CODEWITHSUNDEEP
javascriptjqueryarrays
adhirath wcities
adhirath wcities

Reputation: 71

jquery split array with last element of a chunk being first element of the next chunk

I want to split an array in jquery so that after splitting, the last element for previous array should be the first element of current array.

var arr = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16];

If I do splice then it will splice exactly same size and next array will be next element. so if I am splicing it into 5 my output should be

var firstArr = arr.splice[0,4];
var secondArr = arr.splice[0,4];
var thirdArr = arr.splice[0,4];
var fourthArr = arr.splice[0,4];
var fifthArr = arr.splice[0,4];

Right now I am getting

[1,2,3,4]
[5,6,7,8]
...

What I need is

//firstArr = [1,2,3,4];
//secondArr = [4,5,6,7];
//thirdArr = [7,8,9,10];
//fourthArr = [10,11,12,13];
//fifthArr = [13,14,15,16];

How to achieve this in jQuery ?

Upvotes: 1

Views: 412

Answers (3)

Exodus 4D
Exodus 4D

Reputation: 2822

const chunk = (a1, s) => {
  const a2 = [];
  let i = k = j = 0;
  while (s > 1 && a1[i - k]) {
    j = i % s;
    k = (i - j) / s
    a2[k] ??= [];
    a2[k][j] = a1[i - k];
    i++;
  }
  return a2;
}



console.log(chunk([1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16], 4));
console.log(chunk([1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16], 7));
console.log(chunk([1, 2, 3, 4, 5, 6, 7, 8, 9], 3));
console.log(chunk([1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 3));
console.log(chunk([1, 2, 3], 8));
console.log(chunk([], 10));
console.log(chunk([1, 2, 3, 4], 0));
console.log(chunk(['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i'], 4));

Upvotes: 0

Robin Mackenzie
Robin Mackenzie

Reputation: 19319

I don't know a array function that is preferable to using a plain javascript solution. Below is an example that generically handles your requirement by:

  • taking the number of chunks as the ceil of: the length of input array plus the floor number of duplicated entries, all divided by the size of the output chunk
  • uses slice such that the last element for previous array is the first element of current array.

// chunking function
function chunk(arr, size) {
  if (size === 0) return [];
  var arrCount = Math.ceil((arr.length + Math.floor(arr.length / size)) / size);
  var chunks = [];
  for (let i=0; i<arrCount; ++i) {
    chunks.push(arr.slice(i * (size - 1), i * (size - 1) + size));
  }
  return chunks;
}

// tests
console.log(chunk([1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16], 4));
console.log(chunk([1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16], 7));
console.log(chunk([1,2,3,4,5,6,7,8,9], 3));
console.log(chunk([1,2,3,4,5,6,7,8,9,10], 3));
console.log(chunk([1,2,3], 8));
console.log(chunk([], 10));
console.log(chunk([1, 2, 3, 4], 0));
console.log(chunk(['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i'], 4));
.as-console-wrapper { max-height: 100% !important; top: 0; }

Upvotes: 2

devil1321
devil1321

Reputation: 19

const arr = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16]

let slices = []

let start = 0
let end = 3
for(i = 0; i < 5; i++){
    let slice = arr.slice(start,end)
    slice.push(arr[end])
    start += 3
    end += 3
    slices.push(slice)
}

Upvotes: 0

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