js - order arrays' elements inside an array based on an ordered array [duplicate]

Question

I have an ordered array and 4 arrays inside another array. I want to order each array based on the ordered array.

(this is just an example, I need a code that will work with any kind of elements inside the arrays).

let ordered_array = ['a','b','c','d','e','f','g','h','i','j','k','l'];
  let array = [['d','c','a'],['i','k','b'],['f','e','h'],['l','g','j']];
  //order each array in 'array'
  for(let i = 0; i < array.length; i++){
    for(let l = 0; l < array[i].length; l++){
      document.write(array[i][l]);
    }
    document.write('</br>');
  }
  /*
  expected output (after order):
  acd 
  bik
  efh
  gjl
  */

Nina Scholz · Accepted Answer · 2018-07-08 14:31:08Z

0

You could take a Map and sort by the indices.

var ordered_array = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l'],
    array = [['d', 'c', 'a'], ['i', 'k', 'b'], ['f', 'e', 'h'], ['l', 'g', 'j']],
    order = ordered_array.reduce((m, v, i) => m.set(v, i), new Map);
    
array.forEach(t => t.sort((a, b) => order.get(a) - order.get(b)));

console.log(array);

.as-console-wrapper { max-height: 100% !important; top: 0; }

edited Jul 8, 2018 at 14:31

answered Jul 8, 2018 at 14:30

Nina Scholz

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