Count the number of times a same value appears in a javascript array

Question

I would like to know if there is a native javascript code that does the same thing as this:

function f(array,value){
    var n = 0;
    for(i = 0; i < array.length; i++){
        if(array[i] == value){n++}
    }
    return n;
}

In my honest opinion the for loop you have is a lot cleaner than using reduce. There's no single built-in method to do it, the closest you can get is built-in methods that'll loop the array for you. — SeinopSys
– SeinopSys, Commented May 21, 2016 at 16:41
The function and the loop you provided does what you want it to do. What is the reason you are looking for another method? Is it for performance reasons? Do you have a large array and want to avoid going through the whole array? Knowing what your problem is helps with the answer. — Saeed D.
– Saeed D., Commented May 21, 2016 at 17:55
@Nina, You can also use indexOf to find and count elements. See my working code below. But I like RomanPerekhrest answer and use of filter. — Saeed D.
– Saeed D., Commented May 21, 2016 at 18:21
@SaeedD., yes, but it has two loops, one while and one indexof loop. — Nina Scholz
– Nina Scholz, Commented May 21, 2016 at 18:24

RomanPerekhrest · Accepted Answer · 2023-01-27 14:01:51Z

85

There might be different approaches for such purpose.
And your approach with for loop is obviously not misplaced(except that it looks redundantly by amount of code).
Here are some additional approaches to get the occurrence of a certain value in array:

Using Array.forEach method:

  var arr = [2, 3, 1, 3, 4, 5, 3, 1];

  function getOccurrence(array, value) {
      var count = 0;
      array.forEach((v) => (v === value && count++));
      return count;
  }

  console.log(getOccurrence(arr, 1));  // 2
  console.log(getOccurrence(arr, 3));  // 3

Using Array.filter method:

  function getOccurrence(array, value) {
      return array.filter((v) => (v === value)).length;
  }

  console.log(getOccurrence(arr, 1));  // 2
  console.log(getOccurrence(arr, 3));  // 3

edited Jan 27, 2023 at 14:01

answered May 21, 2016 at 17:58

RomanPerekhrest

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Comments

Nermin · Accepted Answer · 2024-12-02 13:06:45Z

50

Another option is to use Array.filter():

count = myArray.filter(x => x === searchValue).length;

edited Dec 2, 2024 at 13:06

Nermin

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answered Jan 3, 2020 at 5:32

Edgy

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3 Comments

Brettski Over a year ago

I like the simplicity. I suggest using a === comparator though.

Reza Nikfal Over a year ago

It does not work because filter returns an array with just one of the similar elements. So, the length is always one.

Ron Zano Over a year ago

@RezaNikfal That's not true, it works perfectly fine.

omarjmh · Accepted Answer · 2016-05-21 16:36:41Z

33

You could use reduce to get there:

Working example

var a = [1,2,3,1,2,3,4];

var map = a.reduce(function(obj, b) {
  obj[b] = ++obj[b] || 1;
  return obj;
}, {});

answered May 21, 2016 at 16:36

omarjmh

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2 Comments

Redu Over a year ago

My vote goes to this one since I guess this is exactly the fundamental answer to all questions starting with "give me a solution to find the number of times each... bla bla.

ChazUK Over a year ago

This is definitely the best answer here. concise and gives you a data structure which can be manipulated easier.

JJ Cha · Accepted Answer · 2019-10-11 06:28:48Z

9

You can also use forEach

let countObj = {};
let arr = [1,2,3,1,2,3,4];

let countFunc = keys => {
  countObj[keys] = ++countObj[keys] || 1;
}

arr.forEach(countFunc);

