将数组相乘的DRYing函数

Question

I have an array of 100 elements, that I have broken into 10 smaller arrays in order to quadruplicate the original 100 elements into an array of 400.

I am currently using :

function quadulator(arrChunks){
let a = []; 
let b = [];

    for(var i = 0; i< arrChunks.length;++i){
        a.push(arrChunks[i]);
        a.push(arrChunks[i]);
        b.push(arrChunks[i]);
        b.push(arrChunks[i]);
}  

return a.concat(b);

}
let group1 = quadulator(grid100[0]);
let group2 = quadulator(grid100[1]);
let group3 = quadulator(grid100[2]);
let group4 = quadulator(grid100[3]);
let group5 = quadulator(grid100[4]);
let group5 = quadulator(grid100[5])

let newArr = group1.concat(group2,group3,group4,group5);

This does exactly what I want it to do ,but Im looking for a way to eliminate the repetition.

The reason I am splitting it up into multiple parts is because I need the element of index 0 from the original array, to repeat itself in the indexes of 0,1,20, and 21 of the new array, along with every other element to follow this pattern.

Answer 1

You can improve the quadulator function by noticing that a and b are the same. You can create a single array at once that represents both of them with Array.from , and then spread that array twice.

Then, with your group s, you can map grid100 by the quadulator function, and spread it into [].concat to flatten:

function quadulator(arrChunks) {
  const chunksDoubled = Array.from(
    { length: arrChunks.length * 2 },
    (_, i) => arrChunks[Math.floor(i / 2)]
  );
  return [...chunksDoubled, ...chunksDoubled];
}
const newArr = [].concat(...grid100.map(quadulator));

Demo:

 const grid100 = [ [1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16] ]; function quadulator(arrChunks) { const chunksDoubled = Array.from( { length: arrChunks.length * 2 }, (_, i) => arrChunks[Math.floor(i / 2)] ); return [...chunksDoubled, ...chunksDoubled]; } const newArr = [].concat(...grid100.map(quadulator)); console.log(newArr);