanita

Answer 3

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 getOrDefault method gets the value, if key is present or returns (as default) its second argument.

---

Alternatively, the merge method can be used; which conveniently takes a remappingFunction parameter. In this case the Integer::sum function is used.

Map<Character,Integer> frequencies = new HashMap<>();
for (char ch : input.toCharArray())
   frequencies.merge(ch, 1, Integer::sum);

Unclad

Even more concise: frequencies.merge(ch, 1, Integer::sum); it also doesn’t require two map lookups. – Cheep

This is a lot easier to remember than the stream/cast/collect syntax. – Ulcerative

Answer 4

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

Stutz

I need something from the standard libraries honestly. – Molarity

Answer 5

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);
}

Bag

Instead of two nested loops, you can simply use countArray[i - 'a']++; And sorting an array just to find its maximum value, is quite inefficient. You can find the maximum with a simple linear loop and when you do so, you also know its index which corresponds to the letter, making the HashMap obsolete. – Cheep

Answer 6

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

Joung

Answer 7

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

1. Sort the array by characters. Then, counting each character becomes trivial. But you will have to make a copy of the array first.

2. 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]

Sauceda

Answer 8

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;
}

Amphimacer

Answer 9

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.

Galba

Answer 10

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));

Nicholasnichole

Answer 11

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);
           }

Acrobatics

Answer 12

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

Blinking

Answer 13

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).

Intestine

Answer 14

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;
}

Mispleading

Answer 15

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);
        }

Amand

Answer 16

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(); 
        }
        
    }

}

Friar

Answer 17

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++;
        }
 }
}

Hinayana

Answer 18

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;
    }

Epicene

Answer 19

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.

Insinuation

Answer 20

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);

    }
}

Snakebite

Please remember to always add an explanation of your code. – Intangible

Answer 21

*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*

Timmie

Answer 22

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);

Diploma

Answer 23

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));

  }
}

Chaparral

While this code may solve the question, including an explanation 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. – Trio

Answer 24

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);
    }
}

Vocabulary

Answer 25

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);

Calycle

Answer 26

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());
        }
  }
}

Incision

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. – Polysaccharide

Answer 27

#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;
}

Martsen

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. – Indecision

just curious, why would you answer a question tagged for java with a different programming language? – Luby