按 Javascript 数组中的出现次数（计数）排序

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}

Answer 1

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.

Answer 2

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)

Answer 3

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"]

Answer 4

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
}

Answer 5

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.

Answer 6

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);

Answer 7

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);