Get the element with the highest occurrence in an array

Question

I'm looking for an elegant way of determining which element has the highest occurrence (mode) in a JavaScript array.

For example, in

['pear', 'apple', 'orange', 'apple']

the 'apple' element is the most frequent one.

Answer 1

This is just the mode. Here's a quick, non-optimized solution. It should be O(n).

function mode(array)
{
    if(array.length == 0)
        return null;
    var modeMap = {};
    var maxEl = array[0], maxCount = 1;
    for(var i = 0; i < array.length; i++)
    {
        var el = array[i];
        if(modeMap[el] == null)
            modeMap[el] = 1;
        else
            modeMap[el]++;  
        if(modeMap[el] > maxCount)
        {
            maxEl = el;
            maxCount = modeMap[el];
        }
    }
    return maxEl;
}

Answer 2

There have been some developments in javascript since 2009 - I thought I'd add another option. I'm less concerned with efficiency until it's actually a problem so my definition of "elegant" code (as stipulated by the OP) favours readability - which is of course subjective...

function mode(arr){
    return arr.sort((a,b) =>
          arr.filter(v => v===a).length
        - arr.filter(v => v===b).length
    ).pop();
}

mode(['pear', 'apple', 'orange', 'apple']); // apple

In this particular example, should two or more elements of the set have equal occurrences then the one that appears latest in the array will be returned. It's also worth pointing out that it will modify your original array - which can be prevented if you wish with an Array.slice call beforehand.

---

Edit: updated the example with some ES6 fat arrows because 2015 happened and I think they look pretty... If you are concerned with backwards compatibility you can find this in the revision history.

Answer 3

As per George Jempty's request to have the algorithm account for ties, I propose a modified version of Matthew Flaschen's algorithm.

function modeString(array) {
  if (array.length == 0) return null;

  var modeMap = {},
    maxEl = array[0],
    maxCount = 1;

  for (var i = 0; i < array.length; i++) {
    var el = array[i];

    if (modeMap[el] == null) modeMap[el] = 1;
    else modeMap[el]++;

    if (modeMap[el] > maxCount) {
      maxEl = el;
      maxCount = modeMap[el];
    } else if (modeMap[el] == maxCount) {
      maxEl += "&" + el;
      maxCount = modeMap[el];
    }
  }
  return maxEl;
}

This will now return a string with the mode element(s) delimited by a & symbol. When the result is received it can be split on that & element and you have your mode(s).

Another option would be to return an array of mode element(s) like so:

function modeArray(array) {
  if (array.length == 0) return null;
  var modeMap = {},
    maxCount = 1,
    modes = [];

  for (var i = 0; i < array.length; i++) {
    var el = array[i];

    if (modeMap[el] == null) modeMap[el] = 1;
    else modeMap[el]++;

    if (modeMap[el] > maxCount) {
      modes = [el];
      maxCount = modeMap[el];
    } else if (modeMap[el] == maxCount) {
      modes.push(el);
      maxCount = modeMap[el];
    }
  }
  return modes;
}

In the above example you would then be able to handle the result of the function as an array of modes.

Answer 4

Based on Emissary's ES6+ answer, you could use Array.prototype.reduce to do your comparison (as opposed to sorting, popping and potentially mutating your array), which I think looks quite slick.

const mode = (myArray) =>
  myArray.reduce(
    (a,b,i,arr)=>
     (arr.filter(v=>v===a).length>=arr.filter(v=>v===b).length?a:b),
    null)

I'm defaulting to null, which won't always give you a truthful response if null is a possible option you're filtering for, maybe that could be an optional second argument

The downside, as with various other solutions, is that it doesn't handle 'draw states', but this could still be achieved with a slightly more involved reduce function.