// {1: 2, 2: 2, 3: 2, 4: 1}

answered Oct 11, 2019 at 6:28

JJ Cha

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Comments

Maira Diaz Marraffini · Accepted Answer · 2020-07-03 20:09:57Z

9

Here is my solution without using an additional object and using reduce:

const occurrencesOf = (number,numbers) => numbers.reduce((counter, currentNumber)=> (number === currentNumber ? counter+1 : counter),0);
occurrencesOf(1, [1,2,3,4,5,1,1]) // returns 3
occurrencesOf(6, [1,2,3,4,5,1,1]) // returns 0
occurrencesOf(5, [1,2,3,4,5,1,1]) // returns 1

edited Jul 3, 2020 at 20:09

answered Jul 3, 2020 at 20:03

Maira Diaz Marraffini

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Comments

David Ouma · Accepted Answer · 2022-06-08 08:29:04Z

3

You could use the Array filter method and find the length of the new array like this

const count = (arr, value) => arr.filter(val => val === value).length

edited Jun 8, 2022 at 8:29

answered Jun 8, 2022 at 8:26

David Ouma

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1 Comment

Yunnosch Over a year ago

Can you make the difference to stackoverflow.com/a/59573631/7733418 more obvious? I do not get it and you want to avoid the unfavorable impression that might cause.

Jaint Rai · Accepted Answer · 2020-11-09 06:12:20Z

2

let ar = [2,2,3,1,4,9,5,2,1,3,4,4,8,5];

const countSameNumber = (ar: any, findNumber: any) => {
    let count = 0;
    ar.map((value: any) => {
        if(value === findNumber) {
            count = count + 1;
        }
    })
    return count;
}


let set = new Set(ar);
for (let entry of set) {
    console.log(entry+":", countSameNumber(ar, entry));
}

answered Nov 9, 2020 at 6:12

Jaint Rai

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1 Comment

MaxV Over a year ago

Hi Rai. Thanks for your answer. Usually answers with an explanation are more welcomed there. Would you like to add an explanation to your answer?

Steve.g · Accepted Answer · 2025-10-21 11:25:41Z

2

const arr = ["a", "a", "a", "b", "b", "b", "b", "c", "c", "c"];

function countValues(array, countItem) {
  let count = 0;
  array.forEach(itm => {
    if (itm === countItem) count++;
  });
  console.log(`${countItem} ${count}`);
}

countValues(arr, "c");

edited Oct 21 at 11:25

answered Aug 5, 2020 at 8:32

Steve.g

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Comments

Sandra · Accepted Answer · 2022-08-11 03:01:09Z

1

This is how I did mine using just a for loop. It's not as sophisticated as some of the answers above but it worked for me.

function getNumOfTimes(arrayOfNums){
    let found = {}
        for (let i = 0; i < arrayOfNums.length; i++) {
            let keys = arrayOfNums[i].toString()
            found[keys] = ++found[arrayOfNums[i]] || 1
        }

    return found
}

getNumOfTimes([1, 4, 4, 4, 5, 3, 3, 3])

// { '1': 1, '3': 3, '4': 3, '5': 1 }

edited Aug 11, 2022 at 3:01

answered Aug 4, 2022 at 13:34

Sandra

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2 Comments

Donald Duck is with Ukraine Over a year ago

I assume units should be arrayOfNums.

Sandra Over a year ago

@DonaldDuck, yes you're right, thanks for spotting. Have fixed accordingly.

Saeed D. · Accepted Answer · 2016-05-21 17:43:09Z

0

You may want to use indexOf() function to find and count each value in array

function g(array,value){
  var n = -1;
  var i = -1;
  do {
    n++;
    i = array.indexOf(value, i+1);
  } while (i >= 0  );

  return n;
}

answered May 21, 2016 at 17:43

Saeed D.

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Comments

Donald Duck is with Ukraine · Accepted Answer · 2022-04-09 11:52:31Z

0

let countValue = 0;

Array.forEach((word) => {
  if (word === searchValue) {
    return countValue ++;
  }
});
console.log(`The number of word 'asdf': ${countValue}`);

I used this code to count the number of a given word in a text which was previously converted into an array.

edited Apr 9, 2022 at 11:52

Donald Duck is with Ukraine

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answered Apr 8, 2022 at 18:08

Panda

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Comments

Mr.com · Accepted Answer · 2024-02-17 15:58:12Z

-1

var obj = [];
var array = ['a', 'a', 'a', 'b', 'c', 'c', 'd', 'd', 'd', 'd', 'd', 'd'];

array.forEach(element => {
  let count = 1;
  array.forEach(loopo => {
    if(element == loopo)
     {
       obj[loopo] = count++;
     }
  })
});
console.log(obj);

answered Feb 17, 2024 at 15:58

Mr.com

1

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