Count the number of occurrences of a character in a string in Javascript

I have updated this answer. I like the idea of using a match better, but it is slower:

console.log(("str1,str2,str3,str4".match(/,/g) || []).length); //logs 3

console.log(("str1,str2,str3,str4".match(new RegExp("str", "g")) || []).length); //logs 4

Use a regular expression literal if you know what you are searching for beforehand, if not you can use the RegExp constructor, and pass in the g flag as an argument.

match returns null with no results thus the || []

The original answer I made in 2009 is below. It creates an array unnecessarily, but using a split is faster (as of September 2014). I'm ambivalent, if I really needed the speed there would be no question that I would use a split, but I would prefer to use match.

Old answer (from 2009):

If you're looking for the commas:

(mainStr.split(",").length - 1) //3

If you're looking for the str

(mainStr.split("str").length - 1) //4

Both in @Lo's answer and in my own silly performance test split comes ahead in speed, at least in Chrome, but again creating the extra array just doesn't seem sane.

Simply, use the split to find out the number of occurrences of a character in a string.

mainStr.split(',').length // gives 4 which is the number of strings after splitting using delimiter comma

mainStr.split(',').length - 1 // gives 3 which is the count of comma

A quick Google search got this (from http://www.codecodex.com/wiki/index.php?title=Count_the_number_of_occurrences_of_a_specific_character_in_a_string#JavaScript)

String.prototype.count=function(s1) { 
    return (this.length - this.replace(new RegExp(s1,"g"), '').length) / s1.length;
}

Use it like this:

test = 'one,two,three,four'
commas = test.count(',') // returns 3

You can also rest your string and work with it like an array of elements using

• Array.prototype.filter()

const mainStr = 'str1,str2,str3,str4';
const commas = [...mainStr].filter(l => l === ',').length;

console.log(commas);

Or

• Array.prototype.reduce()

const mainStr = 'str1,str2,str3,str4';
const commas = [...mainStr].reduce((a, c) => c === ',' ? ++a : a, 0);

console.log(commas);

Here is a similar solution, but it uses Array.prototype.reduce

function countCharacters(char, string) {
  return string.split('').reduce((acc, ch) => ch === char ? acc + 1: acc, 0)
}

As was mentioned, String.prototype.split works much faster than String.prototype.replace.

It's amazing that in 13 years, this answer hasn't shown up. Intuitively, it seems like it should be fastest:

const s = "The quick brown fox jumps over the lazy dog.";
const oCount = s.length - s.replaceAll('o', '').length;

If there are only two kinds of character in the string, then this is faster still:

const s = "001101001";
const oneCount = s.replaceAll('0', '').length;

I just did a very quick and dirty test on repl.it using Node v7.4. For a single character, the standard for loop is quickest:

Some code:

// winner!
function charCount1(s, c) {
    let count = 0;
    c = c.charAt(0); // we save some time here
    for(let i = 0; i < s.length; ++i) {
        if(c === s.charAt(i)) {
            ++count;
        }
    }
    return count;
}

function charCount2(s, c) {
    return (s.match(new RegExp(c[0], 'g')) || []).length;
}

function charCount3(s, c) {
    let count = 0;
    for(ch of s) {
        if(c === ch) {
            ++count;
        }
    }
    return count;
}

function perfIt() {
    const s = 'Hello, World!';
    const c = 'o';

    console.time('charCount1');
    for(let i = 0; i < 10000; i++) {
        charCount1(s, c);
    }
    console.timeEnd('charCount1');
    
    console.time('charCount2');
    for(let i = 0; i < 10000; i++) {
        charCount2(s, c);
    }
    console.timeEnd('charCount2');
    
    console.time('charCount3');
    for(let i = 0; i < 10000; i++) {
        charCount2(s, c);
    }
    console.timeEnd('charCount3');
}

Results from a few runs:

perfIt()
charCount1: 3.301ms
charCount2: 11.652ms
charCount3: 174.043ms
undefined

perfIt()
charCount1: 2.110ms
charCount2: 11.931ms
charCount3: 177.743ms
undefined

perfIt()
charCount1: 2.074ms
charCount2: 11.738ms
charCount3: 152.611ms
undefined

perfIt()
charCount1: 2.076ms
charCount2: 11.685ms
charCount3: 154.757ms
undefined

