Say I have three arrays depicting some names, number of books read and how awesome these people [in names] are:

let names = ["Mary", "Joe", "Kenan"];
let numberOfBooks = [2, 1, 4];
let awesomenessLevel = ["pretty cool", "meh", "super-reader"];

I'm trying to use .reduce() to bring them together to create an array of objects containing the relevant index in each array, but I am failing miserably:

    let people = [
    {
       name: "Mary",
       noOfBooks: 2,
       awesomeness: "pretty cool"
    },
    {
       name: "Joe",
       noOfBooks: 1,
       awesomeness: "meh"
    },
    {
       name: "Kenan",
       noOfBooks: 4,
       awesomeness: "super-reader"
    }
  ]

I got it with reduce as well:

let arrFinal = [];

 names.reduce(function(all, item, index) {
  arrFinal.push({
    name: item,
    noOfBooks: numberOfBooks[index],
    awesomeness: awesomenessLevel[index]
  })
}, []);

Say I have three arrays depicting some names, number of books read and how awesome these people [in names] are:

let names = ["Mary", "Joe", "Kenan"];
let numberOfBooks = [2, 1, 4];
let awesomenessLevel = ["pretty cool", "meh", "super-reader"];

I'm trying to use .reduce() to bring them together to create an array of objects containing the relevant index in each array, but I am failing miserably:

    let people = [
    {
       name: "Mary",
       noOfBooks: 2,
       awesomeness: "pretty cool"
    },
    {
       name: "Joe",
       noOfBooks: 1,
       awesomeness: "meh"
    },
    {
       name: "Kenan",
       noOfBooks: 4,
       awesomeness: "super-reader"
    }
  ]

I got it with reduce as well:

let arrFinal = [];

 names.reduce(function(all, item, index) {
  arrFinal.push({
    name: item,
    noOfBooks: numberOfBooks[index],
    awesomeness: awesomenessLevel[index]
  })
}, []);

• javascript
• arrays
• reduce
• accumulator

• The way you do it with reducedoes not use the benefit of reduce: note how the callback does not return a value, and the all accumulator is not used, which are typical hallmarks of opting for reduce. You can just replace reduce with forEach in that code (and adjust the arguments -- without all), and it would not make a difference. – trincot Commented Apr 5, 2017 at 14:18
• you're pletely right, I just added them desperately :s. I guess I need to work more on how to use reduce properly – vzR Commented Apr 5, 2017 at 14:21

3 Answers 3

You could do it with map, like this:

let result = names.map( (v, i) => ({
    name: names[i], 
    noOfBooks: numberOfBooks[i],
    awesomenessLevel: awesomenessLevel[i]
}));

let names = ["Mary", "Joe", "Kenan"];
let numberOfBooks = [2, 1, 4];
let awesomenessLevel = ["pretty cool", "meh", "super-reader"];

let result = names.map( (v, i) => ({
    name: names[i], 
    noOfBooks: numberOfBooks[i],
    awesomenessLevel: awesomenessLevel[i]
}));

console.log(result);

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map works better than reduce in this case, because the number of elements you have in the names array (or any of the two others) is the same as the number of elements you need in the output. In that case it is more natural to use map.

Use map to create a 1-to-1 mapping between the input arrays and the output arrays.

let people = names.map(function (e, i) {
    return {name:e, noOfBooks:numberOfBooks[i],awesomeness: awesomenessLevel[i]};
});

let names = ["Mary", "Joe", "Kenan"];
let numberOfBooks = [2, 1, 4];
let awesomenessLevel = ["pretty cool", "meh", "super-reader"];

let people = names.map(function (e, i) {
    return {name:e, noOfBooks:numberOfBooks[i],awesomeness: awesomenessLevel[i]};
});


console.log(people);

You could use a dynamic approach by bining all arrays to one object and use the key names as property names for the result objects in the array

let names = ["Mary", "Joe", "Kenan"],
    numberOfBooks = [2, 1, 4],
    awesomenessLevel = ["pretty cool", "meh", "super-reader"],
    object = { name: names, noOfBooks: numberOfBooks, awesomeness: awesomenessLevel },
    result = Object.keys(object).reduce((r, k) =>
        (object[k].forEach((a, i) =>
            (r[i] = r[i] || {})[k] = a), r), []);

console.log(result);

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