snippet for creating object from destructured array

Question

For example, I had an array with 3 numbers:

var arr = [124, -50, 24];

and I need to convert this array to the object:

{
   x: 124,
   y: -50,
   z: 24
}

I don`t want to use "old-style" syntax for this, for example:

{
  x: arr[0],
  y: arr[1],
  z: arr[2]
}

so for now, I`m using that syntax:

const [x, y, z] = [...arr];
const obj = {x, y, z};

But, is there is any way to do this with a straight dectructuring array to object without need of temporary variables?

off the top of my head `var obj = (([x,y,z]) => ({x,y,z}))(arr);` — Jaromanda X, Sep 26 '17 at 12:35
@Bergi, there's no array -> object in that question ... is there? — Jaromanda X, Sep 26 '17 at 12:37
@JaromandaX It's about "renaming" properties, like from `0` to `x` - whether it's an array object or plain object doesn't make much difference, the choices for destructuring and literal syntax are the same. — Bergi, Sep 26 '17 at 12:38
I think this is similar: https://stackoverflow.com/questions/38242744/destructure-array-to-object-property-keys — llama, Sep 26 '17 at 12:39
@JaromandaX In any case, I answered rather than closing as a duplicate. It's not totally obvious indeed. — Bergi, Sep 26 '17 at 12:42
@JaromandaX thanks, that works! It remains only to understand what is going on there :) — Vlad Povalii, Sep 26 '17 at 12:44
@llama - yes you right! I missed that question :( But the best answer there is getting array values by indexes. I have mentioned in my question that is not what im looking for — Vlad Povalii, Sep 26 '17 at 12:47
`It remains only to understand what is going on there` @VladPovalii it's an IIFE of the arrow function `([x,y,z]) => ({x,y,z})`. But imo you should store this function `const arr2point = ([x,y,z]) => ({x,y,z});` and use that `const obj = arr2point(arr)` instead of using the IIFE inline. — Thomas, Sep 26 '17 at 12:51
Why do people always try to use destructuring to build an object ? :( — Felix Kling, Sep 26 '17 at 14:43

score 1 · Answer 1 · answered Sep 26 '17 at 12:37

1

You can also do

const obj = {};
([obj.x, obj.y, obj.z] = arr);

to avoid the temporary variables, but I'd question whether that's an improvement.

answered Sep 26 '17 at 12:37

Bergi

score 1 · Accepted Answer · edited May 28 '19 at 11:17

1

As it was already mentioned in the comment you can use an Immediately Invoked Function Expression (IIFE) to create the object in a single step, but it's less readable then multiple steps.

const arr = [124, -50, 24];
const obj = (([x, y, z]) => ({ x, y, z }))(arr);

console.log(obj);

edited May 28 '19 at 11:17

Vlad Povalii

answered May 27 '19 at 04:46

Ori Drori

lol. I've changed the sorting order by mistake, and this question popped. I've answered it like it was new (not really looked at the comments), and totally forgot about it. – Ori Drori May 28 '19 at 11:19
1

:) Anyway its ok..finally this question will have the right answer, so thanks! – Vlad Povalii May 28 '19 at 11:25

score 1 · Answer 3 · answered May 27 '19 at 05:02

1

Just use this

let obj = {...arr}

answered May 27 '19 at 05:02

Abdur Rahaman

Not quite right, because you want to control the object keys naming ("x", "y", "z") - but after this keys will be: "0", "1", "2" – Vlad Povalii May 28 '19 at 11:28

3 Answers3