W3CSchools has this example:
var fruits = ["Banana", "Orange", "Apple", "Mango"];
fruits.sort();
fruits.reverse();
Is this the most efficient way to sort strings in descending order in Javascript?
One of the answers is using localeCompare. Just curious whether if we do reverse(), will that work for all locales (Maybe this is a separate question - Just let me know in the comments)?
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]));
localCompare case
(a,b)=>b.localeCompare(a)) not b.localeCompare(a, 'es', {sensitivity: 'base'})) that is for special characters. obj.sort().reverse() doesn't work with special characters
obj.sort((a, b) => b - a) only works for numbers, not strings!Using just sort and reverse a > Z , that is wrong if you want to order lower cases and upper cases strings:
var arr = ["a","b","c","A","B","Z"];
arr.sort().reverse();
console.log(arr)//<-- [ 'c', 'b', 'a', 'Z', 'B', 'A' ] wrong!!!English characters
var arr = ["a","b","c","A","B","Z"];
arr.sort((a,b)=>b.localeCompare(a))
console.log(arr)Special characters using locales, in this example es (spanish)
var arr = ["a", "á", "b","c","A","Á","B","Z"];
arr.sort((a, b) => b.localeCompare(a, 'es', {sensitivity: 'base'}))
console.log(arr)sensitivity in this case is base:
Only strings that differ in base letters compare as unequal. Examples: a ≠ b, a = á, a = A.
sort and reverse could solve the problem that OP is facing using that specific set of data, it is not good as a general solution in my opinion. Also OP said in the title strings, not certain kind of strings. Either way "is wrong" is a bad generalization.
The easiest way to revers the order of sorting is by swapping the operands. In ES2015 that's as easy as [b, a] = [a, b]. A full example:
function compareWithOrder(a, b, shouldReverse = false) {
if (shouldReverse) {
[b, a] = [a, b]
}
return yourComparatorFn(a, b)
}
var arr = ["a","b","c","A","B","Z"];
arr.sort((a,b)=>b.localeCompare(a))
console.log(arr)
I know this is an old question, but an interesting one. This is my solution for a non-special character's input.
var arr = ["a","b","c","A","B","Z"];
console.log(arr.sort((a,b)=> {
const lastCodeIn = b.toLowerCase().charCodeAt();
const lastCode = b.charCodeAt();
const firstCodeIn = a.toLowerCase().charCodeAt();
const firstCode = a.charCodeAt();
if(lastCodeIn - firstCodeIn === 0){
return lastCode - firstCode;
}
return lastCodeIn - firstCodeIn;
})
);//[ 'Z', 'c', 'b', 'B', 'a', 'A' ]
The reason is that ascii code for UPPER case are lower than lower case.
.sort()and.reverse()is already the most efficient way..sort((a, b) => -(a>b)||+(a<b))reverse()doesn't care about the locales, it only modifies the indexes of the array in reverse order