Typescript subtract two boolean

Question

I have an array :

main = [{
           data: x,
           numberField:  1;
 }, {      data: y,
           numberField:  2;
 }, {      data: x,
           numberField:  3;
 },
  {        data:z
           numberField:  4;
 },
    {      data: q
           numberField:  5;
 }]
fixedElements = [ 3, 5]

I need to have :

 fixedElements .includes(a.numberField) - fixedElements .includes(b.numberField) 

Because I need to check whether two coming values are in the array a and not sort them, but sort the rest. when I try to do this typescript throws an error

  left/right hand-side of the argument needs to be of type any or number.

My initial sort function as :

  sort(a,b) {
  if(a.numberField > b.numberField)
       return -1

  if(a.numberField < b.numberField)
     return 1;

    }

So I wanted to check whether the coming a or b are in the fixedElements array, and leave them last. Is there any other way to do this?

Answer 1

You can cast the booleans to numbers by using the + sign (typescript playground demo):

+fixedElements.includes(a.numberField) - +fixedElements.includes(b.numberField)`

And this is the sort logic (typescript playgroud - check console):

const main = [{"data":"x","numberField":1},{"data":"y","numberField":2},{"data":"x","numberField":3},{"data":"z","numberField":4},{"data":"q","numberField":5}]

// I'm using a Set instead of an array, because Set.has is faster than Array.includes
const fixedElements = new Set([3, 5])

main.sort((a, b) => {
  const aF = fixedElements.has(a.numberField)
  const bF = fixedElements.has(b.numberField)
  
  if(!aF && !bF) return b.numberField - a.numberField
  
  return +aF - +bF
})

console.log(main)