Counting the occurrences / frequency of array elements

Question

In Javascript, I'm trying to take an initial array of number values and count the elements inside it. Ideally, the result would be two new arrays, the first specifying each unique element, and the second containing the number of times each element occurs. However, I'm open to suggestions on the format of the output.

For example, if the initial array was:

5, 5, 5, 2, 2, 2, 2, 2, 9, 4

Then two new arrays would be created. The first would contain the name of each unique element:

5, 2, 9, 4

The second would contain the number of times that element occurred in the initial array:

3, 5, 1, 1

Because the number 5 occurs three times in the initial array, the number 2 occurs five times and 9 and 4 both appear once.

I've searched a lot for a solution, but nothing seems to work, and everything I've tried myself has wound up being ridiculously complex. Any help would be appreciated!

Thanks :)

Answer 1

const arr = [2, 2, 5, 2, 2, 2, 4, 5, 5, 9];

function foo (array) {
  let a = [],
    b = [],
    arr = [...array], // clone array so we don't change the original when using .sort()
    prev;

  arr.sort();
  for (let element of arr) {
    if (element !== prev) {
      a.push(element);
      b.push(1);
    }
    else ++b[b.length - 1];
    prev = element;
  }

  return [a, b];
}

const result = foo(arr);
console.log('[' + result[0] + ']','[' + result[1] + ']')
console.log(arr)

Answer 2

You can use an object to hold the results:

const arr = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4];
const counts = {};

for (const num of arr) {
  counts[num] = counts[num] ? counts[num] + 1 : 1;
}

console.log(counts[5], counts[2], counts[9], counts[4]);

So, now your counts object can tell you what the count is for a particular number:

console.log(counts[5]); // logs '3'

If you want to get an array of members, just use the keys() functions

keys(counts); // returns ["5", "2", "9", "4"]

Answer 3

const occurrences = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4].reduce(function (acc, curr) {
  return acc[curr] ? ++acc[curr] : acc[curr] = 1, acc
}, {});

console.log(occurrences) // => {2: 5, 4: 1, 5: 3, 9: 1}

Answer 4

If using underscore or lodash, this is the simplest thing to do:

_.countBy(array);

Such that:

_.countBy([5, 5, 5, 2, 2, 2, 2, 2, 9, 4])
=> Object {2: 5, 4: 1, 5: 3, 9: 1}

As pointed out by others, you can then execute the _.keys() and _.values() functions on the result to get just the unique numbers, and their occurrences, respectively. But in my experience, the original object is much easier to deal with.

Answer 5

One line ES6 solution. So many answers using object as a map but I can't see anyone using an actual Map

const map = arr.reduce((acc, e) => acc.set(e, (acc.get(e) || 0) + 1), new Map());

Use map.keys() to get unique elements

Use map.values() to get the occurrences

Use map.entries() to get the pairs [element, frequency]

var arr = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4]

const map = arr.reduce((acc, e) => acc.set(e, (acc.get(e) || 0) + 1), new Map());

console.info([...map.keys()])
console.info([...map.values()])
console.info([...map.entries()])

Answer 6

Don't use two arrays for the result, use an object:

a      = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4];
result = { };
for(var i = 0; i < a.length; ++i) {
    if(!result[a[i]])
        result[a[i]] = 0;
    ++result[a[i]];
}

Then result will look like:

{
    2: 5,
    4: 1,
    5: 3,
    9: 1
}

Answer 7

How about an ECMAScript2015 option.

const a = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4];

const aCount = new Map([...new Set(a)].map(
    x => [x, a.filter(y => y === x).length]
));

aCount.get(5)  // 3
aCount.get(2)  // 5
aCount.get(9)  // 1
aCount.get(4)  // 1

This example passes the input array to the Set constructor creating a collection of unique values. The spread syntax then expands these values into a new array so we can call map and translate this into a two-dimensional array of [value, count] pairs - i.e. the following structure:

Array [
   [5, 3],
   [2, 5],
   [9, 1],
   [4, 1]
]

The new array is then passed to the Map constructor resulting in an iterable object:

