Sort objects in an array by two different criteria

Question

I have an array that looks something like this:

     var arr = [{user: '3', cash: 2}, 
      {user: 'tim', cash: 3},
      {user: '5', cash: 2}, 
      {user: 'noah', cash: 3}]

I sort it by top earners like this:

arr.sort(function (a, b) {
    return b.tS - a.tS;
});

It works fine, but after I've sorted the guys with the highest cash I want to also sort everyone alphabetically by the user field. Keep in mind that some of the users may have numbers but of type String (not Number).

I can't use libraries and I'd prefer it to work as fast as possible machine-wise.

easy way: sort it twice, making sure the 2nd sort is stable. — dandavis
– dandavis, Commented Nov 11, 2016 at 20:33
I got it to work thanks to Nina, but is there also a way to to arrange them by cash and then by the order they were originally in? (if noah is last index with cash 3 and moves above '3' with cash 2, he should be below tim which is also with cash 3) — user1938653
– user1938653, Commented Nov 11, 2016 at 20:42
yes, look into a "stable sort js"; it's un-intuitively simple. — dandavis
– dandavis, Commented Nov 11, 2016 at 20:44

Nina Scholz · Accepted Answer · 2016-11-11 20:48:06Z

5

You could chain the sorting criteria.

The chaining works for every step where the former delta is zero. Then the next delta, or comparison function is evaluated and early returned, if the value is different from zero.

Here, with only two sort groups is the value returned, but for longer chains, the next comparison is made.

var arr = [{ user: '3', cash: 2 }, { user: 'tim', cash: 3 }, { user: '5', cash: 2 }, { user: 'noah', cash: 3 }];

arr.sort(function (a, b) {
    return b.cash - a.cash || a.user.localeCompare(b.user);
});

console.log(arr);

.as-console-wrapper { max-height: 100% !important; top: 0; }

To get a sort with index, you need to store the index in a temporary array and use sorting with map.

var array = [{ user: '3', cash: 2 }, { user: 'tim', cash: 3 }, { user: '5', cash: 2 }, { user: 'noah', cash: 3 }];

// temporary array holds objects with position and sort-value
var mapped = array.map(function(el, i) {
    return { index: i, cash: el.cash };
});

// sorting the mapped array containing the reduced values
mapped.sort(function(a, b) {
    return  b.cash - a.cash || a.index - b.index;
});

// container for the resulting order
var result = mapped.map(function(el){
    return array[el.index];
});

console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

edited Nov 11, 2016 at 20:48

answered Nov 11, 2016 at 20:21

Nina Scholz

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3 Comments

user1938653 Over a year ago

Is there also a way to arrange them by cash and then by the order they were originally in? (indexes)

user1938653 Over a year ago

I got it to work, but is there a cleaner and shorter way machine-wise? thanks btw

Nina Scholz Over a year ago

short would be to include the original index in each object. i get tim, noah, 3 and 5, which is the original orderm, grouped by cash.

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