Sorting arrays numerically by object property value

Question

How would you sort this array with these objects by distance, so that you have the objects sorted from smallest distance to biggest distance?

[
  { distance: 3388, duration: "6 mins", from: "Lenchen Ave, Centurion 0046, South Africa" },
  { distance: 13564, duration: "12 mins", from: "Lenchen Ave, Centurion 0046, South Africa" },
  { distance: 4046, duration: "6 mins", from: "Lenchen Ave, Centurion 0046, South Africa" },
  { distance: 11970, duration: "17 mins", from: "Lenchen Ave, Centurion 0046, South Africa" }
]

The pattern for [sorting by properties](/q/2466356/4642212) numerically is `myarray.sort((a, b) => a.distance - b.distance)`. To sort [lexicographically](/q/1129216/4642212), use `a.from.localeCompare(b.from)`. To sort descending instead of ascending, negate the return value (e.g. `b.distance - a.distance` instead of `a.distance - b.distance`). To sort [numeric _strings_](/q/979256/4642212), optionally use `Number`. To sort by [multiple properties](/q/6913512/4642212), chain other sorts with `||`, e.g. `b.someNumber - a.someNumber || a.someString.localeCompare(b.someString)`. — Sebastian Simon, Apr 10 '22 at 11:49

Phil · Accepted Answer · 2021-11-22T07:24:19.157

249

Use Array.prototype.sort(), eg

myArray.sort((a, b) => a.distance - b.distance)

The sort() method accepts a comparator function. This function accepts two arguments (both presumably of the same type) and it's job is to determine which of the two comes first.

It does this by returning an integer

Negative (less-than zero): The first argument comes first
Positive (greater-than zero): The second argument comes first
Zero: The arguments are considered equal for sorting

When you're dealing with numeric values, the simplest solution is to subtract the second value from the first which will produce an ascending order result.

edited Nov 22 '21 at 07:24

answered Oct 25 '11 at 12:08

Phil

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Alternative, using [destructuring assignment](https://developer.mozilla.org/en/docs/Web/JavaScript/Reference/Operators/Destructuring_assignment): `myArray.sort(({ distance: a }, { distance: b }) => a - b);`. – Sebastian Simon Apr 10 '22 at 11:46
3

@SebastianSimon that's not an _alternative_ so much as it is the same thing but 10 characters longer syntactically. Potato / tomato – Phil Aug 01 '22 at 05:49

score 76 · Answer 2 · answered Sep 01 '20 at 11:47

76

here's an example with the accepted answer:

 a = [{name:"alex"},{name:"clex"},{name:"blex"}];

For Ascending :

a.sort((a,b)=> (a.name > b.name ? 1 : -1))

output : [{name: "alex"}, {name: "blex"},{name: "clex"} ]

For Decending :

a.sort((a,b)=> (a.name < b.name ? 1 : -1))

output : [{name: "clex"}, {name: "blex"}, {name: "alex"}]

answered Sep 01 '20 at 11:47

Surya Murugan

861
6
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Just in addition to this, don't forget to change data in comparing function to either upper or lower case, so that you get results as in alphabetical order irrespective of ASCII code. – Akash Jul 13 '22 at 13:36

score 26 · Answer 3 · answered Feb 22 '19 at 17:24

26

Here's the same as the current top answer, but in an ES6 one-liner:

myArray.sort((a, b) => a.distance - b.distance);

answered Feb 22 '19 at 17:24

Jon Tonti

533
5
7

Vinod Poorma · Answer 4 · 2018-03-07T07:24:53.843

8

This worked for me

var files=data.Contents;
          files = files.sort(function(a,b){
        return a.LastModified - b. LastModified;
      });

OR use Lodash to sort the array

files = _.orderBy(files,'LastModified','asc');

edited Mar 07 '18 at 07:24

answered Mar 07 '18 at 07:14

Vinod Poorma

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1
6
15

Where did `LastModified` come from? The target property is `distance`. – Sebastian Simon Apr 10 '22 at 11:46
"LastModified" is the key in the object. in my case it is date – Vinod Poorma Apr 30 '22 at 09:08

score 6 · Answer 5 · answered Feb 05 '19 at 18:37

6

Not spectacular different than the answers already given, but more generic is :

sortArrayOfObjects = (arr, key) => {
    return arr.sort((a, b) => {
        return a[key] - b[key];
    });
};

sortArrayOfObjects(yourArray, "distance");

answered Feb 05 '19 at 18:37

Arthur van Acker

472
7
10

1

Can be written with a one liner return statement: `sortArrayOfObjects = (arr, key) => { return arr.sort((a, b) => a[key] - b[key]); };` – user10971804 Feb 28 '20 at 05:25

score 4 · Answer 6 · answered May 12 '21 at 09:31

4

Push all objects in an array myArray then apply sorting using this.

myArray.sort(function(a, b) {
    return a.distance - b.distance;
});

answered May 12 '21 at 09:31

Deepak Kumar

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17

score 4 · Answer 7 · answered Aug 29 '21 at 21:32

4

If you have lowercase and uppercase names, use toLowerCase() function.
Here is an example:

data = [{name:"foo"},{name:"Bar"},{name:"Foo"}];


a.sort((a,b)=> (a.name.toLowerCase() > b.name.toLowerCase() ? 1 : -1))

answered Aug 29 '21 at 21:32

laurisstepanovs

432
5
14

score 0 · Answer 8 · answered Oct 23 '20 at 20:30

0

Here is yet another one-liner for you:

your_array.sort((a, b) => a.distance === b.distance ? 0 : a.distance > b.distance || -1);

answered Oct 23 '20 at 20:30

codemonkey

7,325
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score 0 · Answer 9 · answered Apr 07 '21 at 12:51

0

ES6 arrow function:

const orderArrayBy = (arr, key) => arr.sort((a, b) => a[key] - b[key]);

answered Apr 07 '21 at 12:51

niikoo

11
4

score 0 · Answer 10 · answered Dec 08 '22 at 07:32

You can use Currying for this:

const arr = [
  { distance: 3388, duration: 6},
  { distance: 13564, duration: 12},
  { distance: 4046, duration: 6 },
  { distance: 11970, duration: 17 }
]

const sortByKey = (key) => (a, b) => b[key] - a[key];

const sortByDistance = sortByKey("distance");
const sortByDuration = sortByKey("duration");

console.log(arr.sort(sortByDistance))
console.log(arr.sort(sortByDuration))

Sorting arrays numerically by object property value

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