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.

@muistooshort well default sort() sorts strings alphabeticly, he was looking for the string.length sorting as can be seen in the chosen answer :) — jave.web, Aug 07 '13 at 16:02

Salman A · Accepted Answer · 2018-07-20T23:43:01.750

306

You can use Array.sort method to sort the array. A sorting function that considers the length of string as the sorting criteria can be used as follows:

arr.sort(function(a, b){
  // ASC  -> a.length - b.length
  // DESC -> b.length - a.length
  return b.length - a.length;
});

Note: sorting ["a", "b", "c"] by length of string is not guaranteed to return ["a", "b", "c"]. According to the specs:

The sort is not necessarily stable (that is, elements that compare equal do not necessarily remain in their original order).

If the objective is to sort by length then by dictionary order you must specify additional criteria:

["c", "a", "b"].sort(function(a, b) {
  return a.length - b.length || // sort by length, if equal then
         a.localeCompare(b);    // sort by dictionary order
});

edited Jul 20 '18 at 23:43

answered May 17 '12 at 06:24

Salman A

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I would add that this sorts the array by decreasing length of the items. – davidhq Feb 27 '16 at 20:17
aha it's in the comment :) didn't see it at first – davidhq Feb 27 '16 at 20:17
Just to add an information. This snippet won't work for all the cases. – Arunkumar Srisailapathi Apr 10 '16 at 12:42
Try for arr = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'k', 'l'], it will fail – Arunkumar Srisailapathi Apr 10 '16 at 12:43
@ArunKumar what is the expected output? – Salman A Apr 10 '16 at 19:39
@SalmanA: It should give the sorted elements. But instead will give elements on some random order. You can run and check. At least to my knowledge. – Arunkumar Srisailapathi Apr 19 '16 at 12:07
@ArunKumar I have revised the answer. The sort function is allowed to rearrange two elements if they are considered "equal". – Salman A Apr 19 '16 at 20:57
21

The ES6 way `arr.sort((a, b) => b.length - a.length)` – Fergal Jun 08 '16 at 07:58
this works very well for me, just what i was in search of. – Kingston Fortune Jul 22 '18 at 18:05
@Fergal I think it's unfair to call that "the ES6 way"... might confuse some newer people. For instance if the logic of the function required a block then you need the explicit `return` statement. – dillon.harless Jul 08 '19 at 20:21

score 13 · Answer 2 · edited Jun 20 '20 at 09:12

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'));

score 4 · Answer 3 · answered Aug 04 '21 at 20:00

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));

score 3 · Answer 4 · edited Jun 07 '22 at 02:33

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 );

score 1 · Answer 5 · answered Apr 23 '20 at 17:07

1

#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

answered Apr 23 '20 at 17:07

Rishabh Gupta

11
3

1

Posting the code alone won't help. Give a brief description of what your code does. – coderpc Apr 23 '20 at 17:10

score 1 · Answer 6 · answered Oct 06 '20 at 05:07

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;
}

score 1 · Answer 7 · answered Nov 22 '21 at 06:20

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 }

score 0 · Answer 8 · edited May 23 '17 at 11:47

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!

score 0 · Answer 9 · answered Aug 25 '16 at 22:27

0

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

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

answered Aug 25 '16 at 22:27

this would sort from low length to higher length – Miguel Nov 24 '16 at 16:08

score 0 · Answer 10 · edited Mar 16 '20 at 15:48

0

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'));

edited Mar 16 '20 at 15:48

ToVine

560
5
16

answered Mar 16 '20 at 15:03

Hiền

11

score 0 · Answer 11 · answered Jun 04 '21 at 12:02

0

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

Please the following code

answered Jun 04 '21 at 12:02

Force Bolt

1,117
9
9

score -3 · Answer 12 · answered May 17 '12 at 06:44

<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

What is the sorting comparison function supposed to return? Here's a hint: it isn't a boolean. — mu is too short, Aug 07 '13 at 17:14

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

12 Answers12

ES5 solution

ES6 solution

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