Sorting an array of objects by multiple criteria with null values

Question

Alright, so I have an array of objects that includes null values for a certain property.

The object looks roughly like this for sorting purposes... (40 elements, but this will suffice...).

It needs to be sorted based on roulette descending (with roulette sometimes being null), then novelty, then popularity.

My head is getting a bit crushed.

This works to sort the roulette in descending, but how do I need to extend it to include the other two criteria?

Object:

[
 {
  title: 'one',
  popularity: 4,
  novelty: 3
 },
 {
  title: 'two',
  popularity: 1
  novelty: 4
 },
 {
  title: 'three',
  popularity: 5,
  novelty: 3,
  roulette: 0
 },
 {
  title: 'four',
  popularity: 5,
  novelty: 3,
  roulette: 1
 }
]

Partially working function:

  object.sort(function(a, b) {
    if (a['roulette'] == null) return 1
    if (b['roulette'] == null) return -1
    if (a['roulette'] === b['roulette']) return 0
    return b.roulette > a.roulette ? 1 : -1
  });

please add some more data and the wanted result. i am sorry i can not read what novelty means in sorting. — Nina Scholz
– Nina Scholz, Commented Nov 11, 2016 at 7:04
@NinaScholz according to his partial sort function null values will come at bottom. — Sikorski
– Sikorski, Commented Nov 11, 2016 at 7:06
are there null values in roulette or is roulette just not set? — Nina Scholz
– Nina Scholz, Commented Nov 11, 2016 at 7:23

Nina Scholz · Accepted Answer · 2016-11-11 07:52:31Z

3

An attempt with sorting with priority and groups.

var data = [{ title: 'one', popularity: 4, novelty: 3 }, { title: 'two', popularity: 1, novelty: 4 }, { title: 'three', popularity: 5, novelty: 3, roulette: 0 }, { title: 'four', popularity: 5, novelty: 3, roulette: 1 }, { title: 'five', popularity: 5, novelty: 4, roulette: null }, { title: 'six', popularity: 5, novelty: 5, roulette: undefined }];

data.sort(function (a, b) {
    return (
        (a.roulette === undefined || a.roulette === null) - (b.roulette === undefined || b.roulette === null) ||
        a.roulette - b.roulette ||
        a.novelty - b.novelty ||
        a.popularity - b.popularity
    );       
});

console.log(data);

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

edited Nov 11, 2016 at 7:52

answered Nov 11, 2016 at 7:12

Nina Scholz

388k26 gold badges367 silver badges417 bronze badges

Sign up to request clarification or add additional context in comments.

4 Comments

Ouroborus Over a year ago

Shouldn't the bs be on the left? Otherwise, that's a pretty slick solution.

Ouroborus Over a year ago

Both 'roulette' in {roulette:null} and 'roulette' in {roulette:undefined} are true so those will always come out true unless roulette hasn't been set at all.

Nina Scholz Over a year ago

it not clear from the data, if it is set with null or undefined, not at all inside of the object.

Tulun Over a year ago

This is like... 99.9% there. There is a minor side effect for the first undefined/null data point, but it's minor enough for me not to care. I needed biggest on top, so slight mod: return ( (a.roulette === undefined || a.roulette === null) - (b.roulette === undefined || b.roulette === null) || b.roulette - a.roulette || b.novelty - a.novelty || b.popularity - a.popularity );

kukkuz · Accepted Answer · 2016-11-11 14:23:02Z

Here is the preferential sorting (descending) with roulette, novelty and popularity (in that order)

This handles both null and undefined - check out the demo below:

var object=[{title:"one",popularity:4,novelty:3},{title:"two",popularity:1,novelty:4},{title:"three",popularity:5,novelty:3,roulette:0},{title:"four",popularity:5,novelty:3,roulette:1},{title:"five",roulette:4,novelty:null},{title:"six",popuplarity:7},{title:"seven",novelty:8,roulette:null},{title:"eight",novelty:0},{title:"nine",popularity:10}];

function ifNumber(num) {
  if(num == undefined || num == null)
    return -Infinity;
  else
    return num;
}

var result = object.sort(function(a, b) {
  return (ifNumber(b.roulette) - ifNumber(a.roulette)) 
      || (ifNumber(b.novelty) - ifNumber(a.novelty))
      || (ifNumber(b.popuplarity) - ifNumber(a.popuplarity));
});
console.log(result);

