Javascript Unexpected Sorting Order

Question

Using Chrome Version 66.0.3359.117 (Official Build) (64-bit)

Sorting the below array I'm expecting an item with id = 1 to be at the top, which isn't the case. If I reduce the number of items down to 2, sorting works as expected. Can anyone explain why do I get a non deterministic result here?

     let array = [
          { "id": 1, "path": "01.00.00.00.00.00.00" },
          { "id": 2, "path": "01.02.00.00.00.00.00" },
          { "id": 3, "path": "01.02.03.00.00.00.00" },
          { "id": 4, "path": "01.02.04.00.00.00.00" },
          { "id": 5, "path": "01.02.05.00.00.00.00" },
          { "id": 6, "path": "01.02.06.00.00.00.00" },
          { "id": 7, "path": "01.02.05.07.00.00.00" },
          { "id": 8, "path": "01.02.05.07.08.00.00" },
          { "id": 9, "path": "01.02.05.07.08.09.00" },
          { "id": 10, "path": "01.02.04.10.00.00.00" },
          { "id": 11, "path": "01.02.05.07.08.09.11" },
          { "id": 12, "path": "01.02.04.10.12.00.00" }
        ];
        
        array.sort((f, s) => f.path > s.path);
        console.error("Full", array[0].id);
    
        array = [
          { "id": 1, "path": "01.00.00.00.00.00.00" },
          { "id": 2, "path": "01.02.00.00.00.00.00" }
          //{ "id": 3, "path": "01.02.03.00.00.00.00" },
          //{ "id": 4, "path": "01.02.04.00.00.00.00" },
          //{ "id": 5, "path": "01.02.05.00.00.00.00" },
          //{ "id": 6, "path": "01.02.06.00.00.00.00" },
          //{ "id": 7, "path": "01.02.05.07.00.00.00" },
          //{ "id": 8, "path": "01.02.05.07.08.00.00" },
          //{ "id": 9, "path": "01.02.05.07.08.09.00" },
          //{ "id": 10, "path": "01.02.04.10.00.00.00" },
          //{ "id": 11, "path": "01.02.05.07.08.09.11" },
          //{ "id": 12, "path": "01.02.04.10.12.00.00" }
        ];
    
        array.sort((f, s) => f.path > s.path);
        console.error("Reduced", array[0].id);

If it's worth anything, on Firefox 59 the results are as you are expecting them, i.e. the first item does have `id: 1`. — devius, May 04 '18 at 13:28
Doesn't `sort()` expect you to return a number instead of a boolean? You return true or false, which will be cast to 0 or 1. So you never return -1 or a different negative number, so records will not be sorted correctly if one record belongs before another. — Shilly, May 04 '18 at 13:31
Possible duplicate of [How to sort strings in JavaScript](https://stackoverflow.com/questions/51165/how-to-sort-strings-in-javascript) — vbriand, May 04 '18 at 13:40
Please specify if you want to sort by id or by path value. AS I mention before, you need to return a number inside a sort, so if you sort on id, just subtract them. If you want to sort by path though, you need another function to calculate the value, since number order isn't the same as string order. Compare `1000 > 999` and `'1000' > '999'`. So if you need to compare paths, you need to cast the path back to some numerical value. ( I usually use spluit on the dot and multiply the index, so items that start with 01 will be a quadrillion something, etc... ) — Shilly, May 04 '18 at 13:40

Rihards Fridrihsons · Answer 1 · 2018-05-04T13:53:07.743

2

You have to provide 3 options

array.sort((f, s) => {
  if(f.path < s.path){return -1}
  if(f.path > s.path){return 1} 
  return 0;
});

Basically js comparing function needs to return an integer, since '>' returns a boolean, then true == 1 and false == 0, so when false is returned it is interpreted as the elements are equal, instead of -1 one is maller than the other.

edited May 04 '18 at 13:53

answered May 04 '18 at 13:40

Rihards Fridrihsons

854
5
11

Should be the same as `return s.path - f.path`, no? – mplungjan May 04 '18 at 14:40
@mplungjan no, js cant convert "00.20.10" to number, therefore it returns NaN – Rihards Fridrihsons May 04 '18 at 14:43
Ah - I can test < or > since that is in ascii order, I actually thought it would work with "subtraction" too, but never had to try it. Works on ID though – mplungjan May 04 '18 at 14:47

score 0 · Answer 2 · answered May 04 '18 at 13:30

0

First, you are comparing two string as like they are numbers, so I think the compiler transforms the string to numbers and then compare them when you do :

f.path > s.path

answered May 04 '18 at 13:30

TBenchikhi

11
2

hehe - why does it do differently for 2 arrays?:) – Simple.Js May 04 '18 at 13:31
this is the logic you are comparing two strings as they were numbers, so you will see some unexpected behaviors – TBenchikhi May 04 '18 at 13:34

score 0 · Answer 3 · answered May 04 '18 at 13:30

0

You are sorting by path, not id.

Try array.sort((f, s) => f.id - s.id);

answered May 04 '18 at 13:30

fethe

677
6
12

score 0 · Accepted Answer · answered May 04 '18 at 13:33

You need to change your sort condition to f.id - s.id which is incorrect in your code.

let array = [
  { "id": 1, "path": "01.00.00.00.00.00.00" },
  { "id": 2, "path": "01.02.00.00.00.00.00" },
  { "id": 3, "path": "01.02.03.00.00.00.00" },
  { "id": 4, "path": "01.02.04.00.00.00.00" },
  { "id": 5, "path": "01.02.05.00.00.00.00" },
  { "id": 6, "path": "01.02.06.00.00.00.00" },
  { "id": 7, "path": "01.02.05.07.00.00.00" },
  { "id": 8, "path": "01.02.05.07.08.00.00" },
  { "id": 9, "path": "01.02.05.07.08.09.00" },
  { "id": 10, "path": "01.02.04.10.00.00.00" },
  { "id": 11, "path": "01.02.05.07.08.09.11" },
  { "id": 12, "path": "01.02.04.10.12.00.00" }
];

array.sort((f, s) => f.id - s.id);
console.error("Full", array[0].id);

that works, still curious'ed as to why do I get different results for the same inputs? — Simple.Js, May 04 '18 at 13:36

Javascript Unexpected Sorting Order

4 Answers4