Alphanumerical / Natural Sort an Array with leading zeroes in Javascript

Question

I need to sort through an array of strings that contain letters / numbers / and leading zeros. If null it must go to the top of the list and appear as an empty string. I've looked and researched many different ways but nothing that sorts the way I'm expecting it to.

Initial Array:

input = [
    { value: null },
    { value: '' },
    { value: '-' },
    { value: '0' },
    { value: 'A1' },
    { value: '0001' },
    { value: '01' },
    { value: '1' },
    { value: '01.008A' },
    { value: '01.008' },
    { value: '02' },
    { value: 'A' },
    { value: '10' },
    { value: '100A1' },
    { value: '10-1' },
    { value: '100' },
    { value: '001' },
    { value: '2' },
  ]

Expected Output:

" "
" "
"-"
"0"
"0001"
"001"
"01"
"01.008"
"01.008A"
"1"
"02"
"2"
"10"
"10-1"
"100"
"100A1"
"A"
"A1"

EDIT: (My Original Code)

function sortAlphanumeric<T>(this: Array<T>, property: string): Array<T> {
  var collator = new Intl.Collator(undefined, { numeric: true, sensitivity: 'base' });

  function dynamicSort(property) {
    var sortOrder = 1;

    if (property[0] === "-") {
      sortOrder = -1;
      property = property.substr(1);
    }

    return function (a, b) {
      if (sortOrder == -1) {
        return b[property].localeCompare(a[property]);
      } else {
        return a[property].localeCompare(b[property]);
      }
    }
  }
  return this.sort(dynamicSort(property)).sort((a, b) => {
    var nullToEmptyString = '';

    if (a[property] === null) {
      a[property] = nullToEmptyString;
    }

    if (b[property] === null) {
      b[property] = nullToEmptyString;
    }
    return collator.compare(a[property], b[property]);
  });
}

None of the methods in [this thread](https://stackoverflow.com/questions/2802341/javascript-natural-sort-of-alphanumerical-strings) worked? — Jeremy Thille, Jan 09 '20 at 16:02
you need to add a replacement for `null`, like `''`. then the duplicate tartget works. — Nina Scholz, Jan 09 '20 at 16:03
*and appear as an empty string*, that is not sorting, that is mapping. — Titus, Jan 09 '20 at 16:07
Sorry first time posting here. I edited in my original code. — Revolver_FOX, Jan 09 '20 at 16:43

score 0 · Accepted Answer · answered Jan 09 '20 at 16:11

0

This should do it:

input.sort(({value:a}, {value:b}) => {
  const ai = parseInt(a, 10), bi = parseInt(b, 10);
  return (b == null) - (a == null)
    || (ai - bi)
    || (a > b) - (b > a);
}).map(x => x.value || "")

answered Jan 09 '20 at 16:11

Bergi

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Why not map the array before sorting it ? Also, for string comparison there is [.localeCompare(..)](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/localeCompare) – Titus Jan 09 '20 at 16:20
@Titus The `map` was more of an afterthought for "*null must appear as an empty string*", I consider it optional. I'm not using `localeCompare` because the comparison under some locale is rather unpredictable, I value the simple standard string comparison. – Bergi Jan 09 '20 at 16:29
Amazing solution. This has been breaking my head. – Revolver_FOX Jan 09 '20 at 16:33

Alphanumerical / Natural Sort an Array with leading zeroes in Javascript

1 Answers1