Sort null values to last in Array of Objects

Question

I have an array of objects as follows:

c = [{a: null}, {a: 12}, {a: 1}, {a: 50}, {a: 2}, {a: null}]

I want to kind-of sort them by the objects having values first then objects with null.

What I tried is:

c.sort(function(b) { return b.a ? -1 : 1 })

OUTPUT

[{a: 2}, {a: 50}, {a: 1}, {a: 12}, {a: null}, {a: null}]

EXPECTED OUTPUT

[{a: 12}, {a: 1}, {a: 50}, {a: 2}, {a: null}, {a: null}]

How can I achieve this?

@GokulanPH Yes, I can. I just haven't used it in my question, so didn't tag it — Abhi, Jun 28 '17 at 12:47
The [Array#sort](https://developer.mozilla.org/en/docs/Web/JavaScript/Reference/Global_Objects/Array/sort) function has an arity of 2, it has an `a` and `b` argument for comparison. — Xotic750, Jun 28 '17 at 12:48
Also, beware of sort stability: https://stackoverflow.com/questions/1517793/stability-in-sorting-algorithms — Xotic750, Jun 28 '17 at 12:54

georg · Answer 1 · 2017-06-28T14:39:51.183

6

This will put nulls and other falsy values to the end of the list:

c = [{a: null}, {a: 12}, {a: 1}, {a: 50}, {a: 2}, {a: null}];
c.sort((x, y) => !!y.a - !!x.a);

console.log(c);

However, since you don't really sort anything, you can just split the list into two parts an rejoin them:

c = [{a: null}, {a: 12}, {a: 1}, {a: 50}, {a: 2}, {a: null}];

r = [
  ...c.filter(x => x.a !== null),
  ...c.filter(x => x.a === null)
]

console.log(r)

This also doesn't rely on the sort function being stable.

edited Jun 28 '17 at 14:39

answered Jun 28 '17 at 12:49

georg

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1

it moves zero values to the bottom. – Nina Scholz Jun 28 '17 at 12:50
The `filter` option is much better than using `sort` in this use case, it avoids any stability problems. – Xotic750 Jun 28 '17 at 13:00

Nina Scholz · Accepted Answer · 2017-06-28T12:58:38.737

You could test the value. If null, then take the delta of the comparison.

var c = [{ a: null }, { a: 12 }, { a: 1 }, { a: 50 }, { a: 2 }, { a: null }];

c.sort(function (a, b) {
    return (a.a === null) - (b.a === null);
});

console.log(c);

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

For a stable sort, you could use sorting with map and use the indices as second sort option.

// the array to be sorted
var list = [{ a: null }, { a: 12 }, { a: 1 }, { a: 50 }, { a: 2 }, { a: null }];

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

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

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

console.log(result);

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

Yordan Nikolov · Answer 3 · 2017-06-28T12:54:01.080

1

That way ?

c=[{a: null}, {a: 12}, {a: 1}, {a: 50}, {a: 2}, {a: null}];
c.sort(function(a,b){ return a.a===null ? 1:-1 })
console.log(c);

Edited

edited Jun 28 '17 at 12:54

answered Jun 28 '17 at 12:49

Yordan Nikolov

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This sorts the objects, I just want to push objects with null values to last – Abhi Jun 28 '17 at 12:50

score 1 · Answer 4 · answered Jun 28 '17 at 12:50

1

const c = [{a: null}, {a: 12}, {a: 1}, {a: 50}, {a: 2}, {a: null}];

let result = [...c.filter(_=>_["a"]!==null),...c.filter(_=>_["a"]===null)];

console.log(result);

answered Jun 28 '17 at 12:50

Weedoze

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vsr · Answer 5 · 2017-06-28T13:24:03.437

1

var c = [{
  a: null
}, {
  a: 12
}, {
  a: 1
}, {
  a: 50
}, {
  a: 2
}, {
  a: null
}];

c.sort(function(a, b) {
  return (a.a !== null) ? 0 : 1;
});
console.log(c);

Returning 0 in the sort function will keep the order as it is.

If compareFunction(a, b) returns 0, leave a and b unchanged with respect to each other, but sorted with respect to all different elements.

https://developer.mozilla.org/en/docs/Web/JavaScript/Reference/Global_Objects/Array/sort

edited Jun 28 '17 at 13:24

answered Jun 28 '17 at 12:51

vsr

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Although @Xotic750's comment **"The Array#sort function has an arity of 2, it has an a and b argument for comparison."** is right, but this answer worked for me. – Abhi Jun 28 '17 at 12:59
maybe i see wrong, but in edge, i see a null value at top and one at bottom. it heavily rely on just one side of the compairing between two vaules. – Nina Scholz Jun 28 '17 at 13:07
1

@vsr, still the same problem. – Nina Scholz Jun 28 '17 at 13:36

score 1 · Answer 6 · answered Jun 28 '17 at 12:52

1

Try with return !a.a - !b.a .valid object goes first

var c = [{a: null}, {a: 12}, {a: 1}, {a: 50}, {a: 2}, {a: null}];

c.sort(function (a, b) {
    return !a.a - !b.a
});
console.log(c);

answered Jun 28 '17 at 12:52

prasanth

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Jan Ciołek · Answer 7 · 2017-06-28T12:58:34.903

It is not perfect but still works,cheers!

var c = [{a: null}, {a: 12}, {a: 1}, {a: 50}, {a: 2}, {a: null}]

c.sort(function(object1,object2){

  
  if(object1.a === null && object2.a !== null){return 1}
  
  if([object1.a,object2.a].every((a)=>a === null) ||      
     [object1.a,object2.a].every((a)=>a !== null)
     ){return 0}
  
  if(object1.a !== null && object2.a === null){return -1}
  


})

console.log(c);

score 0 · Answer 8 · edited May 07 '19 at 12:12

0

Here's a case of sorting a Date with null value at the end if exists:

userSortList = users.sort((a, b) => {
    if (b.lastConnectionDate === null || a.lastConnectionDate) {
        return -1
    } else {
        return (
            new Date(b.lastConnectionDate) - new Date(a.lastConnectionDate)
        )
    }
})

edited May 07 '19 at 12:12

Matt

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answered May 07 '19 at 11:44

Erwan A. R. Riou

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Sort null values to last in Array of Objects

8 Answers8