Re-order 2d array based on another 2d array

Question

The following function re-orders an array based on the declared sortOrder.

const originalArray = ['Apple', 'Cat', 'Fan', 'Goat', 'Van', 'Zebra'];
const sortOrder = ['Zebra', 'Van'];
const sorter = (a, b) => {
   if(sortOrder.includes(a)){
      return -1;
   };
   if(sortOrder.includes(b)){
      return 1;
   };
   return 0;
};
originalArray.sort(sorter);
console.log(originalArray);

How can do correctly compare 2d arrays, given the following:

const originalArray = [['Apple',1], ['Cat',2], ['Fan',3], ['Goat',4], ['Van',5], ['Zebra',6]];
const sortOrder = [['Zebra',7], ['Van',8]];

The Bomb Squad · Accepted Answer · 2021-02-16 02:10:54Z

1

I used the Array.some method.

It returns true if any time the argument function's executed and it returns true

Therefore, if my argument function in the Array.some method asks for a comparison in the string value parts of the 2d array(array[index][0]) that means it would work JUST LIKE includes for what you have :D

const originalArray = [['Apple',1], ['Cat',2], ['Fan',3], ['Goat',4], ['Van',5], ['Zebra',6]];
const sortOrder = [['Zebra',7], ['Van',8]];
const sorter = (a, b) => {
   //the has function would return true if only ONCE the originalArray[index][0]==sortOrder[someIndex][0]
   function has(x){return sortOrder.some(e=>e[0]==x[0])}
   if(has(a)){
      return -1;
   };
   if(has(b)){
      return 1;
   };
   return 0;
};
originalArray.sort(sorter);
console.log(originalArray);

answered Feb 16, 2021 at 2:10

The Bomb Squad

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