Sort an array of objects

Question

Is this the correct way to sort an array of objects?:

class Alarm{
    constructor(data){
        this.a = data.a;
        this.b = data.b;
    }
}

const a1 = new Alarm({a:4,b:1});
const a2 = new Alarm({a:1,b:4});
const a3 = new Alarm({a:2,b:3});
const a4 = new Alarm({a:3,b:2});

let a = [a1,a2,a3,a4];

a.sort((x,y)=>x.a>y.a);

for(const i of a){
    console.log(i)
}

a.sort((x,y)=>x.b>y.b);

for(const i of a){
    console.log(i)
}

When I look at the documentation:

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/sort

They are doing this instead, which seems extremely awkward:

a.sort((x,y)=>
    {
        if(x.a>y.a) return 1;
        if(x.a<y.a) return -1;
        return 0;
    });

sort should return 3 values. 1 for greater than, 0 for no change and -1 for less than. — Rajesh Dixit
– Rajesh Dixit, Commented Nov 2, 2016 at 9:03
@Keith for string values, I'm not sure how - would work. For numeric values, it will work, but as OP is asking for better approach, I guess we should consider both situations — Rajesh Dixit
– Rajesh Dixit, Commented Nov 2, 2016 at 9:06
It's not awkward. It's how the comparator function works. By the way, what is your question? — user663031
– user663031, Commented Nov 2, 2016 at 9:14

Nina Scholz · Accepted Answer · 2016-11-02 09:16:28Z

5

You could use the difference for sorting

a.sort((x, y) => x.a - y.a);

Taken from Array#sort

If compareFunction is supplied, the array elements are sorted according to the return value of the compare function. If a and b are two elements being compared, then:

If compareFunction(a, b) is less than 0, sort a to a lower index than b, i.e. a comes first.

If compareFunction(a, b) returns 0, leave a and b unchanged with respect to each other, but sorted with respect to all different elements. Note: the ECMAscript standard does not guarantee this behaviour, and thus not all browsers (e.g. Mozilla versions dating back to at least 2003) respect this.

If compareFunction(a, b) is greater than 0, sort b to a lower index than a.

compareFunction(a, b) must always return the same value when given a specific pair of elements a and b as its two arguments. If inconsistent results are returned then the sort order is undefined.

class Alarm{
    constructor(data){
        this.a = data.a;
        this.b = data.b;
    }
}

const a1 = new Alarm({a:4,b:1});
const a2 = new Alarm({a:1,b:4});
const a3 = new Alarm({a:2,b:3});
const a4 = new Alarm({a:3,b:2});

let a = [a1, a2, a3, a4];

a.sort((x,y) => x.a - y.a);

console.log(JSON.stringify(a, 0, 4));

edited Nov 2, 2016 at 9:16

answered Nov 2, 2016 at 9:03

Nina Scholz

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5 Comments

Baz Over a year ago

Is there a logical reason as to why the sort lambda should return 1, -1, and 0 instead of a boolean value. Seams really weird to me. Is there an alternative way of sorting where you can return booleans? I'm new to JS and ES6 :)

user663031 Over a year ago

@Baz Because that's how it works. What would a boolean mean? The return value needs to indicate whether the first parameter comes before or after the second, or at the same location.

Keith Over a year ago

The sort needs to know if a value is equal, greater than, or less than. A boolean does not give you that.

Keith Over a year ago

To understand why <=> are needed, a good idea is to look at how sort functions are done. Look for bubble sort / quick sort etc. Wiki most likely will have good docs on this.

Nina Scholz Over a year ago

great! the spaceship operator :)

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