sort by number of occurrence(count) in Javascript array

Question

I am new to Jquery and Javascript. Can someone please help me with Jquery sorting based on number of occurrence(count) in array. I tried various sorting methods but none of them worked.

I have an array in Javascript which is

allTypesArray = ["4", "4","2", "2", "2", "6", "2", "6", "6"]

// here  2 is printed four times, 6 is printed thrice, and 4 is printed twice

I need output like this

newTypesArray = ["2","6","4"]

I tried

function array_count_values(e) {
var t = {}, n = "",
    r = "";
var i = function (e) {
    var t = typeof e;
    t = t.toLowerCase();
    if (t === "object") {
        t = "array"
    }
    return t
};
var s = function (e) {
    switch (typeof e) {
    case "number":
        if (Math.floor(e) !== e) {
            return
        };
    case "string":
        if (e in this && this.hasOwnProperty(e)) {
            ++this[e]
        } else {
            this[e] = 1
        }
    }
};
r = i(e);
if (r === "array") {
    for (n in e) {
        if (e.hasOwnProperty(n)) {
            s.call(t, e[n])
        }
    }
}
return t
}
6: 3
}

output is {4: 2, 2: 6, 6:3}

I hope the below post might help you to achieve the same.. stackoverflow.com/questions/19464440/… — J.P
– J.P, Commented Feb 25, 2014 at 10:00
That is not sorting (since you change the contents of the array) — Gabriele Petrioli
– Gabriele Petrioli, Commented Feb 25, 2014 at 10:06
@GabyakaG.Petrioli, Yes it's sorting by occurrence count for each element. — MaveRick
– MaveRick, Commented Feb 25, 2014 at 10:11

Denys Séguret · Accepted Answer · 2014-02-25 10:06:58Z

8

I don't think there's a direct solution in one step and of course it's not just a sort (a sort doesn't remove elements). A way to do this would be to build an intermediary map of objects to store the counts :

var allTypesArray = ["4", "4","2", "2", "2", "6", "2", "6", "6"];
var s = allTypesArray.reduce(function(m,v){
  m[v] = (m[v]||0)+1; return m;
}, {}); // builds {2: 4, 4: 2, 6: 3} 
var a = [];
for (k in s) a.push({k:k,n:s[k]});
// now we have [{"k":"2","n":4},{"k":"4","n":2},{"k":"6","n":3}] 
a.sort(function(a,b){ return b.n-a.n });
a = a.map(function(a) { return a.k });

Note that you don't need jQuery here. When you don't manipulate the DOM, you rarely need it.

answered Feb 25, 2014 at 10:06

Denys Séguret

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2 Comments

elclanrs Over a year ago

FWIW you could save a couple chars return m[v] = ++m[v]||0, m

Denys Séguret Over a year ago

@elclanrs did it look like I was trying to golf ? ^^

adiga · Accepted Answer · 2020-05-17 15:52:33Z

7

Just adding my idea as well (a bit too late)

var allTypesArray = ["4", "4", "2", "2", "2", "6", "2", "6", "6"];
var map = allTypesArray.reduce(function(p, c) {
  p[c] = (p[c] || 0) + 1;
  return p;
}, {});

var newTypesArray = Object.keys(map).sort(function(a, b) {
  return map[b] - map[a];
});

console.log(newTypesArray)

edited May 17, 2020 at 15:52

adiga

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answered Feb 25, 2014 at 10:28

Gabriele Petrioli

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Comments

lnNoam · Accepted Answer · 2016-11-17 13:17:55Z

I don't think jquery is needed here.

There are several great answers to this question already, but I have found reliability to be an issue in some browsers (namely Safari 10 -- though there could be others).

