How to sort an array by value

Question

I have an array which includes some numbers and i wanna sort them by value. For example imagine i have this array:

let myArray = [ 10 , 50 , 30 , 20 , 40]

I wanna sort them by the value from biggest to smallest with for loop or if/else or anything else which i can get result like that:

myArray = [ 50 , 40 , 30 , 20 , 10 ]

or maybe check values and input them in a new array :

newArray = [ 50 , 40 , 30 , 20 , 10 ]

It looks simple but i can't do it,Thanks

Please mention the language you're using, if `JS` just call`sort()` — K-Galalem, Dec 20 '22 at 18:46
@K-Galalem That won't work in JavaScript for something like `[2, 10]`. See the link in David's comment for the reason. — Andrew Morton, Dec 20 '22 at 19:17

Engineer Patterson · Answer 1 · 2022-12-20T19:47:11.043

There are multiple ways you can achieve this. For what you are wanting at first you would write this:

const myArray = [10, 50, 30, 20, 40];
    myArray.sort((a, b) => b - a);
    console.log(myArray);

to check array value and add it to another array you would write this:

const myArray = [10, 50, 30, 20, 40];
      let newArray = []; //creates an empty array 

      /*this sorts thru the array and for each item in 
      the array it will push it to the newArray from lowest to highest */
      myArray.sort((a, b) => b - a).forEach((item) => 
      newArray.push(item));
      myArray.sort((a, b) => a - b);
      
      console.log(newArray); 
      console.log(myArray);

And if you want to take it a step further you could also sort the values in separate arrays based on if the number is even or odd like so:

       const myArray2 = [2, 4, 11, 8, 28, 23, 26, 13, 30, 33, 81, 37];
       const evenArray = [];
       const oddArray = [];
       myArray2.sort((a, b) => a - b).forEach((item) => {
        if (item % 2 === 0) {
            evenArray.push(item);
        } else {
            oddArray.push(item);
        }
        }
       );
       console.log(evenArray);
       console.log(oddArray);

Create 2 arrays one for even numbers and one for odd.
then sort thru the array
for each number in array, if it is even push it to evenArray
if it is not an even number push it to oddArray then return or in this case console.log()

Hope this helps you understand it a little better!

How about object in array? for example: const item1 = { value: 2, name: chris } const item2 = { value: 1, name: alex } const array = [item1 , item2] if i want to sort items by item.value how i can do this? — Alireza, Dec 21 '22 at 11:52
To sort objects it is the same as above. anytime you need to sort thru an array you just call the "sort() function which is a built in JS function. — Engineer Patterson, Dec 21 '22 at 15:39

Zapwoi · Answer 2 · 2022-12-21T06:38:31.690

0

In javascript you can user function

Sort numbers in ascending order:

const points = [40, 100, 1, 5, 25, 10];
points.sort((a, b)=>{return a-b});

Sort numbers in descending order:

const points = [40, 100, 1, 5, 25, 10];
points.sort((a, b)=>{return b-a});

default function is ascending

Ref..https://www.w3schools.com/js/js_array_sort.asp

edited Dec 21 '22 at 06:38

answered Dec 20 '22 at 18:46

Zapwoi

51
1
3

1

Please take a look at https://stackoverflow.com/help/formatting – Gilles Quénot Dec 20 '22 at 18:47

score 0 · Answer 3 · answered Dec 20 '22 at 18:48

0

You can do it in one line:

Ascending: myArray.sort() === [10, 20, 30, 40, 50]

Descending: myArray.sort((a,b)=> b-a) === [50, 40, 30, 20, 10]

answered Dec 20 '22 at 18:48

Niv Ohayon

19
2

Sash Sinha · Answer 4 · 2022-12-20T18:55:51.527

Consider utilizing sort() to sort the numbers in descending order in place:

const myArray = [10, 50, 30, 20, 40];
myArray.sort((a, b) => b - a);
console.log(`myArray: [${myArray}]`);

You can sort without mutation of the original array by making a copy before sorting:

const myArray = [10, 50, 30, 20, 40];
const newArray = [...myArray].sort((a, b) => b - a);
console.log(`newArray: [${newArray}]`);
console.log(`myArray: [${myArray}]`);

How to sort an array by value

4 Answers4