Answer 5

a=['pear', 'apple', 'orange', 'apple'];
b={};
max='', maxi=0;
for(let k of a) {
  if(b[k]) b[k]++; else b[k]=1;
  if(maxi < b[k]) { max=k; maxi=b[k] }
}

Answer 6

As I'm using this function as a quiz for the interviewers, I post my solution:

const highest = arr => (arr || []).reduce( ( acc, el ) => {
  acc.k[el] = acc.k[el] ? acc.k[el] + 1 : 1
  acc.max = acc.max ? acc.max < acc.k[el] ? el : acc.max : el
  return acc  
}, { k:{} }).max

const test = [0,1,2,3,4,2,3,1,0,3,2,2,2,3,3,2]
console.log(highest(test))

Answer 7

Trying out a declarative approach here. This solution builds an object to tally up the occurrences of each word. Then filters the object down to an array by comparing the total occurrences of each word to the highest value found in the object.

const arr = ['hello', 'world', 'hello', 'again'];

const tally = (acc, x) => { 

  if (! acc[x]) { 
    acc[x] = 1;
    return acc;
  } 

  acc[x] += 1;
  return acc;
};

const totals = arr.reduce(tally, {});

const keys = Object.keys(totals);

const values = keys.map(x => totals[x]);

const results = keys.filter(x => totals[x] === Math.max(...values));

Answer 8

This solution has O(n) complexity:

function findhighestOccurenceAndNum(a) {
  let obj = {};
  let maxNum, maxVal;
  for (let v of a) {
    obj[v] = ++obj[v] || 1;
    if (maxVal === undefined || obj[v] > maxVal) {
      maxNum = v;
      maxVal = obj[v];
    }
  }
  console.log(maxNum + ' has max value = ' + maxVal);
}

findhighestOccurenceAndNum(['pear', 'apple', 'orange', 'apple']);

Answer 9

Here’s the modern version using built-in maps (so it works on more than things that can be converted to unique strings):

'use strict';

const histogram = iterable => {
    const result = new Map();

    for (const x of iterable) {
        result.set(x, (result.get(x) || 0) + 1);
    }

    return result;
};

const mostCommon = iterable => {
    let maxCount = 0;
    let maxKey;

    for (const [key, count] of histogram(iterable)) {
        if (count > maxCount) {
            maxCount = count;
            maxKey = key;
        }
    }

    return maxKey;
};

console.log(mostCommon(['pear', 'apple', 'orange', 'apple']));

Answer 10

For the sake of really easy to read, maintainable code I share this:

function getMaxOcurrences(arr = []) {
  let item = arr[0];
  let ocurrencesMap = {};

  for (let i in arr) {
    const current = arr[i];

    if (ocurrencesMap[current]) ocurrencesMap[current]++;
    else ocurrencesMap[current] = 1;

    if (ocurrencesMap[item] < ocurrencesMap[current]) item = current;
  }

  return { 
    item: item, 
    ocurrences: ocurrencesMap[item]
  };
}

Hope it helps someone ;)!

Answer 11

Time for another solution:

function getMaxOccurrence(arr) {
    var o = {}, maxCount = 0, maxValue, m;
    for (var i=0, iLen=arr.length; i<iLen; i++) {
        m = arr[i];

        if (!o.hasOwnProperty(m)) {
            o[m] = 0;
        }
        ++o[m];

        if (o[m] > maxCount) {
            maxCount = o[m];
            maxValue = m;
        }
    }
    return maxValue;
}

If brevity matters (it doesn't), then:

function getMaxOccurrence(a) {
    var o = {}, mC = 0, mV, m;
    for (var i=0, iL=a.length; i<iL; i++) {
        m = a[i];
        o.hasOwnProperty(m)? ++o[m] : o[m] = 1;
        if (o[m] > mC) mC = o[m], mV = m;
    }
    return mV;
}

If non–existent members are to be avoided (e.g. sparse array), an additional hasOwnProperty test is required:

function getMaxOccurrence(a) {
    var o = {}, mC = 0, mV, m;
    for (var i=0, iL=a.length; i<iL; i++) {
        if (a.hasOwnProperty(i)) {
            m = a[i];
            o.hasOwnProperty(m)? ++o[m] : o[m] = 1;
            if (o[m] > mC) mC = o[m], mV = m;
        }
    }
    return mV;
}

getMaxOccurrence([,,,,,1,1]); // 1

Other answers here will return undefined.

