Javascript - Sort object by multiple values

Question

var data = {
  "input": [{
      "countA": 1,
      "countB": 10
    },
    {
      "countA": 15,
      "countB": 13
    },
    {
      "countA": 26,
      "countB": 24
    },
    {
      "countA": 6,
      "countB": 25
    },
    {
      "countA": 15,
      "countB": 20
    }
  ]
};

var sorted = data.input.sort(function(a, b) {
  return a['countB'] < b['countB'] ? 1 : -1;
});

console.log(sorted);

The outcome after the first sorting should be after another sorting:

[
    {
        "countA": 6,
        "countB": 25
    },
    {
        "countA": 15,
        "countB": 20
    },
    {
        "countA": 1,
        "countB": 10
    }
    {
        "countA": 26,
        "countB": 24
    },
    {
        "countA": 15,
        "countB": 13
    }
]

So, it should be the highest of "countB" and then descending as long as "countB" is higher than "countA". So far I tried multiple ways, but there's no outcome so far.

Thanks for any help!

*"... as long as "countB" is higher than "countA"."* which `"countA"` and `"countB"`? `a`'s or `b`'s? And then, what should be the comparison? — Cid, Aug 30 '21 at 17:20
Telling us your use case or your exercise statement will help us to understand what you're trying to achieve — Cid, Aug 30 '21 at 17:24
The integer of "countB" always has to be higher than the integer of "countA", that's the idea behind my post. So, whenever "countA" is higher than "countB", "countA" should be not relevant for the sorting. — Hols1990, Aug 30 '21 at 17:25
What to do with the values then when `"countb" < "countA"`? should they be randomly placed at the end of the array? — Cid, Aug 30 '21 at 17:25
@Cid: That's exactly the question...That's why I think that another sorting after the first sorting as shown "return a['countB'] < b['countB'] ? 1 : -1;" is needed. But I can't find the solution for it. — Hols1990, Aug 30 '21 at 17:27
Does this answer your question? [How to sort an array of objects by multiple fields?](https://stackoverflow.com/questions/6913512/how-to-sort-an-array-of-objects-by-multiple-fields) — pilchard, Aug 31 '21 at 09:28

score 1 · Accepted Answer · answered Aug 30 '21 at 17:37

1

You could sort by the result of the comparison of countB > countA and then by the value of countB.

const
    data = [{ countA: 1, countB: 10 }, { countA: 15, countB: 13 }, { countA: 26, countB: 24 }, { countA: 6, countB: 25 }, { countA: 15, countB: 20 }];

data.sort((a, b) =>
    (b.countB > b.countA) - (a.countB > a.countA) ||
    b.countB - a.countB
);

console.log(data);

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

answered Aug 30 '21 at 17:37

Nina Scholz

376,160
25
347
392

Thanks @Nina Scholz, that's the answer that helped me the most to understand the sorting! – Hols1990 Aug 30 '21 at 17:53

score 0 · Answer 2 · answered Aug 30 '21 at 17:37

You can first extract the values of the array when "countb" >= "countA", sort that array then add the remaining values at the end (note that for order is kept for when "countb" < "countA") :

var data = {
  "input": [{
      "countA": 1,
      "countB": 10
    },
    {
      "countA": 15,
      "countB": 13
    },
    {
      "countA": 26,
      "countB": 24
    },
    {
      "countA": 6,
      "countB": 25
    },
    {
      "countA": 15,
      "countB": 20
    }
  ]
};

const ElementsToSort = data.input.filter(elem => elem.countA <= elem.countB);
const RemainingElements = data.input.filter(elem => ElementsToSort.indexOf(elem) < 0);

// sort as you did
const PartiallySorted = ElementsToSort.sort(function(a, b) {
  return a['countB'] < b['countB']
          ? 1
          : a['countB'] > b['countB']
            ? -1
            : 0;
});


//add the remaining values
const sorted = PartiallySorted.concat(RemainingElements);
console.log(sorted);

Javascript - Sort object by multiple values

2 Answers2