javascript sort and sort equals on result. how?

Question

I have a problem sorting objects of structure {"name", "count"}:

1. name => "aaa", count => 1
2. name => "bbb", count => 2
3. name => "ccc", count => 3
4. name => "ddd", count => 1
5. name => "eee", count => 1

I need to sort this on "count" (small to big, then big to small).

My sort function:

mass.sort(compareElements);

function compareElements(a, b)
{
 if(a < b) return -1;
 else if(a > b) return 1;
 else return 0;
}

On first call (small to big) I get one result 'res1'
on second call (big to small) I get result 'res2'
on third call (small to big again) I get 'res3' !== 'res1' !!

The order of the elements in res1 and res3 is not the same. I need them to be the same.

score 4 · Accepted Answer · answered Aug 28 '13 at 13:36

4

To make you sort stable, you will need to compare the "equal" items by their index:

// mass = [{name:…, count:…}, {name:…, count:…}, …]
for (var i=0; i<mass.length; i++)
    mass[i].index = i;
mass.sort(function(a, b) {
    return compareElements(a, b) || a.index - b.index;
});
function compareElements(a, b) {
    // something
    return a.count - b.count;
}

answered Aug 28 '13 at 13:36

Bergi

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mass.sort((a, b) => { return a.count.toString().localeCompare(b.count.toString()) }) Isn't this much simpler? – Vimal Maheedharan Nov 13 '18 at 04:52
@VimalMahi No, that does a completely different thing - and is not stable – Bergi Nov 13 '18 at 05:25
OK. I will dig more deep into that – Vimal Maheedharan Nov 13 '18 at 08:49

score 2 · Answer 2 · answered Aug 28 '13 at 13:30

2

If your objects are like:

var arrObj = {
   name: "aaa",
   count: 1
};

Use this function for comparing:

function compareElements(a, b)
{
  return a.count - b.count;
}

answered Aug 28 '13 at 13:30

Artyom Neustroev

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javascript sort and sort equals on result. how?

2 Answers2