If you consider
obj.sort().reverse();
VS
obj.sort((a, b) => (a > b ? -1 : 1))
VS
obj.sort((a, b) => b.localeCompare(a) )
The performance winner is : obj.sort().reverse()
.
Testing with an array of 10.000 elements,
obj.sort().reverse()
is faster thanobj.sort( function )
(except on chrome), andobj.sort( function )
(usinglocalCompare
).
Performance test here :
var results = [[],[],[]]
for(let i = 0; i < 100; i++){
const randomArrayGen = () => Array.from({length: 10000}, () => Math.random().toString(30));
const randomArray = randomArrayGen();
const copyArray = x => x.slice();
obj = copyArray(randomArray);
let t0 = performance.now();
obj.sort().reverse();
let t1 = performance.now();
obj = copyArray(randomArray);
let t2 = performance.now();
obj.sort((a, b) => (a > b ? -1 : 1))
let t3 = performance.now();
obj = copyArray(randomArray);
let t4 = performance.now();
obj.sort((a, b) => b.localeCompare(a))
let t5 = performance.now();
results[0].push(t1 - t0);
results[1].push(t3 - t2);
results[2].push(t5 - t4);
}
const calculateAverage = x => x.reduce((a,b) => a + b) / x.length ;
console.log("obj.sort().reverse(): " + calculateAverage(results[0]));
console.log("obj.sort((a, b) => (a > b ? -1 : 1)): " + calculateAverage(results[1]));
console.log("obj.sort((a, b) => b.localeCompare(a)): " + calculateAverage(results[2]));