在最接近的解决方案之后排序数组

Question

I have an unsorted array with results of an calculation:

const solution = 955

const unsortedArray = 
[
  [["solution result cab"],[955.709]],
  [["solution result abc"],[951.11]],
  [["solution result cba"],[954.709]],
  [["solution result bac"],[957.709]]
]

I want to sort the array after +/- the closest deviation of the solution:

[
  [["solution result cab"],[955.709]],
  [["solution result cba"],[954.709]],
  [["solution result bac"],[957.709]],
  [["solution result abc"],[951.11]]
]

Is there some way to apply a range comparison into sort() or is there a better solution?

Answer 1

You can make a compare function that will compare the two elements' absolute differences from the solution. A positive number means the first element is bigger. A negative one means the second is bigger.

function compare(a, b) {
  const aAbs = Math.abs(a[1][0] - solution);
  const bAbs = Math.abs(b[1][0] - solution);
  return aAbs - bAbs;
}

Then you can run unsortedArray.sort(compare) .

Answer 2

Per the documentation of Array.prototype.sort , you can supply a sorting function:

const compareDeviationFrom = (solution) => {
  (resA, resB) => Math.abs(resA[1][0] - solution) - Math.abs(resB[1][0] - solution)
}

unsortedArray.sort(compareDeviationFrom(solution))