Can I use .sort() to sort an array of objects based on a number property?

Question

If I had an array of numbers, I can quickly sort them using .sort() function like this:

let array = [1, 5, 3, 2, 9, 6];
let sortedArray = array.sort();

However, I am wondering if I can even use .sort() if my array is made up of objects who have a number property. For example, I'm wondering if I could turn this:

let array = [
    { name: "Random1", number: "1" },
    { name: "Random2", number: "5" },
    { name: "Random3", number: "3" },
    { name: "Random4", number: "2" },
    { name: "Random5", number: "9" },
    { name: "Random6", number: "6" },
]

Into this based on the value of the number property.

let array = [
    { name: "Random1", number: "1" },
    { name: "Random4", number: "2" },
    { name: "Random3", number: "3" },
    { name: "Random2", number: "5" },
    { name: "Random6", number: "6" },
    { name: "Random5", number: "9" },    
]

score 1 · Answer 1 · answered Feb 01 '20 at 05:08

1

The sort method accepts a comparison function. You can read more about it here: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/sort

array.sort((first, second) => first.number - second.number);

answered Feb 01 '20 at 05:08

kingkupps

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Mamun · Accepted Answer · 2020-02-01T05:47:47.927

1

You can do that by passing the compareFunction as the parameter to sort().

compareFunction | Optional

Specifies a function that defines the sort order. If omitted, the array elements are converted to strings, then sorted according to each character's Unicode code point value.

let array = [
    { name: "Random1", number: "1" },
    { name: "Random2", number: "5" },
    { name: "Random3", number: "3" },
    { name: "Random4", number: "2" },
    { name: "Random5", number: "9" },
    { name: "Random6", number: "6" },
]

array.sort((a, b) => (a.number > b.number) ? 1 : -1);
console.log(array);

If the number property is always number in the form of string then use the following:

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

edited Feb 01 '20 at 05:47

answered Feb 01 '20 at 05:09

Mamun

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May I ask what is the difference between your example, and dev_junwen's example? They're similar yet yours has an extra ternary if statement. – Onyx Feb 01 '20 at 05:16
@Bobimaru, the difference is for the type actually---string and number:) – Mamun Feb 01 '20 at 05:19

score 1 · Answer 3 · answered Feb 01 '20 at 05:10

You can make use of the array sort function callback to compare the object property number.

const array = [
    { name: "Random1", number: "1" },
    { name: "Random2", number: "5" },
    { name: "Random3", number: "3" },
    { name: "Random4", number: "2" },
    { name: "Random5", number: "9" },
    { name: "Random6", number: "6" },
];

const sortedArray = array.sort((a, b) => a.number - b.number);
console.log(sortedArray);

Can I use .sort() to sort an array of objects based on a number property?

3 Answers3