Chunk array in javascript by adding only 1 element from next chunk

Question

I am looking to divide my array in JavaScript into chunk of 3. But I only want to increment the chunk by 1 element. Example below:

[1,2,3,4,5,6,7,8] => [1,2,3], [2,3,4], [3,4,5], [4,5,6], [5,6,7], [6,7,8]

All the resulted arrays should be of size 3. So only the one element is swiped to the left.

I have following code which divides the array into chunk of 3:

_.chunk($scope.substituteItems, 3);

This code divides the array as: [1,2,3,4,5,6,7,8] => [1,2,3], [3,4,5], [5,6,7], [8]

Obviously this code is only dividing the array into equal chunks which is not what I want.

Eddie · Accepted Answer · 2018-10-26 03:04:46Z

2

You can use reduce() to loop through the array. Use slice() to shallow copy the array into chunks.

let arr = [1, 2, 3, 4, 5, 6, 7, 8];
let numOfChunk = 3;

let result = arr.reduce((c, v, i, a) => {
  if (i + numOfChunk <= a.length) c.push(a.slice(i, i + numOfChunk));
  return c;
}, [])


console.log(result);

You can also use concat() instead of push() if you prefer one line of code.

let arr = [1, 2, 3, 4, 5, 6, 7, 8];
let numOfChunk = 3;

let result = arr.reduce((c, v, i, a) => i + numOfChunk <= a.length ? c.concat([a.slice(i, i + numOfChunk)]) : c, [])

console.log(result);

edited Oct 26, 2018 at 3:04

answered Oct 26, 2018 at 2:57

Eddie

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Comments

chiliNUT · Accepted Answer · 2018-10-26 02:58:34Z

2

using native js... If your array of n elements is to be chunked into chunks of size c sequentially as you describe, there will be n-c+1 total chunks, in your case 8-3+1=6, map the first n-c+1 elements of the array to the chunk containing itself + the next 2:

let c=3, arr=[1,2,3,4,5,6,7,8];
let result=arr.slice(0,arr.length-c+1).map((v,k)=>arr.slice(k,k+c));
console.log(result);

answered Oct 26, 2018 at 2:58

chiliNUT

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2 Comments

Arun Kumar Over a year ago

Also the code has to be compatible with IE. IE doesn't support let command.

chiliNUT Over a year ago

just use var instead of let and it should work exactly the same

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