Count number of occurrences for each char in a string

Question

I want to count the number of occurrences of each character in a given string using JavaScript.

For example:

var str = "I want to count the number of occurances of each char in this string";

Output should be:

h = 4;
e = 4; // and so on 

I tried searching Google, but didn't find any answer. I want to achieve something like this; order doesn't matter.

Answer 1

This is really, really simple in JavaScript (or any other language that supports maps):

// The string
var str = "I want to count the number of occurances of each char in this string";

// A map (in JavaScript, an object) for the character=>count mappings
var counts = {};

// Misc vars
var ch, index, len, count;

// Loop through the string...
for (index = 0, len = str.length; index < len; ++index) {
    // Get this character
    ch = str.charAt(index); // Not all engines support [] on strings

    // Get the count for it, if we have one; we'll get `undefined` if we
    // don't know this character yet
    count = counts[ch];

    // If we have one, store that count plus one; if not, store one
    // We can rely on `count` being falsey if we haven't seen it before,
    // because we never store falsey numbers in the `counts` object.
    counts[ch] = count ? count + 1 : 1;
}

Now counts has properties for each character; the value of each property is the count. You can output those like this:

for (ch in counts) {
    console.log(ch + " count: " + counts[ch]);
}

Answer 2

Shorter answer, with reduce:

let s = 'hello';
var result = [...s].reduce((a, e) => { a[e] = a[e] ? a[e] + 1 : 1; return a }, {}); 
console.log(result); // {h: 1, e: 1, l: 2, o: 1}

Answer 3

let str = "atul kumar srivastava";
let obj ={};
for(let s of str)if(!obj[s])obj[s] = 1;else obj[s] = obj[s]  + 1;
console.log(obj)

Answer 4

I iterated over each character and put it in a nested object along with the count. If the character already exists in the object, I simply increment the count. This is what myObj looks like:

myObj = {
char1 = { count : <some num> },
char2 = { count : <some num> },
....
}

Here is the code:

function countChar(str) {
    let myObj= {};
    for (let s of str) {
        if ( myObj[s] ? myObj[s].count ++ : myObj[s] = { count : 1 } );
    }
    return myObj;
}

var charCount = countChar('abcceddd');

Answer 5

This worked well for me :

function Char_Count(str1) {
    var chars = {};
    str1.replace(/\S/g, function(l){chars[l] = (isNaN(chars[l]) ? 1 : chars[l] + 1);});
    return chars;
}

var myString = "This is my String";
console.log(Char_Count(myString));

Answer 6

You can use the maps in ES6 in Javascript. Provides a cleaner and concise code in my opinion. Here is how I would go about

function countChrOccurence ('hello') {
 let charMap = new Map();
 const count = 0;
  for (const key of str) {
   charMap.set(key,count); // initialize every character with 0. this would make charMap to be 'h'=> 0, 'e' => 0, 'l' => 0, 
  }

  for (const key of str) {
    let count = charMap.get(key);
    charMap.set(key, count + 1);
  }
// 'h' => 1, 'e' => 1, 'l' => 2, 'o' => 1

  for (const [key,value] of charMap) {
    console.log(key,value);
  }
// ['h',1],['e',1],['l',2],['o',1]
}  

Answer 7

I am giving you very very simple code.

 // Converts String To Array
        var SampleString= Array.from("saleem");

        // return Distinct count as a object
        var allcount = _.countBy(SampleString, function (num) {
            return num;
        });

        // Iterating over object and printing key and value
        _.map(allcount, function(cnt,key){
            console.log(key +":"+cnt);
        });

        // Printing Object
        console.log(allcount);

<script src="https://cdnjs.cloudflare.com/ajax/libs/underscore.js/1.8.3/underscore-min.js"></script>

    <p>Set the variable to different value and then try...</p>
    

Answer 8

str = "aaabbbccccdefg";

words = str.split("");

var obj = [];

var counter = 1, jump = 0;

for (let i = 0; i < words.length; i++) {
    if (words[i] === words[i + 1]) {
        counter++;
        jump++;
    }
    else {
        if (jump > 0) {
            obj[words[i]] = counter;
            jump = 0;
            counter=1
        }
        else
            obj[words[i]] = 1;
    }

}
console.log(obj);

Answer 9

A 1-liner ES6 way:

const some_string = 'abbcccdddd';
const charCountIndex = [ ...some_string ].reduce( ( a, c ) => ! a[ c ] ? { ...a, [ c ]: 1 } : { ...a, [ c ]: a[ c ] + 1 }, {} );
console.log( charCountIndex )

Answer 10

You can use an object for a task.

Step 1 - create an object

step 2 - traverse through a string

Step 3 - add character as key and character count as value in the object

var obj={}

function countWord(arr)
{
for(let i=0;i<arr.length;i++)
{
if(obj[arr[i]]) //check if character is present in the obj as key
{
    obj[arr[i]]=obj[arr[i]]+1; //if yes then update its value
}
else
{
    obj[arr[i]]=1; //initialise it with a value 1

}
}
}

Answer 11

    function cauta() {

        var str = document.form.stringul.value;
        str = str.toLowerCase();
        var tablou = [];

        k = 0;
        //cautarea caracterelor unice
        for (var i = 0, n = 0; i < str.length; i++) {
            for (var j = 0; j < tablou.length; j++) {
                if (tablou[j] == str[i]) k = 1;
            }
            if (k != 1) {
                if (str[i] != ' ')
                    tablou[n] = str[i]; n++;
            }
            k = 0;
        }
        //numararea aparitilor
        count = 0;
        for (var i = 0; i < tablou.length; i++) {
            if(tablou[i]!=null){
            char = tablou[i];
            pos = str.indexOf(char);
            while (pos > -1) {
                ++count;
                pos = str.indexOf(char, ++pos);