Map {
    5 => 3,
    2 => 5,
    9 => 1,
    4 => 1
}

The great thing about a Map object is that it preserves data-types - that is to say aCount.get(5) will return 3 but aCount.get("5") will return undefined. It also allows for any value / type to act as a key meaning this solution will also work with an array of objects.

function frequencies(/* {Array} */ a){
    return new Map([...new Set(a)].map(
        x => [x, a.filter(y => y === x).length]
    ));
}

let foo = { value: 'foo' },
    bar = { value: 'bar' },
    baz = { value: 'baz' };

let aNumbers = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4],
    aObjects = [foo, bar, foo, foo, baz, bar];

frequencies(aNumbers).forEach((val, key) => console.log(key + ': ' + val));
frequencies(aObjects).forEach((val, key) => console.log(key.value + ': ' + val));

Answer 8

I think this is the simplest way how to count occurrences with same value in array.

var a = [true, false, false, false];
a.filter(function(value){
    return value === false;
}).length

Answer 9

const data = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4]

function count(arr) {
  return arr.reduce((prev, curr) => (prev[curr] = ++prev[curr] || 1, prev), {})
}

console.log(count(data))

Answer 10

If you favour a single liner.

arr.reduce(function(countMap, word) {countMap[word] = ++countMap[word] || 1;return countMap}, {});

Edit (6/12/2015): The Explanation from the inside out. countMap is a map that maps a word with its frequency, which we can see the anonymous function. What reduce does is apply the function with arguments as all the array elements and countMap being passed as the return value of the last function call. The last parameter ({}) is the default value of countMap for the first function call.

Answer 11

ES6 version should be much simplifier (another one line solution)

let arr = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4];
let acc = arr.reduce((acc, val) => acc.set(val, 1 + (acc.get(val) || 0)), new Map());

console.log(acc);
// output: Map { 5 => 3, 2 => 5, 9 => 1, 4 => 1 }

A Map instead of plain Object helping us to distinguish different type of elements, or else all counting are base on strings

Answer 12

2021's version

The more elegant way is using Logical nullish assignment (x ??= y) combined with Array#reduce() with O(n) time complexity.

The main idea is still using Array#reduce() to aggregate with output as object to get the highest performance (both time and space complexity) in terms of searching & construct bunches of intermediate arrays like other answers.

const arr = [2, 2, 2, 2, 2, 4, 5, 5, 5, 9];
const result = arr.reduce((acc, curr) => {
  acc[curr] ??= {[curr]: 0};
  acc[curr][curr]++;
  
  return acc;
}, {});

console.log(Object.values(result));

Clean & Refactor code

Using Comma operator (,) syntax.

The comma operator (,) evaluates each of its operands (from left to right) and returns the value of the last operand.

const arr = [2, 2, 2, 2, 2, 4, 5, 5, 5, 9];
const result = arr.reduce((acc, curr) => (acc[curr] = (acc[curr] || 0) + 1, acc), {});
console.log(result);

Output

{
  "2": 5,
  "4": 1,
  "5": 3,
  "9": 1
}

Answer 13

Based on answer of @adamse and @pmandell (which I upvote), in ES6 you can do it in one line:

• 2017 edit: I use || to reduce code size and make it more readable.

var a=[7,1,7,2,2,7,3,3,3,7,,7,7,7];
alert(JSON.stringify(

a.reduce((r,k)=>{r[k]=1+r[k]||1;return r},{})

));

---

It can be used to count characters:

var s="ABRACADABRA";
alert(JSON.stringify(

s.split('').reduce((a, c)=>{a[c]++?0:a[c]=1;return a},{})

));

Answer 14

Edit 2020: this is a pretty old answer (nine years). Extending the native prototype will always generate discussion. Although I think the programmer is free to choose her own programming style, here's a (more modern) approach to the problem without extending Array.prototype:

{
  // create array with some pseudo random values (1 - 5)
  const arr = Array.from({length: 100})
    .map( () => Math.floor(1 + Math.random() * 5) );
  // frequencies using a reducer
  const arrFrequencies = arr.reduce((acc, value) => 
      ({ ...acc, [value]: acc[value] + 1 || 1}), {} )
  console.log(arrFrequencies);    
  console.log(`Value 4 occurs ${arrFrequencies[4]} times in arrFrequencies`);

