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?

Answer 1

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) => !!ya - !!xa); 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 => xa !== null), ...c.filter(x => xa === null) ] console.log(r) 

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

Answer 2

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 (aa === null) - (ba === 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; } 

Answer 3

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

Answer 4

 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); 

Answer 5

 var c = [{ a: null }, { a: 12 }, { a: 1 }, { a: 50 }, { a: 2 }, { a: null }]; c.sort(function(a, b) { return (aa !== 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

Answer 6

Try with return !aa - !ba .valid object goes first

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

Answer 7

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); 

Answer 8

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)
        )
    }
})