Answer 12

Here is another ES6 way of doing it with O(n) complexity

const result = Object.entries(
    ['pear', 'apple', 'orange', 'apple'].reduce((previous, current) => {
        if (previous[current] === undefined) previous[current] = 1;
        else previous[current]++;
        return previous;
    }, {})).reduce((previous, current) => (current[1] >= previous[1] ? current : previous))[0];
console.log("Max value : " + result);

Answer 13

This solution can return multiple elements of an array in case of a tie. For example, an array

arr = [ 3, 4, 3, 6, 4, ];

has two mode values: 3 and 6.

Here is the solution.

function find_mode(arr) {
    var max = 0;
    var maxarr = [];
    var counter = [];
    var maxarr = [];

    arr.forEach(function(){
       counter.push(0);
    });

    for(var i = 0;i<arr.length;i++){
       for(var j=0;j<arr.length;j++){
            if(arr[i]==arr[j])counter[i]++; 
       }
    } 


    max=this.arrayMax(counter);   
  
    for(var i = 0;i<arr.length;i++){
         if(counter[i]==max)maxarr.push(arr[i]);
    }

    var unique = maxarr.filter( this.onlyUnique );
    return unique;

  };


function arrayMax(arr) {
      var len = arr.length, max = -Infinity;
      while (len--) {
              if (arr[len] > max) {
              max = arr[len];
              }
      }
  return max;
 };

 function onlyUnique(value, index, self) {
       return self.indexOf(value) === index;
 }

Answer 14

function mode(arr){
  return arr.reduce(function(counts,key){
    var curCount = (counts[key+''] || 0) + 1;
    counts[key+''] = curCount;
    if (curCount > counts.max) { counts.max = curCount; counts.mode = key; }
    return counts;
  }, {max:0, mode: null}).mode
}

Answer 15

// O(n)
var arr = [1, 2, 3, 2, 3, 3, 5, 6];
var duplicates = {};
max = '';
maxi = 0;
arr.forEach((el) => {
    duplicates[el] = duplicates[el] + 1 || 1;
  if (maxi < duplicates[el]) {
    max = el;
    maxi = duplicates[el];
  }
});
console.log(max);

Answer 16

Another JS solution from: https://www.w3resource.com/javascript-exercises/javascript-array-exercise-8.php

Can try this too:

let arr =['pear', 'apple', 'orange', 'apple'];

function findMostFrequent(arr) {
  let mf = 1;
  let m = 0;
  let item;

  for (let i = 0; i < arr.length; i++) {
    for (let j = i; j < arr.length; j++) {
      if (arr[i] == arr[j]) {
        m++;
        if (m > mf) {
          mf = m;
          item = arr[i];
        }
      }
    }
    m = 0;
  }

  return item;
}

findMostFrequent(arr); // apple

Answer 17

    const frequence = (array) =>
      array.reduce(
        (acc, item) =>
          array.filter((v) => v === acc).length >=
          array.filter((v) => v === item).length
            ? acc
            : item,
        null
      );

frequence([1, 1, 2])

Answer 18

var array = [1, 3, 6, 6, 6, 6, 7, 7, 12, 12, 17],
    c = {}, // counters
    s = []; // sortable array

for (var i=0; i<array.length; i++) {
    c[array[i]] = c[array[i]] || 0; // initialize
    c[array[i]]++;
} // count occurrences

for (var key in c) {
    s.push([key, c[key]])
} // build sortable array from counters

s.sort(function(a, b) {return b[1]-a[1];});

var firstMode = s[0][0];
console.log(firstMode);

Answer 19

Here is my solution to this problem but with numbers and using the new 'Set' feature. Its not very performant but i definitely had a lot of fun writing this and it does support multiple maximum values.

const mode = (arr) => [...new Set(arr)]
  .map((value) => [value, arr.filter((v) => v === value).length])
  .sort((a,b) => a[1]-b[1])
  .reverse()
  .filter((value, i, a) => a.indexOf(value) === i)
  .filter((v, i, a) => v[1] === a[0][1])
  .map((v) => v[0])

mode([1,2,3,3]) // [3]
mode([1,1,1,1,2,2,2,2,3,3,3]) // [1,2]

By the way do not use this for production this is just an illustration of how you can solve it with ES6 and Array functions only.