  // bonus: restore Array from frequencies
  const arrRestored = Object.entries(arrFrequencies)
    .reduce( (acc, [key, value]) => acc.concat(Array(value).fill(+key)), [] );
  console.log(arrRestored.join());  
}

.as-console-wrapper { top: 0; max-height: 100% !important; }

The old (2011) answer: you could extend Array.prototype, like this:

{
  Array.prototype.frequencies = function() {
    var l = this.length,
      result = {
        all: []
      };
    while (l--) {
      result[this[l]] = result[this[l]] ? ++result[this[l]] : 1;
    }
    // all pairs (label, frequencies) to an array of arrays(2)
    for (var l in result) {
      if (result.hasOwnProperty(l) && l !== 'all') {
        result.all.push([l, result[l]]);
      }
    }
    return result;
  };

  var freqs = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4].frequencies();
  console.log(`freqs[2]: ${freqs[2]}`); //=> 5
  
  // or
  var freqs = '1,1,2,one,one,2,2,22,three,four,five,three,three,five'
    .split(',')
    .frequencies();
    
  console.log(`freqs.three: ${freqs.three}`); //=> 3
  
// Alternatively you can utilize Array.map:

    Array.prototype.frequencies = function() {
      var freqs = {
        sum: 0
      };
      this.map(function(a) {
        if (!(a in this)) {
          this[a] = 1;
        } else {
          this[a] += 1;
        }
        this.sum += 1;
        return a;
      }, freqs);
      return freqs;
    }
}

.as-console-wrapper { top: 0; max-height: 100% !important; }

Answer 15

If you are using underscore you can go the functional route

a = ['foo', 'foo', 'bar'];

var results = _.reduce(a,function(counts,key){ counts[key]++; return counts },
                  _.object( _.map( _.uniq(a), function(key) { return [key, 0] })))

so your first array is

_.keys(results)

and the second array is

_.values(results)

most of this will default to native javascript functions if they are available

demo : http://jsfiddle.net/dAaUU/

Answer 16

So here's how I'd do it with some of the newest javascript features:

First, reduce the array to a Map of the counts:

let countMap = array.reduce(
  (map, value) => {map.set(value, (map.get(value) || 0) + 1); return map}, 
  new Map()
)

By using a Map, your starting array can contain any type of object, and the counts will be correct. Without a Map, some types of objects will give you strange counts. See the Map docs for more info on the differences.

This could also be done with an object if all your values are symbols, numbers, or strings:

let countObject = array.reduce(
  (map, value) => { map[value] = (map[value] || 0) + 1; return map },
  {}
)

Or slightly fancier in a functional way without mutation, using destructuring and object spread syntax:

let countObject = array.reduce(
  (value, {[value]: count = 0, ...rest}) => ({ [value]: count + 1, ...rest }),
  {}
)

At this point, you can use the Map or object for your counts (and the map is directly iterable, unlike an object), or convert it to two arrays.

For the Map:

countMap.forEach((count, value) => console.log(`value: ${value}, count: ${count}`)

let values = countMap.keys()
let counts = countMap.values()

Or for the object:

Object
  .entries(countObject) // convert to array of [key, valueAtKey] pairs
  .forEach(([value, count]) => console.log(`value: ${value}, count: ${count}`)

let values = Object.keys(countObject)
let counts = Object.values(countObject)

Answer 17

var array = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4];

function countDuplicates(obj, num){
  obj[num] = (++obj[num] || 1);
  return obj;
}

var answer = array.reduce(countDuplicates, {});
// answer => {2:5, 4:1, 5:3, 9:1};

If you still want two arrays, then you could use answer like this...

var uniqueNums = Object.keys(answer);
// uniqueNums => ["2", "4", "5", "9"];

var countOfNums = Object.keys(answer).map(key => answer[key]);
// countOfNums => [5, 1, 3, 1];

Or if you want uniqueNums to be numbers

var uniqueNums = Object.keys(answer).map(key => +key);
// uniqueNums => [2, 4, 5, 9];

