How to sort an array based on the length of each element?

Question

I have an array like this:

arr = []
arr[0] = "ab"
arr[1] = "abcdefgh"
arr[2] = "abcd"

After sorting, the output array should be:

arr[0] = "abcdefgh"
arr[1] = "abcd"
arr[2] = "ab"  

I want in the descending order of the length of each element.

Answer 1

You can use Array.sort method to sort the array. The callback function should use the length of item as the sorting criteria:

// sort ascending - shorter items first
arr.sort((a, b) => a.length - b.length);

// sort descending - longer items first
arr.sort((a, b) => b.length - a.length);

You can specify additional criteria if length of two items are same:

// sort by length
// if equal then sort by dictionary order
["c", "a", "b"].sort((a, b) => a.length - b.length || a.localeCompare(b));

Answer 2

Here is the sort, depending on the length of a string with javascript using Bubble sort as you asked:

var arr = ['1234', '12', '12345', '1'];

bubbleSort(arr );

function bubbleSort(a) {
    var swapped;
    do {
        swapped = false;
        for (var i = 0; i < a.length - 1; i++) {
            if (a[i].length < a[i + 1].length) {
                var temp = a[i];
                a[i] = a[i + 1];
                a[i + 1] = temp;
                swapped = true;
            }
        }
    } while (swapped);
}

console.log(arr );

Answer 3

let arr  = [5,2,100,1,20,3];
arr.sort((a,b)=>{
  return a-b 
})

console.log(arr) //[1, 2, 3, 5, 20, 100]

on the return value, the sort method will perform the functionality of swapping of an elements

return < 0  { i.e -ve number then  a comes before b}
return > 0  { i.e +ve number then  b comes before a}
return == 0 { order of a and b remains same }

Answer 4

With modern JavaScript you can do like this:

Descending order

const arr = [
  "ab",
  "abcdefgh",
  "abcd",
  "abcdefghijklm"
];

arr.sort((a, b) => b.length - a.length);

console.log(JSON.stringify(arr, null, 2));

Ascending Order - Just switch the a with b

const arr = [
  "ab",
  "abcdefgh",
  "abcd",
  "abcdefghijklm"
];

arr.sort((a, b) => a.length - b.length);

console.log(JSON.stringify(arr, null, 2));

Answer 5

let array = [`ab`, `abcdefgh`, `abcd`];
let newArray = array.sort((a,b) => {
    return b.length - a.length
})
console.log(newArray);

Please the following code

Answer 6

If you want to preserve the order of the element with the same length as the original array, use bubble sort.

Input = ["ab","cdc","abcd","de"];

Output  = ["ab","cd","cdc","abcd"]

Function:

function bubbleSort(strArray){
  const arrayLength = Object.keys(strArray).length;
    var swapp;
    var newLen = arrayLength-1;
    var sortedStrArrByLenght=strArray;
    do {
        swapp = false;
        for (var i=0; i < newLen; i++)
        {
            if (sortedStrArrByLenght[i].length > sortedStrArrByLenght[i+1].length)
            {
               var temp = sortedStrArrByLenght[i];
               sortedStrArrByLenght[i] = sortedStrArrByLenght[i+1];
               sortedStrArrByLenght[i+1] = temp;
               swapp = true;
            }
        }
        newLen--;
    } while (swap);
  return sortedStrArrByLenght;
}

Answer 7

#created a sorting function to sort by length of elements of list
def sort_len(a):
    num = len(a)
    d = {}
    i = 0
    while i<num:
        d[i] = len(a[i])
        i += 1
    b = list(d.values())
    b.sort()
    c = []
    for i in b:
        for j in range(num):
            if j in list(d.keys()):
                if d[j] == i:
                    c.append(a[j])
                    d.pop(j)
    return c

Answer 8

This code should do the trick:

var array = ["ab", "abcdefgh", "abcd"];

array.sort(function(a, b){return b.length - a.length});

console.log(JSON.stringify(array, null, '\t'));

Answer 9

We can use Array.sort method to sort this array.

ES5 solution

var array = ["ab", "abcdefgh", "abcd"];

array.sort(function(a, b){return b.length - a.length});

console.log(JSON.stringify(array, null, '\t'));

For ascending sort order: a.length - b.length

For descending sort order: b.length - a.length

ES6 solution

Attention: not all browsers can understand ES6 code!

In ES6 we can use an arrow function expressions.

let array = ["ab", "abcdefgh", "abcd"];

array.sort((a, b) => b.length - a.length);

console.log(JSON.stringify(array, null, '\t'));

Answer 10

I adapted @shareef's answer to make it concise. I use,

.sort(function(arg1, arg2) { return arg1.length - arg2.length })

Answer 11

Based on Salman's answer, I've written a small function to encapsulate it:

function sortArrayByLength(arr, ascYN) {
        arr.sort(function (a, b) {           // sort array by length of text
            if (ascYN) return a.length - b.length;              // ASC -> a - b
            else return b.length - a.length;                    // DESC -> b - a
        });
    }

then just call it with

sortArrayByLength( myArray, true );

Note that unfortunately, functions can/should not be added to the Array prototype, as explained on this page.

Also, it modified the array passed as a parameter and doesn't return anything. This would force the duplication of the array and wouldn't be great for large arrays. If someone has a better idea, please do comment!

Answer 12

<script>
         arr = []
         arr[0] = "ab"
         arr[1] = "abcdefgh"
         arr[2] = "sdfds"
         arr.sort(function(a,b){
            return a.length<b.length
         })
         document.write(arr)

</script>

The anonymous function that you pass to sort tells it how to sort the given array.hope this helps.I know this is confusing but you can tell the sort function how to sort the elements of the array by passing it a function as a parameter telling it what to do