Answer 20

const mode = (str) => {
  return str
    .split(' ')
    .reduce((data, key) => {
      let counter = data.map[key] + 1 || 1
      data.map[key] = counter

      if (counter > data.counter) {
        data.counter = counter
        data.mode = key
      }

      return data
    }, {
      counter: 0,
      mode: null,
      map: {}
    })
    .mode
}

console.log(mode('the t-rex is the greatest of them all'))

Answer 21

Here is my solution :-

function frequent(number){
    var count = 0;
    var sortedNumber = number.sort();
    var start = number[0], item;
    for(var i = 0 ;  i < sortedNumber.length; i++){
      if(start === sortedNumber[i] || sortedNumber[i] === sortedNumber[i+1]){
         item = sortedNumber[i]
      }
    }
    return item
  
}

   console.log( frequent(['pear', 'apple', 'orange', 'apple']))

Answer 22

Try it too, this does not take in account browser version.

function mode(arr){
var a = [],b = 0,occurrence;
    for(var i = 0; i < arr.length;i++){
    if(a[arr[i]] != undefined){
        a[arr[i]]++;
    }else{
        a[arr[i]] = 1;
    }
    }
    for(var key in a){
    if(a[key] > b){
        b = a[key];
        occurrence = key;
    }
    }
return occurrence;
}
alert(mode(['segunda','terça','terca','segunda','terça','segunda']));

Please note that this function returns latest occurence in the array when 2 or more entries appear same number of times!

Answer 23

With ES6, you can chain the method like this:

    function findMostFrequent(arr) {
      return arr
        .reduce((acc, cur, ind, arr) => {
          if (arr.indexOf(cur) === ind) {
            return [...acc, [cur, 1]];
          } else {
            acc[acc.indexOf(acc.find(e => e[0] === cur))] = [
              cur,
              acc[acc.indexOf(acc.find(e => e[0] === cur))][1] + 1
            ];
            return acc;
          }
        }, [])
        .sort((a, b) => b[1] - a[1])
        .filter((cur, ind, arr) => cur[1] === arr[0][1])
        .map(cur => cur[0]);
    }
    
    console.log(findMostFrequent(['pear', 'apple', 'orange', 'apple']));
    console.log(findMostFrequent(['pear', 'apple', 'orange', 'apple', 'pear']));

If two elements have the same occurrence, it will return both of them. And it works with any type of element.

Answer 24

I came up with a shorter solution, but it's using lodash. Works with any data, not just strings. For objects can be used:

const mostFrequent = _.maxBy(Object.values(_.groupBy(inputArr, el => el.someUniqueProp)), arr => arr.length)[0];

This is for strings:

const mostFrequent = _.maxBy(Object.values(_.groupBy(inputArr, el => el)), arr => arr.length)[0];

Just grouping data under a certain criteria, then finding the largest group.

Answer 25

Here is my way to do it so just using .filter.

var arr = ['pear', 'apple', 'orange', 'apple'];

function dup(arrr) {
    let max = { item: 0, count: 0 };
    for (let i = 0; i < arrr.length; i++) {
        let arrOccurences = arrr.filter(item => { return item === arrr[i] }).length;
        if (arrOccurences > max.count) {
            max = { item: arrr[i], count: arrr.filter(item => { return item === arrr[i] }).length };
        }
    }
    return max.item;
}
console.log(dup(arr));

Answer 26

Easy solution !

function mostFrequentElement(arr) {
    let res = [];
    for (let x of arr) {
        let count = 0;
        for (let i of arr) {
            if (i == x) {
                count++;
            }
        }
        res.push(count);
    }
    return arr[res.indexOf(Math.max(...res))];
}
array = [13 , 2 , 1 , 2 , 10 , 2 , 1 , 1 , 2 , 2];
let frequentElement = mostFrequentElement(array);
console.log(`The frequent element in ${array} is ${frequentElement}`);

Loop on all element and collect the Count of each element in the array that is the idea of the solution

