Counting frequency of characters in a string using JavaScript

Question

I need to write some kind of loop that can count the frequency of each letter in a string.

For example: "aabsssd"

output: a:2, b:1, s:3, d:1

Also want to map same character as property name in object. Any good idea how to do this?

I am not sure how to do it.

This is where I am so far:

var arr = ["a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z"];

function counter(x) {
  var count = 0,
    temp = [];
  x = x.split('');
  console.log(x);
  for (var i = 0, len = x.length; i < len; i++) {
    if (x[i] == "a") {
      count++;
    }
  }
  return count;
}
var a = "aabbddd";
console.log(counter(a));

Answer 1

Here you go:

function getFrequency(string) {
    var freq = {};
    for (var i=0; i<string.length;i++) {
        var character = string.charAt(i);
        if (freq[character]) {
           freq[character]++;
        } else {
           freq[character] = 1;
        }
    }

    return freq;
};

Answer 2

some ES6 syntax with reduce:

let counter = str => {
  return str.split('').reduce((total, letter) => {
    total[letter] ? total[letter]++ : total[letter] = 1;
    return total;
  }, {});
};

counter("aabsssd"); // => { a: 2, b: 1, s: 3, d: 1 }

Answer 3

Another solution:

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

Answer 4

With some ES6 features and short-circuiting:

const counter = s => [...s].reduce((a, c) => (a[c] = ++a[c] || 1) && a, {})

console.log(
  counter("hello") // {h: 1, e: 1, l: 2, o: 1}
)  

Answer 5

Here's another way:

const freqMap = s => [...s].reduce((freq,c) => {freq[c] = -~freq[c]; return freq} ,{})

Or, if you prefer a "for" loop:

function freqMap(s) { 
   freq={}; 
   for (let c of s) 
      freq[c]=-~freq[c]; 
   return freq;
}

e.g. freqMap("MaMaMia") returns Object{M : 3, a : 3, i : 1}

This method leverages the fact that in javascript, bitwise not on "undefined" gives -1, (whereas "undefined+1" gives NaN). So, -~undefined is 1, -~1 is 2, -~2 is 3 etc.

We can thus iterate over the characters of the string, and simply increment freq[c] without any "if". The first time we encounter a character c, freq[c] will be undefined, so we set it to -~freq[c] which is 1. If we subsequently encounter c again, we again set freq[c] to -~freq[c], which will now be 2, etc.

Simple, elegant, concise.

Answer 6

More declarative way to get a word histogram will be to utilise reduce to iterate through letters and come up with a new object that contains letters as keys and frequencies as values.

function getFrequency(str) {
  return str.split('').reduce( (prev, curr) => {
    prev[curr] = prev[curr] ? prev[curr] + 1 : 1;
    return prev;
  }, {});
};

console.log(getFrequency('test')); // => {t: 2, e: 1, s: 1}

Answer 7

a leaner, functional solution:

using ES6 Arrows && Logical Operators:

const buildFreqDict = string =>
  string.split('').reduce((freqDict, char) => {
    freqDict[char] = (freqDict[char] || 0) + 1;
    return freqDict;
  }, {})

console.log(buildFreqDict("banana"))

Explained

• split string into array of characters.
  • and then feed it into a reduce method (using method.chaining()).
• if char is already logged in countDict then add 1 to it.
  • or if character not found in countDict then set it to 1.
• return new values back up to reduce's accumulator object
• NB: don't forget about including the third argument of .reduce(): in this case it is a {} (object literal) that serves to initialize the freqDict object.

for more info see Counting instances of values in an object half way down the page here: MDN Reduce
and for more info about using logical operators please see here: MDN Logical Operators

Answer 8

An easy way. In addition, its get you an alphabetically sorted list. It loops throught an arrray and evaluate if the character is already in the object: if false, the character is added to the object, if true, its frequency increase a unit.

const text= "Lorem ipsum dolor sit amet consectetur adipiscing"
const textAsArray = text.split('').sort()
let charactersList = {}

for (char of textAsArray) {

   if (!charactersList[char]) {
    charactersList[char]=1;
   }
   else {
    charactersList[char]++
  }
}

console.log(charactersList)   

Answer 9

I have reviewed and I think that this adapts very well to the need they pose. I would like it to be in a single line but I don't know how to generate the object dynamically.

const uniqueCount=(arr)=>{
let rs ={};
arr.sort().join("").match(/(.)(\1*)/g).map(i=>rs[i[0]]=i.length);
return rs;
};
console.log(uniqueCount(["a","b","c","d","d","e","a","b","c","f","g","h","h","h","e","a"]));
//{ a: 3, b: 2, c: 2, d: 2, e: 2, f: 1, g: 1, h: 3 }

I find it very successful to use .match() and regex /(.)(\1*)/g as explained above.

If it is just a string, you just need to add a .split("") before and that's it.

Answer 10

One more version with sorting by alphabetically. This function works for both.