            }

            document.getElementById("rezultat").innerHTML += tablou[i] + ":" + count + '\n';
            count = 0;
        }
        }

    }

This function will put each unique char in array, and after will find the appearances of each char in str. In my Case, i get and put data into

Answer 12

 // Converts String To Array
        var SampleString= Array.from("saleem");

        // return Distinct count as a object
        var allcount = _.countBy(SampleString, function (num) {
            return num;
        });

        // Iterating over object and printing key and value
        _.map(allcount, function(cnt,key){
            console.log(key +":"+cnt);
        });

        // Printing Object
        console.log(allcount);

<script src="https://cdnjs.cloudflare.com/ajax/libs/underscore.js/1.8.3/underscore-min.js"></script>

    <p>Set the variable to different value and then try...</p>
    

Answer 13

Hope this helps someone

function getNoOfOccurences(str){
    var temp = {};
    for(var oindex=0;oindex<str.length;oindex++){
        if(typeof temp[str.charAt(oindex)] == 'undefined'){
            temp[str.charAt(oindex)] = 1;
        }else{
            temp[str.charAt(oindex)] = temp[str.charAt(oindex)]+1;
        }
    }
    return temp;
}

Answer 14

    package com.company;

import java.util.HashMap;


 public class Main {

    public static void main(String[] args) {
    // write your code here
    HashMap<Character, Integer> sHashMap = new HashMap();  // using hashMap<key , value > here key = character and  value = count

    String arr = "HelloWorld";

    for (int i = 0; i < arr.length(); i++) {
        boolean flag = sHashMap.containsKey(arr.charAt(i));  // check if char is already  present 

    if (flag == true)
        {
            int Count = sHashMap.get(arr.charAt(i)); // get the char count
            sHashMap.put(arr.charAt(i), ++Count); //   increment the count and update in hashMap
        } 
        else 
        {
            sHashMap.put(arr.charAt(i), 1); //if  char not present then insert into hashMap
        }
    }

     System.out.println(sHashMap);
    //OutPut would be like ths {r=1, d=1, e=1, W=1, H=1, l=3, o=2}

}

}

Answer 15

I have used Map object , The map object doesn't let you set any duplicate key and that makes our job easy . I am checking if the key already exists in map , if not I am inserting and setting the count to 1 , if it already exists I am getting the value and then incrementing

const str = "Hello H"
    const strTrim = str.replace(/\s/g,'') // HelloH
    const strArr=strTrim.split('')

    let myMap = new Map(); // Map object 

    strArr.map(ele=>{
    let count =0
    if(!myMap.get(ele)){
    myMap.set(ele,++count)
    }else {
    let cnt=myMap.get(ele)
    myMap.set(ele,++cnt)
    }
    console.log("map",myMap)
    })

Answer 16

let newStr= "asafasdhfasjkhfweoiuriujasfaksldjhalsjkhfjlkqaofadsfasasdfas";
       
function checkStringOccurnace(newStr){
    let finalStr = {};
    let checkArr = [];
    let counterArr = [];
    for(let i = 0; i < newStr.length; i++){
        if(checkArr.indexOf(newStr[i]) == -1){
            checkArr.push(newStr[i])
            let counter = 0;
            counterArr.push(counter + 1)
            finalStr[newStr[i]] = 1;
        }else if(checkArr.indexOf(newStr[i]) > -1){
            let index = checkArr.indexOf(newStr[i])
            counterArr[index] = counterArr[index] + 1;
            finalStr[checkArr[index]] = counterArr[index];
        }
    }
    return finalStr;
}

let demo = checkStringOccurnace(newStr);
console.log(" finalStr >> ", demo);

Answer 17

I tried it with chceking "empty space" as well as "special characters":

function charCount(str){
    const requiredString = str.toLowerCase();

    const leng = str.length;

    let output = {};

    for(let i=0; i<leng; i++){
        const activeCharacter = requiredString[i];
        if(/[a-z0-9]/.test(activeCharacter)){
            output.hasOwnProperty(activeCharacter) ? output[activeCharacter]++ : output[activeCharacter] = 1;
        }
    }
    return output;
}

Answer 18

Split the string with the spread ... operator instead of .split(''):

''.split('')
//=> ["\ud83c", "\udf2f", "\ud83c", "\udf2f", "\ud83c", "\udf63", "\ud83c", "\udf7b"]

vs

[...'']
//=> ["", "", "", ""]

vs

''.charAt(0)
//=> "\ud83c"

Then reduce:

[...''].reduce((m, c) => (m[c] = (m[c] || 0) + 1, m), {})
//=> {'': 2, '': 1, '': 1}

Answer 19

To check and avoid spaces and convert capital letters to small and count we can do like below.

function CountNumberOfCharacters(str) {
  let countObject = {};
  const lowerCaseString = str.toLowerCase();
  for (let char of lowerCaseString) {
    if (!countObject[char] && char !== ' ') {
      countObject[char] = 1;
    } else if (char !== ' ') {
      countObject[char] = countObject[char] + 1;
    }
  }
  return countObject;
}