return reordered version of array b based on reordered version of a

January 4, 2022

I have two arrays a and b like this:

const a = [1,2,3,4]
const b = ['1','2','3','4'] // could be 'a','b','c','d'

const reordered_a = [4,1,3,2] // based on this reordering 

function reorder_b() {
    // should return ['4','1','3','2']
}

How can I return reordered version of b based on reordered positions in reordered_a?

>Solution :

You could take an object with the indices for the values and map the pattern with the values of the indices.

const a = [1, 2, 3, 4]
const b = ['1', '2', '3', '4'] // could be 'a','b','c','d'

const reordered_a = [4, 1, 3, 2] // based on this reordering 

function reorder_b() {
   const references = Object.fromEntries(Object.entries(a).map(a => a.reverse()));
   return reordered_a.map(k => b[references[k]]);
}

console.log(reorder_b()); // 4 1 3 2