1. Frequency of characters by alphabetically sorted
2. Frequency of characters in order of occurrence

Caveat: Only works if whole string is in lowercase

function freqWithAlphabetTable(str, doNeedToSort) {
    let cnt = new Array(26).fill(0), firstLowerCase = 97, output = {}
    for (let i = 0; i < str.length; i++)
        cnt[str[i].charCodeAt(0) - firstLowerCase]++ // filling the array with count at it's index
    if (doNeedToSort) {
        for (let i = 0; i < cnt.length; i++) {
            if (cnt[i] !== 0)
                output[String.fromCharCode(firstLowerCase)] = cnt[i]
            firstLowerCase++;
        }
    } else {
        for (let i = 0; i < str.length; i++) {
            let letterIndexVal = cnt[str[i].charCodeAt(0) - firstLowerCase];
            if (letterIndexVal != 0 ) {
                output[str[i]] = letterIndexVal
                letterIndexVal = 0 // replacing it with zero to avoid repetition
            }
        }
    }
    console.log(output);
    return output;
}

Answer 11

Here's another option using underscore.js:

function charCount(str) {
    return _(str.split('')).countBy(function(char) {
        return char.toLowerCase();
    });
}

charCount('aaabbbbdd') outputs Object {a: 3, b: 4, d: 2}

Answer 12

for(i = strlen(string)var string = 'aabsssd';
var chars = new Array();
for(var i = 0; i < string.length; i++){
    var char = string.charAt(i);
    if(chars[char] == undefined){
        chars[char] = 0;
    }
    chars[char]++;
}
console.log(chars);

Answer 13

 const recorrences = ['a', 'b', 'c', 'a', 'b','a']
                .map(i => !!~i.indexOf('a'))
                .filter(i => i)
                .length;
console.log(`recorrences ${recorrences}`) 
//recorrences 3

Answer 14

// Count frequency of characters in a string
// input: 'Hello, I'm Paul!'
// result: {
//      H: 1,
//      E: 1,
//      L: 3,
//      ... and so on ...
// }

const countChars = (string) => {
    let charStats = {};
    string = string.replace(' ', '').toUpperCase().split('');

    string.forEach((char) => {
        if (charStats[char]) {
            charStats[char]++;
        } else {
            charStats[char] = 1;
        }
    });

    return charStats;
};

Answer 15

Another Solution

    function maxChar(str) {

        const charMap = {};
        let max = 0;
        let maxChar = '';

        for(let char of str){
            if(charMap[char]){
                charMap[char]++;
            }else{
                charMap[char] = 1;
            }
        }

        for(let char in charMap){
            if(charMap[char] > max){
                max = charMap[char];
                maxChar = char;
            }
        }

        return maxChar; 
}

===>

 maxChar('355385') 
  "5"

Answer 16

var str = 'abcccdddd';

function maxCharCount(target) {
    const chars = {};

    let maxChar = '';
    let maxValue = 1;

    for (let char of target) {
        chars[char] = chars[char] + 1 || 1;
    }

    return chars;
}

console.log(maxCharCount(str));

Answer 17

The same solution but refactored. So cool how we can solve this problem with so many different answers :)

function getFrequency(string) {

    var freq = {};

    for (let character in string) {

         let char = string[character];        
         (freq[char]) ? freq[char]++ : freq[char] = 1

    }

    return freq;

};

Answer 18

You can use this. Just pass the string and it will return object with all the character frequency.

function buildCharMap(string) {
  const charMap = {};
  string.replace(/[^\w]/g, '').toLowerCase();
  for (let char of string) {
    charMap[char] = charMap[char] + 1 || 1;
  }
  return charMap;
}

Answer 19

cheat code to count frequency of a char in a string is

let target = "e";
let string = " i want to see that person that came in here last";
let frequency = string.split(target).length - 1;

or all in one line

console.log(string.split("e").length - 1)

Answer 20

 [...str].map( char => map.get(char) ? map.set( char, map.get(char) + 1) : map.set(char,1) ) 

Answer 21

Everyone using split and reduce are over-complicating things.

string is an iterator so you can use a for/of loop to go over each letter - there's no need to split it into an array so reduce can use it. reduce is very useful for lots of things but it often seems like: "when all you have is a hammer everything looks like a nail". I think its used unnecessarily in many places.

Anyway...

1. Create a new object.
2. Loop over the string.
3. If there is no key in the object that corresponds to the current letter, add it and set it it to zero.
4. Increment it.

function counter(str) {

  // Create an object
  const obj = {};

  // Loop through the string
  for (const letter of str) {

    // If the object doesn't have a `letter`
    // property create one and set it to 0;
    obj[letter] ??= 0;

    // Increment the value
    ++obj[letter];

  }

  // Finally return the object
  return obj;

}

const str = 'aabbddd';
console.log(counter(str));

Additional documentation

• for/of

• Logical nullish assignment