Answer 27

Here is my solution :-

 const arr = [
2, 1, 10, 7, 10, 3, 10, 8, 7, 3, 10, 5, 4, 6, 7, 9, 2, 2, 2, 6, 3, 7, 6, 9, 8,
9, 10, 8, 8, 8, 4, 1, 9, 3, 4, 5, 8, 1, 9, 3, 2, 8, 1, 9, 6, 3, 9, 2, 3, 5, 3,
2, 7, 2, 5, 4, 5, 5, 8, 4, 6, 3, 9, 2, 3, 3, 10, 3, 3, 1, 4, 5, 4, 1, 5, 9, 6,
2, 3, 10, 9, 4, 3, 4, 5, 7, 2, 7, 2, 9, 8, 1, 8, 3, 3, 3, 3, 1, 1, 3,
];

function max(arr) {
let newObj = {};

arr.forEach((d, i) => {
    if (newObj[d] != undefined) {
        ++newObj[d];
    } else {
        newObj[d] = 0;
    }
});
let nwres = {};
for (let maxItem in newObj) {
    if (newObj[maxItem] == Math.max(...Object.values(newObj))) {
        nwres[maxItem] = newObj[maxItem];
    }
}
return nwres;
}


console.log(max(arr));

Answer 28

var mode = 0;
var c = 0;
var num = new Array();
var value = 0;
var greatest = 0;
var ct = 0;

Note: ct is the length of the array.

function getMode()
{
    for (var i = 0; i < ct; i++)
    {
        value = num[i];
        if (i != ct)
        {
            while (value == num[i + 1])
            {
                c = c + 1;
                i = i + 1;
            }
        }
        if (c > greatest)
        {
            greatest = c;
            mode = value;
        }
        c = 0;
    }
}

Answer 29

I guess you have two approaches. Both of which have advantages.

Sort then Count or Loop through and use a hash table to do the counting for you.

The hashtable is nice because once you are done processing you also have all the distinct elements. If you had millions of items though, the hash table could end up using a lot of memory if the duplication rate is low. The sort, then count approach would have a much more controllable memory footprint.

Answer 30

You can try this:

 // using splice()   
 // get the element with the highest occurence in an array
    function mc(a) {
      var us = [], l;
      // find all the unique elements in the array
      a.forEach(function (v) {
        if (us.indexOf(v) === -1) {
          us.push(v);
        }
      });
      l = us.length;
      while (true) {
        for (var i = 0; i < l; i ++) {
          if (a.indexOf(us[i]) === -1) {
            continue;
          } else if (a.indexOf(us[i]) != -1 && a.length > 1) {
            // just delete it once at a time
            a.splice(a.indexOf(us[i]), 1);
          } else {
            // default to last one
            return a[0];
          }
        }
      }
    }

// using string.match method
function su(a) {
    var s = a.join(),
            uelms = [],
            r = {},
            l,
            i,
            m;

    a.forEach(function (v) {
        if (uelms.indexOf(v) === -1) {
            uelms.push(v);
        }
    });

    l = uelms.length;

    // use match to calculate occurance times
    for (i = 0; i < l; i ++) {
        r[uelms[i]] = s.match(new RegExp(uelms[i], 'g')).length;
    }

    m = uelms[0];
    for (var p in r) {
        if (r[p] > r[m]) {
            m = p;
        } else {
            continue;
        }
    }

    return m;
}

Answer 31

You could solve it in O(n) complexity

var arr = [1,3,54,56,6,6,1,6];
var obj = {};

/* first convert the array in to object with unique elements and number of times each element is repeated */
for(var i = 0; i < arr.length; i++)
{
   var x = arr[i];
   if(!obj[x])
     obj[x] = 1;
   else 
     obj[x]++;
}

console.log(obj);//just for reference

/* now traverse the object to get the element */
var index = 0;
var max = 0;

for(var obIndex in obj)
{
  if(obj[obIndex] > max)
  {
    max = obj[obIndex];
    index = obIndex;
  }
}
console.log(index+" got maximum time repeated, with "+ max +" times" );

Just copy and paste in chrome console to run the above code.

