Reputation: 4261
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?
Upvotes: 5
Views: 7030
Reputation: 403
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)
)
}
})
Upvotes: 0
Reputation: 214949
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.
Upvotes: 7
Reputation: 3198
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
Upvotes: 1
Reputation: 386560
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; }
Upvotes: 3
Reputation: 1816
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);
Upvotes: 0
Reputation: 2678
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
Upvotes: 1
Reputation: 22490
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);
Upvotes: 1
Reputation: 13943
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);
Upvotes: 1