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

Rajesh Dixit · Accepted Answer · 2016-11-11 06:56:46Z

You can try sorting based on weighted rank:

var data=[{title:"one",popularity:4,novelty:3},{title:"two",popularity:1,novelty:4},{title:"three",popularity:5,novelty:3,roulette:0},{title:"four",popularity:5,novelty:3,roulette:1}];

data.sort(function(a, b) {
  var r1 = a.roulette === undefined ? -1 : a.roulette;
  var r2 = b.roulette === undefined ? -1 : b.roulette;
  var n1 = a.novelty === undefined ? -1 : a.novelty;
  var n2 = b.novelty === undefined ? -1 : b.novelty;
  var p1 = a.popularity === undefined ? -1 : a.popularity;
  var p2 = b.popularity === undefined ? -1 : b.popularity;

  var r_rank = r1 > r2 ? -100 : r1 < r2 ? 100 : 0;
  var n_rank = n1 > n2 ? -10 : n1 < n2 ? 10 : 0;
  var p_rank = p1 > p2 ? -1 : p1 < p2 ? 1 : 0;

  return r_rank + n_rank + p_rank;
})

var r_rank = r1 > r2 ? -100 : r1 < r2 ? 100 : 0;
var n_rank = n1 > n2 ? -10 : n1 < n2 ? 10 : 0;
var p_rank = p1 > p2 ? -1 : p1 < p2 ? 1 : 0;
return r_rank + n_rank + p_rank;
})

console.log(data)

Oriol · Accepted Answer · 2016-11-11 07:03:01Z

0

Just include more conditionals:

var data = [{"title":"one","popularity":4,"novelty":3},{"title":"two","popularity":1,"novelty":4},{"title":"three","popularity":5,"novelty":3,"roulette":0},{"title":"four","popularity":5,"novelty":3,"roulette":1}];
data.sort(function(a,b) {
  if (a.roulette < b.roulette || a.roulette == null) return +1;
  if (a.roulette > b.roulette || b.roulette == null) return -1;
  if (a.novelty < b.novelty || a.novelty == null) return +1;
  if (a.novelty > b.novelty || b.novelty == null) return -1;
  if (a.popularity < b.popularity || a.popularity == null) return +1;
  if (a.popularity > b.popularity || b.popularity == null) return -1;
  return 0;
})
console.log(data);

answered Nov 11, 2016 at 7:03

Oriol

291k71 gold badges459 silver badges535 bronze badges

1 Comment

Oriol Over a year ago

Maybe something like a.roulette < b.roulette || a.roulette == null && b.roulette != null or (a.roulette != null ? a.roulette : -Infinity) < b.roulette would be more appropriate.

Sikorski · Accepted Answer · 2016-11-11 07:21:24Z

0

You just have to keep on breaking ties if one parameter is same.

 obj.sort(function(a, b) {
    var rouletteDiff = compare(a.roulette, b.roulette);
    if(rouletteDiff != 0) return rouletteDiff;
    var noveltyDiff = compare(a.novelty, b.novelty);
    if(noveltyDiff != 0) return noveltyDiff;
    return compare(a.popularity, b.popularity);
  });

  function compare(x,y){
    if(x == undefined) return 1;
    if(y == undefined) return -1;
    if(x === y){
        return 0;
    }else{
        return x > y ? -1 : 1
    }
  }

edited Nov 11, 2016 at 7:21

answered Nov 11, 2016 at 7:04

Sikorski

2,6913 gold badges27 silver badges48 bronze badges

Collectives™ on Stack Overflow

Sorting an array of objects by multiple criteria with null values

5 Answers 5

4 Comments

Comments

Comments

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

5 Answers 5

4 Comments

Comments

Comments

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related