Answer 32

This function is generic function for every type of info. It counts the occurrence of the elements and then returns array with maximum occurring elements.

function mode () {
  var arr = [].slice.call(arguments);
  if ((args.length == 1) && (typeof args[0] === "object")) {
    args = args[0].mode();
  }

  var obj = {};
  for(var i = 0; i < arr.length; i++) {
    if(obj[arr[i]] === undefined) obj[arr[i]] = 1;
    else obj[arr[i]]++;
  }

  var max = 0;
  for (w in obj) {
    if (obj[w] > max) max = obj[w];
  }

  ret_val = [];
  for (w in obj) {
    if (obj[w] == max) ret_val.push(w);
  }

  return ret_val;
}

Answer 33

function mode(){
  var input = $("input").val().split(",");
  var mode = [];
  var m = [];
  var p = [];
    for(var x = 0;x< input.length;x++){
      if(m.indexOf(input[x])==-1){
        m[m.length]=input[x];
    }}
  for(var x = 0; x< m.length;x++){
    p[x]=0;
    for(var y = 0; y<input.length;y++){
      if(input[y]==m[x]){
      p[x]++; 
 }}}
 for(var x = 0;x< p.length;x++){
   if(p[x] ==(Math.max.apply(null, p))){
     mode.push(m[x]);
 }} 
$("#output").text(mode);}

Answer 34

function mode(array){
    var set = Array.from(new Set(array));
    var counts = set.map(a=>array.filter(b=>b==a).length);
    var indices = counts.map((a,b)=>Math.max(...counts)===a?b:0).filter(b=>b!==0);
    var mode = indices.map(a=>set[a]);
    return mode;
}

Answer 35

Here is my way. I try to group data fist.

const _ = require("underscore")

var test  = [ 1, 1, 2, 1 ];
var groupResult = _.groupBy(test, (e)=> e);

The groupResult should be

{
  1: [1, 1, 1]
  2: [2] 
}

Then find the property which has the longest array

function findMax(groupResult){
   var maxArr = []
   var max;
   for(var item in groupResult){
     if(!max) { 
        max = { value:item, count: groupResult[item].length } ; 
        maxArr.push(max); 
        continue;
     }
     if(max.count < groupResult[item].length){ 
        maxArr = [];
        max = { value:item, count: groupResult[item].length }
        maxArr.push(max)
     } else if(max === groupResult[item].length)
        maxArr.push({ value:item, count: groupResult[item].length })
   }
   return maxArr;
}

The complete code looks like

const _ = require("underscore")

var test  = [ 1, 1, 2, 1 ];
var groupResult= _.groupBy(test, (e)=> e);
console.log(findMax(groupResult)[0].value);

function findMax(groupResult){
   var maxArr = []
   var max;
   for(var item in groupResult){
     if(!max) { 
        max = { value:item, count: groupResult[item].length } ; 
        maxArr.push(max); 
        continue;
     }
     if(max.count < groupResult[item].length){ 
        maxArr = [];
        max = { value:item, count: groupResult[item].length }
        maxArr.push(max)
     } else if(max === groupResult[item].length)
        maxArr.push({ value:item, count: groupResult[item].length })
   }
   return maxArr;
}

Answer 36

var cats = ['Tom','Fluffy','Tom','Bella','Chloe','Tom','Chloe'];
var counts = {};
var compare = 0;
var mostFrequent;
(function(array){
   for(var i = 0, len = array.length; i < len; i++){
       var word = array[i];

       if(counts[word] === undefined){
           counts[word] = 1;
       }else{
           counts[word] = counts[word] + 1;
       }
       if(counts[word] > compare){
             compare = counts[word];
             mostFrequent = cats[i];
       }
    }
  return mostFrequent;
})(cats);

Answer 37

Can try :

var arr = [10,3,4,5,3,4,3,8,3,6,3,5,1];
var temp = {};

for(let i=0;i<arr.length;i++){
    if(temp[arr[i]]==undefined){
       temp[arr[i]]=1;
    }else{
        temp[arr[i]]+=1;
    }
}

var max=0, maxEle;

for(const i in temp){
    if(temp[i]>max){
        max = temp[i];
        maxEle=i;
    }
}

console.log(`most occurred element is ${maxEle} and number of times is ${max}`);`

