Reputation: 917
I have an array that could be either just a list of strings or it could be just a list of strings of numbers, i.e array could be like below
let array = ['abc','def','aef','gfh']
or
let array = ['123','456','192','412']
I want to have sort function that can handle the natural sort in either of these case, below code doesn't seem to handle the string of numbers, is there a way to handle this?
array.sort((a,b) => {
if(a > b) return 1;
if(a < b) return -1;
return 0;
});
Upvotes: 0
Views: 78
Reputation: 73896
You can do this using String#localeCompare()
method like:
let array = ['abc', 'def', 'aef', 'gfh']
array.sort((a, b) => a.localeCompare(b, undefined, { numeric: true }));
console.log(array)
array = ['123', '456', '192', '412', '5']
array.sort((a, b) => a.localeCompare(b, undefined, { numeric: true }));
console.log(array)
.as-console-wrapper { max-height: 100% !important; top: 0; }
As mentioned in the docs, for numeric sorting we just need to use {numeric: true}
option.
// by default, "2" > "10"
console.log(["2", "10"].sort((a,b) => a.localeCompare(b))); // ["10", "2"]
// numeric using options:
console.log(["2", "10"].sort((a,b) => a.localeCompare(b, undefined, {numeric: true})));
// ["2", "10"]
Upvotes: 2
Reputation: 16908
You can check whether the element in the array
is a number or a string in the sort
function.
The isNaN(string)
check would tell you if it is not a number:
function sortNumOrStr(array) {
return array.sort((a, b) => isNaN(a) ? a.localeCompare(b) : +a - b);
}
let array = ['abc', 'def', 'aef', 'gfh']
console.log(sortNumOrStr(array));
array = ['123', '456', '-90', '192', '412'];
console.log(sortNumOrStr(array));
Upvotes: 2