Sort one array based on the order of a separate array, when some items are missing?-CodePudding

I have an array that stores a prefered order, and a larger array that I need to sort:

const months = ['March', 'Jan', 'Feb', 'Dec'];
const prefOrder = ['Feb', 'March']
months.sort((a, b) => (prefOrder.indexOf(a) - prefOrder.indexOf(b)));
// desired output: ['Feb', 'March', 'Dec', 'Jan']
// or: ['Feb', 'March', 'Jan', 'Dec' ]
console.log(months);

However, when two items being compared are not in the list (i.e. (prefOrder.indexOf(a) - prefOrder.indexOf(b)) gives (-1) - (-1) = 0)) rather than them being at the end, they end up at the start... even though if you a not in the prefered order list, and b in the list, you'd get (-1) - positive = negative so a should be sorted after b.

How can I get use the preferred order list to put unlisted item at the end?

CodePudding user response：

You could add one for getting a falsy value for unknown values and take a large value instead to move this items to the end of the known items.

const
    months = ['March', 'Jan', 'Feb', 'Dec'],
    prefOrder = ['Feb', 'March'];

months.sort((a, b) =>
    (prefOrder.indexOf(a)   1 || Number.MAX_VALUE) -
    (prefOrder.indexOf(b)   1 || Number.MAX_VALUE)
);

console.log(months);

Instead of using an array, take an object with wanted order and move unknown items to the end with a large default value.

const
    months = ['March', 'Jan', 'Feb', 'Dec'],
    order = { Feb: 1, March: 2 };

months.sort((a, b) =>
    (order[a] || Number.MAX_VALUE) -
    (order[b] || Number.MAX_VALUE)
);

console.log(months);

CodePudding user response：

I will do that this way...

const
  months    = ['March', 'Jan', 'Feb', 'Dec']
, prefOrder = ['Feb', 'March']
  ;

months.sort((a, b) =>
  { 
  if (!prefOrder.includes(a)) return  1
  if (!prefOrder.includes(b)) return -1
  return (prefOrder.indexOf(a) - prefOrder.indexOf(b))  
  })

console.log( JSON.stringify( months ))