A somewhat ugly, but seemingly reliable, way to solve this is as follows:

function uniqueCountPreserve(inputArray){
    //Sorts the input array by the number of time
    //each element appears (largest to smallest)

    //Count the number of times each item
    //in the array occurs and save the counts to an object
    var arrayItemCounts = {};
    for (var i in inputArray){
        if (!(arrayItemCounts.hasOwnProperty(inputArray[i]))){
            arrayItemCounts[inputArray[i]] = 1
        } else {
            arrayItemCounts[inputArray[i]] += 1
        }
    }

    //Sort the keys by value (smallest to largest)
    //please see Markus R's answer at: http://stackoverflow.com/a/16794116/4898004
    var keysByCount = Object.keys(arrayItemCounts).sort(function(a, b){
        return arrayItemCounts[a]-arrayItemCounts[b];
    });

    //Reverse the Array and Return
    return(keysByCount.reverse())
}

Test

uniqueCountPreserve(allTypesArray)
//["2", "6", "4"]

Sid · Accepted Answer · 2020-11-08 02:34:42Z

0

This is the function i use to do this kind of stuff:

function orderArr(obj){
    const tagsArr = Object.keys(obj)
    const countArr = Object.values(obj).sort((a,b)=> b-a)
  const orderedArr = []
  countArr.forEach((count)=>{
    tagsArr.forEach((tag)=>{
        if(obj[tag] == count && !orderedArr.includes(tag)){
        orderedArr.push(tag)
      }
    })
  })
  return orderedArr
}

answered Nov 8, 2020 at 2:34

Sid

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Comments

Zach Nugent · Accepted Answer · 2020-12-29 20:07:08Z

0

const allTypesArray = ["4", "4","2", "2", "2", "6", "2", "6", "6"]

const singles = [...new Set(allTypesArray)]
const sortedSingles = singles.sort((a,b) => a - b)
console.log(sortedSingles)

Set objects are collections of values. A value in the Set may only occur once; it is unique in the Set's collection.

The singles variable spreads all of the unique values from allTypesArray using the Set object with the spread operator inside of an array.

The sortedSingles variable sorts the values of the singles array in ascending order by comparing the numbers.

answered Dec 29, 2020 at 20:07

Zach Nugent

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1 Comment

Gabriele Petrioli Over a year ago

the OP wants to sort based on the number each value appears in the original array.

Airofoil · Accepted Answer · 2022-06-21 17:22:51Z

Not sure if there's enough neat answers here, this is what I came up with:

Fill an object with counts for each of the elements:

let array = ['4', '4', '2', '2', '2', '6', '2', '6', '6'];
let arrayCounts = {}

for (j in array) arrayCounts[array[j]] ? arrayCounts[array[j]].count++ : arrayCounts[array[j]] = { val: array[j], count: 1 };

/* arrayCounts = {
  '2': { val: '2', count: 4 },
  '6': { val: '4', count: 2 },
  '4': { val: '6', count: 3 }
} */

For the values in that new object, sort them by .count, and map() them into a new array (with just the values):

let sortedArray = Object.values(arrayCounts).sort(function(a,b) { return b.count - a.count }).map(({ val }) => val);

/* sortedArray = [ '2', '6', '4' ] */

Altogether:

let arrayCounts = {}

for (j in array) arrayCounts[array[j]] ? arrayCounts[array[j]].count++ : arrayCounts[array[j]] = { val: array[j], count: 1 };
    
let sortedArray = Object.values(arrayCounts)
    .sort(function(a,b) { return b.count - a.count })
    .map(({ val }); => val);

Tyler2P · Accepted Answer · 2022-12-05 16:03:20Z

0

var number = [22,44,55,11,33,99,77,88];

for (var i = 0;i<number.length;i++) {
  for (var j=0;j<number.length;j++){
    if (number[j]>number[j+1]) {
      var primary = number[j];
      number[j] =  number[j+1];
      number[j+1] = primary;
    }
  }
}
document.write(number);

edited Dec 5, 2022 at 16:03

Tyler2P

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answered Dec 2, 2022 at 16:59

Rejwan Sharif

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2 Comments

Rejwan Sharif Over a year ago

Here are sorting an Array using 2 loops. First One is for showing output and the second one is for comparing values between arrays.

Tyler2P Over a year ago

Your answer could be improved by adding more information on what the code does and how it helps the OP.

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