Answer 18

Here's just something light and easy for the eyes...

function count(a,i){
 var result = 0;
 for(var o in a)
  if(a[o] == i)
   result++;
 return result;
}

Edit: And since you want all the occurences...

function count(a){
 var result = {};
 for(var i in a){
  if(result[a[i]] == undefined) result[a[i]] = 0;
  result[a[i]]++;
 }
 return result;
}

Answer 19

Solution using a map with O(n) time complexity.

var arr = [2, 2, 2, 2, 2, 4, 5, 5, 5, 9];

const countOccurrences = (arr) => {
    const map = {};
    for ( var i = 0; i < arr.length; i++ ) {
        map[arr[i]] = ~~map[arr[i]] + 1;
    }
    return map;
}

Demo: http://jsfiddle.net/simevidas/bnACW/

Answer 20

My solution with ramda:

const testArray = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4]

const counfFrequency = R.compose(
  R.map(R.length),
  R.groupBy(R.identity),
)

counfFrequency(testArray)

Link to REPL.

Answer 21

Using MAP you can have 2 arrays in the output: One containing the occurrences & the other one is containing the number of occurrences.

const dataset = [2,2,4,2,6,4,7,8,5,6,7,10,10,10,15];
let values = [];
let keys = [];

var mapWithOccurences = dataset.reduce((a,c) => {
  if(a.has(c)) a.set(c,a.get(c)+1);
  else a.set(c,1);
  return a;
}, new Map())
.forEach((value, key, map) => {
  keys.push(key);
  values.push(value);
});


console.log(keys)
console.log(values)

Answer 22

Using Lodash

const values = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4];
const frequency = _.map(_.groupBy(values), val => ({ value: val[0], frequency: val.length }));
console.log(frequency);

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.15/lodash.min.js"></script>

Answer 23

I know this question is old but I realized there are too few solutions where you get the count array as asked with a minimal code so here is mine

// The initial array we want to count occurences
var initial = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4];  

// The count array asked for
var count = Array.from(new Set(initial)).map(val => initial.filter(v => v === val).length);  

// Outputs [ 3, 5, 1, 1 ]

Beside you can get the set from that initial array with

var set = Array.from(new Set(initial));  

//set = [5, 2, 9, 4]  

Answer 24

A shorter version use reduce and tilde (~) operator.

const data = [2, 2, 2, 2, 2, 4, 5, 5, 5, 9];

function freq(nums) {
  return nums.reduce((acc, curr) => {
    acc[curr] = -~acc[curr];
    return acc;
  }, {});
}

console.log(freq(data));

Answer 25

Check out the code below.

<html>
<head>
<script>
// array with values
var ar = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4];

var Unique = []; // we'll store a list of unique values in here
var Counts = []; // we'll store the number of occurances in here

for(var i in ar)
{
    var Index = ar[i];
    Unique[Index] = ar[i];
    if(typeof(Counts[Index])=='undefined')  
        Counts[Index]=1;
    else
        Counts[Index]++;
}

// remove empty items
Unique = Unique.filter(function(){ return true});
Counts = Counts.filter(function(){ return true});

alert(ar.join(','));
alert(Unique.join(','));
alert(Counts.join(','));

var a=[];

for(var i=0; i<Unique.length; i++)
{
    a.push(Unique[i] + ':' + Counts[i] + 'x');
}
alert(a.join(', '));

</script>
</head>
<body>

</body>
</html>

Answer 26

There is a much better and easy way that we can do this using ramda.js. Code sample here

const ary = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4]; R.countBy(r=> r)(ary) countBy documentation is at documentation

Answer 27

Given the array supplied below:

const array = [ 'a', 'b', 'b', 'c', 'c', 'c' ];

You can use this simple one-liner to generate a hash map which links a key to the number of times it appears in the array:

const hash = Object.fromEntries([ ...array.reduce((map, key) => map.set(key, (map.get(key) || 0) + 1), new Map()) ]);
// { a: 1, b: 2, c: 3 }

Expanded & Explained:

// first, we use reduce to generate a map with values and the amount of times they appear
const map = array.reduce((map, key) => map.set(key, (map.get(key) || 0) + 1), new Map())