Update 2021-Feb-10: Fixed typo in repl.it demo

Update 2020-Oct-24: Still the case with Node.js 12 (play with it yourself here)

Here is my solution. Lots of solution already posted before me. But I love to share my view here.

const mainStr = 'str1,str2,str3,str4';

const commaAndStringCounter = (str) => {
  const commas = [...str].filter(letter => letter === ',').length;
  const numOfStr = str.split(',').length;

  return `Commas: ${commas}, String: ${numOfStr}`;
}

// Run the code
console.log(commaAndStringCounter(mainStr)); // Output: Commas: 3, String: 4

Here you find my REPL

The fastest method seems to be via the index operator:

function charOccurances (str, char)
{
  for (var c = 0, i = 0, len = str.length; i < len; ++i)
  {
    if (str[i] == char)
    {
      ++c;
    }
  }
  return c;
}

console.log( charOccurances('example/path/script.js', '/') ); // 2

Or as a prototype function:

String.prototype.charOccurances = function (char)
{
  for (var c = 0, i = 0, len = this.length; i < len; ++i)
  {
    if (this[i] == char)
    {
      ++c;
    }
  }
  return c;
}

console.log( 'example/path/script.js'.charOccurances('/') ); // 2

function len(text,char){

return text.innerText.split(string).length
}

console.log(len("str1,str2,str3,str4",","))

This is a very short function.

I know I am late to the party here but I was rather baffled no one answered this with the most basic of approaches. A large portion of the answers provided by the community for this question are iteration based but all are moving over strings on a per-character basis which is not really efficient.

When dealing with a large string that contains thousands of characters walking over each character to get the occurance count can become rather extraneous not to mention a code-smell. The below solutions take advantage of slice, indexOf and the trusted traditional while loop. These approaches prevent us having to walk over each character and will greatly speed up the time it takes to count occurances. These follow similar logic to that you'd find in parsers and lexical analyzers that require string walks.

Using with Slice

In this approach we are leveraging slice and with every indexOf match we will move our way through the string and eliminate the previous searched potions. Each time we call indexOf the size of the string it searches will be smaller.

function countChar (char: string, search: string): number {
  
  let num: number = 0;
  let str: string = search;
  let pos: number = str.indexOf(char);
  
  while(pos > -1) {
    str = str.slice(pos + 1);
    pos = str.indexOf(char);
    num++;
  }

  return num;

}

// Call the function
countChar('x', 'foo x bar x baz x') // 3

Using with IndexOf from position

Similar to the first approach using slice but instead of augmenting the string we are searching it will leverage the from parameter in indexOf method.

function countChar (char: string, str: string): number {
  
  let num: number = 0;
  let pos: number = str.indexOf(char);
  
  while(pos > -1) {
    pos = str.indexOf(char, pos + 1);
    num++;
  }

  return num;

}

// Call the function
countChar('x', 'foo x bar x baz x') // 3

Personally, I go for the second approach over the first, but both are fine and performant when dealing with large strings but also smaller sized ones too.

The following uses a regular expression to test the length. testex ensures you don't have 16 or greater consecutive non-comma characters. If it passes the test, then it proceeds to split the string. counting the commas is as simple as counting the tokens minus one.

var mainStr = "str1,str2,str3,str4";
var testregex = /([^,]{16,})/g;
if (testregex.test(mainStr)) {
  alert("values must be separated by commas and each may not exceed 15 characters");
} else {
  var strs = mainStr.split(',');
  alert("mainStr contains " + strs.length + " substrings separated by commas.");
  alert("mainStr contains " + (strs.length-1) + " commas.");
}

And there is:

function character_count(string, char, ptr = 0, count = 0) {
    while (ptr = string.indexOf(char, ptr) + 1) {count ++}
    return count
}

Works with integers too!

var mainStr = "str1,str2,str3,str4";
var splitStr = mainStr.split(",").length - 1; // subtracting 1 is important!
alert(splitStr);

Splitting into an array gives us a number of elements, which will always be 1 more than the number of instances of the character. This may not be the most memory efficient, but if your input is always going to be small, this is a straight-forward and easy to understand way to do it.

If you need to parse very large strings (greater than a few hundred characters), or if this is in a core loop that processes large volumes of data, I would recommend a different strategy.