Answer 38

There are a lot of answers already but just want to share with you what I came up with :) Can't say this solution counts on any edge case but anyway )

const getMostFrequentElement = ( arr ) => {
  const counterSymbolKey = 'counter'
  const mostFrequentSymbolKey = 'mostFrequentKey'

  const result = arr.reduce( ( acc, cur ) => {
    acc[ cur ] = acc[ cur ] ? acc[ cur ] + 1 : 1

    if ( acc[ cur ] > acc[ Symbol.for( counterSymbolKey ) ] ) {
      acc[ Symbol.for( mostFrequentSymbolKey ) ] = cur
      acc[ Symbol.for( counterSymbolKey ) ] = acc[ cur ]
    }

    return acc
  }, {
    [ Symbol.for( mostFrequentSymbolKey ) ]: null,
    [ Symbol.for( counterSymbolKey ) ]: 0
  } )

  return result[ Symbol.for( mostFrequentSymbolKey ) ]
}

Hope it will be helpful for someone )

Answer 39

//const arr = [1, 2, 4, 3, 5, 1, 2, 3, 3];
const arr = ['pear', 'apple', 'orange', 'apple'];

// init max occurance element
let maxOcc = {'element': null, occured: 0};

// to find occurances
const res = arr.reduce((acc, el) => {
    acc[el] = acc[el] ? acc[el]+1 : 1;
    if(acc[el]> maxOcc.occured){
        maxOcc = { 'element': el, occured: acc[el] };
    }
    return acc;
}, {});

console.log(maxOcc);

Answer 40

function getData(arr){
  let obj = {}
  let maxElementCount = 0
  let maxEle = ''
  for(let i = 0 ;i<arr.length;i++){
    if(!obj[arr[i]]){
        obj[arr[i]] = 1
    }else{
        obj[arr[i]] += 1
      if(maxElementCount < obj[arr[i]]){
        maxElementCount = obj[arr[i]]
        maxEle = arr[i]
      }
    }
  }
  console.log(maxElementCount, maxEle)
  return obj
}

You can use this simple method to get max count of element

Answer 41

const data = ['x','y','x','z',5,2,4,5,2,3,2,'x', { x: 1 }, (x) => x];

function getModeData(data) {
  return data.reduce((a,c) => {
    if(typeof a[c] === "undefined") {
      a[c] = 1;
    } else {
      a[c]++;
    }
    if(
      typeof a.mode === "undefined" ||
      (typeof a.mode !== "undefined") && a.mode.occurrences < a[c]
    ) {
      a.mode = {
        elem: c,
        occurrences: a[c]
      }
    }
    return a;
  }, { mode: undefined });
}

const { mode: { elem, occurrences }, ...totals } = getModeData(data);

console.log(`The mode is ${elem} with ${occurrences} occurrences`);

console.log('The totals are:');
console.log(totals)

Answer 42

I saw that a lot of answers had a lot of recursions through the array, which I wanted to avoid. In addition to this, I wanted to have Typescript checking the result and inferring the right type.

So this is my version:

function findhighestOccurence<Type extends string | number>(myArray: Type[]) {
    const countOccorrencies = myArray.reduce<{ [key in Type]: number }>(
        (acc, curr) => ({ ...acc, [curr]: acc[curr] ? acc[curr] + 1 : 1 }),
        {} as { [K in Type]: number }
    )

    return (Object.entries(countOccorrencies) as [Type, number][]).reduce<{
        values: Type[] // there might be multiple "highest occurences" values
        occurrences: number // how many times it has/they have occurred
    }>(
        (acc, [value, occurrences]) => {
            // new highest occurrence
            if (occurrences > acc.occurrences)
                return {
                    ...acc,
                    occurrences: occurrences,
                    values: [value],
                }
            // new value with same highest amount of occurrences
            else if (occurrences === acc.occurrences)
                return { ...acc, values: [...acc.values, value] }
            return acc
        },
        { values: [], occurrences: 0 }
    )
}

Succint version:

function findhighestOccurenceShort<T extends string | number>(myArray: T[]) {
    return (
        Object.entries(
            myArray.reduce<{ [key in T]: number }>(
                (acc, cur) => ({ ...acc, [cur]: acc[cur] ? acc[cur] + 1 : 1 }),
                {} as { [K in T]: number }
            )
        ) as [T, number][]
    ).reduce<{ val: T[]; occ: number }>(
        (acc, [val, occ]) =>
            occ > acc.occ
                ? { occ, val: [val] }
                : occ === acc.occ
                ? { occ, val: [...acc.val, val] }
                : acc,
        { val: [], occ: 0 }
    )
}