CODEWITHSUNDEEP
javascriptunderscore.js
Kalashir
Kalashir

Reputation: 1127

Group array values in group of 3 objects in each array using underscore.js

Below is an array in which I have to group 3 values in each object:

var xyz = {"name": ["hi","hello","when","test","then","that","now"]};

Output should be below array:

[["hi","hello","when"],["test","then","that"],["now"]]

Upvotes: 15

Views: 18712

Answers (8)

Avare Kodcu
Avare Kodcu

Reputation: 396

Pure javascript code:

function groupArr(data, n) {
    const group = [];
    for (let i = 0, j = 0; i < data.length; i++) {
        if (i >= n && i % n === 0)
            j++;
        group[j] = group[j] || [];
        group[j].push(data[i])
    }
    return group;
}

groupArr([1,2,3,4,5,6,7,8,9,10,11,12], 3);

Upvotes: 16

ontanj
ontanj

Reputation: 1619

Here is another simple oneliner, quite similar to the solution of gtournie.

  1. We create an array of the desired length array.length / n.
  2. We map portions of the initial array to the new array elements.
const group = (array, n) => 
  [...Array(Math.ceil(array.length / n))]
    .map((el, i) => array.slice(i * n, (i + 1) * n));

var xyz = {"name": ["hi","hello","when","test","then","that","now"]};
group(xyz.name, 3)

gives

[["hi","hello","when"],["test","then","that"],["now"]]

Upvotes: 3

Penny Liu
Penny Liu

Reputation: 17408

This can be covered by lodash _.chunk:

var xyz = {"name": ["hi","hello","when","test","then","that","now"]},size = 3;
console.log(_.chunk(xyz.name, size));
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.js"></script>

Upvotes: 7

Geza Kerecsenyi
Geza Kerecsenyi

Reputation: 1218

Here's a short and simple solution abusing the fact that .push always returns 1 (and 1 == true):

const arr = [0, 1, 2, 3, 4, 5, 6]
const n = 3

arr.reduce((r, e, i) =>
    (i % n ? r[r.length - 1].push(e) : r.push([e])) && r
, []); // => [[0, 1, 2], [3, 4, 5], [6]]

Plus, this one requires no libraries, in case someone is looking for a one-liner pure-JS solution.

Upvotes: 17

Charles Loder
Charles Loder

Reputation: 85

I ran into this same problem and came up with solution using vanilla js and recursion

const groupArr = (arr, size) => {
    let testArr = [];
    const createGroup = (arr, size) => {
        // base case
        if (arr.length <= size) {
            testArr.push(arr);
        } else {
            let group = arr.slice(0, size);
            let remainder = arr.slice(size);
            testArr.push(group);
            createGroup(remainder, size);
        }
    }
    createGroup(arr, size);
    return testArr;
}

let data = [1, 2, 3, 4, 5, 6, 7, 8, 9];
console.log(groupArr(data, 3));
>>> [[1, 2, 3], [4, 5, 6], [7, 8, 9]]

Upvotes: 1

Fauxsaurus
Fauxsaurus

Reputation: 520

Here's a curry-able version that builds off Avare Kodcu's Answer.

function groupBy(groupSize,rtn,item,i)
{
    const j=Math.floor(i/groupSize)

    !rtn[j]?rtn[j]=[item]:
            rtn[j].push(item)

    return rtn
}

arrayOfWords.reduce(curry(groupBy,3),[])

Upvotes: 1

gtournie
gtournie

Reputation: 4382

You may use:

function groupBy(arr, n) {
  var group = [];
  for (var i = 0, end = arr.length / n; i < end; ++i)
    group.push(arr.slice(i * n, (i + 1) * n));
  return group;
}

console.log(groupBy([1, 2, 3, 4, 5, 6, 7, 8], 3));

Upvotes: 4

Gayathri Mohan
Gayathri Mohan

Reputation: 2962

Hi please refer this https://plnkr.co/edit/3LBcBoM7UP6BZuOiorKe?p=preview. for refrence Split javascript array in chunks using underscore.js

using underscore you can do

JS

 var data = ["a1", "a2", "a3", "a4", "a5", "a6", "a7", "a8", "a9", "a10", "a11", "a12", "a13"];
var n = 3;
var lists = _.groupBy(data, function(element, index){
  return Math.floor(index/n);
});
lists = _.toArray(lists); //Added this to convert the returned object to an array.
console.log(lists);

or

Using the chain wrapper method you can combine the two statements as below:

var data = ["a1", "a2", "a3", "a4", "a5", "a6", "a7", "a8", "a9", "a10", "a11", "a12", "a13"];
var n = 3;
var lists = _.chain(data).groupBy(function(element, index){
  return Math.floor(index/n);
}).toArray()
.value();

Upvotes: 3

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