The fifth method in Leo Sauers answer fails, if the character is on the beginning of the string. e.g.

var needle ='A',
  haystack = 'AbcAbcAbc';

haystack.split('').map( function(e,i){ if(e === needle) return i;} )
  .filter(Boolean).length;

will give 2 instead of 3, because the filter funtion Boolean gives false for 0.

Other possible filter function:

haystack.split('').map(function (e, i) {
  if (e === needle) return i;
}).filter(function (item) {
  return !isNaN(item);
}).length;

one more answer:

function count(string){
  const count={}
  
  string.split('').forEach(char=>{
    count[char] = count[char] ? (count[char]+1) : 1;
  })
  
  return count
}

console.log(count("abfsdfsddsfdfdsfdsfdsfda"))

I know this might be an old question but I have a simple solution for low-level beginners in JavaScript.

As a beginner, I could only understand some of the solutions to this question so I used two nested FOR loops to check each character against every other character in the string, incrementing a count variable for each character found that equals that character.

I created a new blank object where each property key is a character and the value is how many times each character appeared in the string(count).

Example function:-

function countAllCharacters(str) {
  var obj = {};
  if(str.length!==0){
    for(i=0;i<str.length;i++){
      var count = 0;
      for(j=0;j<str.length;j++){
        if(str[i] === str[j]){
          count++;
        }
      }
      if(!obj.hasOwnProperty(str[i])){
        obj[str[i]] = count;
      }
    }
  }
  return obj;
}

I believe you will find the below solution to be very short, very fast, able to work with very long strings, able to support multiple character searches, error proof, and able to handle empty string searches.

function substring_count(source_str, search_str, index) {
    source_str += "", search_str += "";
    var count = -1, index_inc = Math.max(search_str.length, 1);
    index = (+index || 0) - index_inc;
    do {
        ++count;
        index = source_str.indexOf(search_str, index + index_inc);
    } while (~index);
    return count;
}

Example usage:

console.log(substring_count("Lorem ipsum dolar un sit amet.", "m "))

function substring_count(source_str, search_str, index) {
    source_str += "", search_str += "";
    var count = -1, index_inc = Math.max(search_str.length, 1);
    index = (+index || 0) - index_inc;
    do {
        ++count;
        index = source_str.indexOf(search_str, index + index_inc);
    } while (~index);
    return count;
}

The above code fixes the major performance bug in Jakub Wawszczyk's that the code keeps on looks for a match even after indexOf says there is none and his version itself is not working because he forgot to give the function input parameters.

My solution with ramda js:

const testString = 'somestringtotest'

const countLetters = R.compose(
  R.map(R.length),
  R.groupBy(R.identity),
  R.split('')
)

countLetters(testString)

Link to REPL.

The function takes string str as parameter and counts occurrence of each unique characters in the string. The result comes in key - value pair for each character.

var charFoundMap = {};//object defined
    for (var i = 0; i < str.length; i++) {

       if(!charFoundMap[ str[i] ])  {
        charFoundMap[ str[i] ]=1;
       } 
       else
       charFoundMap[ str[i] ] +=1;
       //if object does not contain this 
    }
    return charFoundMap;

} 

let str = "aabgrhaab"
let charMap = {}

for(let char of text) {
   if(charMap.hasOwnProperty(char)){
      charMap[char]++
   } else {
     charMap[char] = 1
   }
}

console.log(charMap); //{a: 4, b: 2, g: 1, r: 1, h: 1}

There is a very tricky way, but it is in reverse:

const sampleStringText = "/john/dashboard/language";

Assume the above sample, for counting the number of forward-slashs you can do like this:

console.log( sampleStringText.split('/') - 1 );

So I recommended to use a function for it (TypeScript):

const counter = (sentence: string, char: string): number => sentence.split(char) - 1;

This below is the simplest logic, which is very easy to understand

  //Demo string with repeat char 
  let str = "Coffee"
  //Splitted the str into an char array for looping
  let strArr = str.split("")
  //This below is the final object which holds the result
  let obj = {};
  //This loop will count char (You can also use traditional one for loop)
  strArr.forEach((value,index)=>{
      //If the char exists in the object it will simple increase its value
      if(obj[value] != undefined)
      {
          obj[value] = parseInt(obj[value]) + 1;
      }//else it will add the new one with initializing 1
      else{
          obj[value] =1;
      }      
  });

  console.log("Char with Count:",JSON.stringify(obj)); //Char with Count:{"C":1,"o":1,"f":2,"e":2}