// next, we spread this map into an array
const table = [ ...map ];

// finally, we use Object.fromEntries to generate an object based on this entry table
const result = Object.fromEntries(table);

credit to @corashina for the array.reduce code

Answer 28

Try this:

Array.prototype.getItemCount = function(item) {
    var counts = {};
    for(var i = 0; i< this.length; i++) {
        var num = this[i];
        counts[num] = counts[num] ? counts[num]+1 : 1;
    }
    return counts[item] || 0;
}

Answer 29

I was solving a similar problem on codewars and devised the following solution which worked for me.

This gives the highest count of an integer in an array and also the integer itself. I think it can be applied to string array as well.

To properly sort Strings, remove the function(a, b){return a-b} from inside the sort() portion

function mostFrequentItemCount(collection) {
    collection.sort(function(a, b){return a-b});
    var i=0;
    var ans=[];
    var int_ans=[];
    while(i<collection.length)
    {
        if(collection[i]===collection[i+1])
        {
            int_ans.push(collection[i]);
        }
        else
        {
            int_ans.push(collection[i]);
            ans.push(int_ans);
            int_ans=[];
        }
        i++;
    }

    var high_count=0;
    var high_ans;

    i=0;
    while(i<ans.length)
    {
        if(ans[i].length>high_count)
        {
            high_count=ans[i].length;
            high_ans=ans[i][0];
        }
        i++;
    }
    return high_ans;
}

Answer 30

Here is a way to count occurrences inside an array of objects. It also places the first array's contents inside a new array to sort the values so that the order in the original array is not disrupted. Then a recursive function is used to go through each element and count the quantity property of each object inside the array.

var big_array = [
  { name: "Pineapples", quantity: 3 },
  { name: "Pineapples", quantity: 1 },
  { name: "Bananas", quantity: 1 },
  { name: "Limes", quantity: 1 },
  { name: "Bananas", quantity: 1 },
  { name: "Pineapples", quantity: 2 },
  { name: "Pineapples", quantity: 1 },
  { name: "Bananas", quantity: 1 },
  { name: "Bananas", quantity: 1 },
  { name: "Bananas", quantity: 5 },
  { name: "Coconuts", quantity: 1 },
  { name: "Lemons", quantity: 2 },
  { name: "Oranges", quantity: 1 },
  { name: "Lemons", quantity: 1 },
  { name: "Limes", quantity: 1 },
  { name: "Grapefruit", quantity: 1 },
  { name: "Coconuts", quantity: 5 },
  { name: "Oranges", quantity: 6 }
];

function countThem() {
  var names_array = [];
  for (var i = 0; i < big_array.length; i++) {
    names_array.push( Object.assign({}, big_array[i]) );
  }

  function outerHolder(item_array) {
    if (item_array.length > 0) {
      var occurrences = [];
      var counter = 0;
      var bgarlen = item_array.length;
      item_array.sort(function(a, b) { return (a.name > b.name) ? 1 : ((b.name > a.name) ? -1 : 0); });

      function recursiveCounter() {
        occurrences.push(item_array[0]);
        item_array.splice(0, 1);
        var last_occurrence_element = occurrences.length - 1;
        var last_occurrence_entry = occurrences[last_occurrence_element].name;
        var occur_counter = 0;
        var quantity_counter = 0;
        for (var i = 0; i < occurrences.length; i++) {
          if (occurrences[i].name === last_occurrence_entry) {
            occur_counter = occur_counter + 1;
            if (occur_counter === 1) {
              quantity_counter = occurrences[i].quantity;
            } else {
              quantity_counter = quantity_counter + occurrences[i].quantity;
            }
          }
        }

        if (occur_counter > 1) {
          var current_match = occurrences.length - 2;
          occurrences[current_match].quantity = quantity_counter;
          occurrences.splice(last_occurrence_element, 1);
        }

        counter = counter + 1;

        if (counter < bgarlen) {
          recursiveCounter();
        }
      }

      recursiveCounter();

      return occurrences;
    }
  }
  alert(JSON.stringify(outerHolder(names_array)));
}