How to convert multiple arrays into one array of multiple objects?

Question

Given the following arrays:

let x = [a, b, c, d];
let y = [e, f, g, h];
let w = [i, j, k, l];

How to generated a new array of objects that look like that:

let z = [
    {x: a, y: e, w: i},
    {x: b, y: f, w: j},
    {x: c, y: g, w: k},
    {x: d, y: h, w: l}
];

This is what I came up so far:

for(var i; i < x.length; i++) {

    x = x[i];
    y = y[i];
    w = w[i];

    obj = {
        x: x,
        y: y,
        w: w
    };

    z = [];
    z.push(obj);
}

Thanks!

Since all arrays have equal length: "for `index` from 0 to amount of items in one of my arrays, insert an object into array `z` such that `z[index].x` is equal to `x[index]`, `z[index].y` is equal to `y[index]` and `z[index].w` is equal to `w[index]`" — nem035, Dec 05 '17 at 16:44
Who is upvoting this? It is the definition of "lacks effort" — Liam, Dec 05 '17 at 16:47
This is a solid and interesting question to solve. But please show us what you've already tried that hasn't worked for you, since SO is not a code-writing service. :-) — jeffdill2, Dec 05 '17 at 16:47
I agree on *lack of effort*, this is not a kind of problem which requires solid algorithm and data-structure understanding. OP could have made some effort to produce an ugly version. OP hasn't even replied to inquiries from other experts. — gurvinder372, Dec 05 '17 at 16:55
After your edit, move `z = [];` *before* the loop and you're done. — Federico klez Culloca, Dec 05 '17 at 16:55
@FedericoklezCulloca the duplicate link makes no sense at all. — ibrahim mahrir, Dec 05 '17 at 16:59
Why are you telling *me* that? I voted to close as "questions seeking debugging help..." — Federico klez Culloca, Dec 05 '17 at 17:00
@FedericoklezCulloca Ah, I just saw your name first. BTW _"questions seeking debugging help..."_ also makes no sense. — ibrahim mahrir, Dec 05 '17 at 17:02

Suren Srapyan · Answer 1 · 2017-12-05T16:44:25.947

4

Use Array#map function and get also the index, which you will use to get the items from the second and third array. I used also || if first array has more items that the others.

let x = ['a', 'b', 'c',' d'];
let y = ['e', 'f', 'g', 'h'];
let w = ['i', 'j', 'k', 'l'];

let mapped = x.map((item, index) => ({ x: item, y: y[index] || '', w: w[index] || '' }));

console.log(mapped);

edited Dec 05 '17 at 16:44

answered Dec 05 '17 at 16:42

Suren Srapyan

66,568
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5

Sometimes I wonder why low quality questions get answers by high-reputation users that should vote to close them instead. – Federico klez Culloca Dec 05 '17 at 16:44
@FedericoklezCulloca Questions lacking attempts should not be closed. Downvoted maybe, but not closed. There is no category/reason for them in the _why should this question be closed?_ box. – ibrahim mahrir Dec 05 '17 at 16:56
@ibrahimmahrir do you know if there's ever been a discussion about this kind of questions in meta? – Federico klez Culloca Dec 05 '17 at 16:57
@FedericoklezCulloca No I don't. Do you? – ibrahim mahrir Dec 05 '17 at 16:58
@ibrahimmahrir No, I don't, that's why I was asking :) – Federico klez Culloca Dec 05 '17 at 16:59
@FedericoklezCulloca wait, I'll check. After all maybe I'm wrong. – ibrahim mahrir Dec 05 '17 at 17:00
@ibrahimmahrir I found [this](https://meta.stackoverflow.com/questions/260828/do-we-need-a-close-reason-for-zero-effort-questions) which is rather long, I'll read later. – Federico klez Culloca Dec 05 '17 at 17:03
@FedericoklezCulloca Well done! You have found the right meta question. I advise you to read it. The question is exactly what you are saying, the accepted answer is exactly what I'm saying (exactly my point). – ibrahim mahrir Dec 05 '17 at 17:19

score 1 · Answer 2 · answered Dec 05 '17 at 16:42

Try

var z = x.map( (s,i) => ({ x : x[i], y : y[i], w : w[i] }) );

Explanation

iterate x using map
for each index of x, return an object having keys as x, y and z with values from their respective index.

Demo

var x = ['a', 'b', 'c', 'd'];
var y = ['e', 'f', 'g', 'h'];
var w = ['i', 'j', 'k', 'l'];

var z = x.map((s, i) => ({
  x: x[i],
  y: y[i],
  w: w[i]
}));

console.log(z);

score 0 · Answer 3 · answered Dec 05 '17 at 17:12

let items = [ "a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l" ];
for (let i = 0; i < items.length; i++)
  Object.defineProperty(window, items[i], {
    value: items[i],
    writable: false
  });
//The code above there is just to define the variables in your arrays.

let x = [a, b, c, d];
let y = [e, f, g, h];
let w = [i, j, k, l];

let keys = [ "x", "y", "w" ]; //This array contains the name for accessing to your arrays.
let dimension = Math.max(...keys.map(function(array) {
  return eval(array).length;
}));
let z = new Array(dimension);
for (let i = 0; i < dimension; i++) {
  let obj = {};
  for (let ii = 0; ii < keys.length; ii++) {
    let key = keys[ii];
    obj[key] = eval(key)[i];
  }
  z[i] = obj;
}

console.log(JSON.stringify(z));

You can also look at this fiddle.

score 0 · Answer 4 · answered Dec 05 '17 at 17:21

0

You can use forEach as following :

var result = [];
a.forEach((currenValue, index) => {
    result.push({x:x[index], y:b[index], w:c[index]});
});
console.log(result);

answered Dec 05 '17 at 17:21

Renaud is Not Bill Gates

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score 0 · Answer 5 · answered Dec 05 '17 at 17:25

You could use an array with the keys for the object and reduce the array with the arrays.

var x = ['a', 'b', 'c', 'd'],
    y = ['e', 'f', 'g', 'h'],
    w = ['i', 'j', 'k', 'l'],
    keys = ['x', 'y', 'w'],
    result = [x, y, w].reduce(function (r, a, i) {
        a.forEach(function (v, j) {
            r[j] = r[j] || {}
            r[j][keys[i]] = v;
        });
        return r;
    }, []);

console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

How to convert multiple arrays into one array of multiple objects?

5 Answers5