Sorting strings in descending order in Javascript (Most efficiently)?

问题

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?

Update

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)?

回答1:

If you consider

obj.sort().reverse();

obj.sort((a, b) => (a > b ? -1 : 1))

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 about 100 times faster than obj.sort( function ), and obj.sort( function ) (using localCompare)

Performance test here : https://jsperf.com/reverse-string-sort/1

回答2:

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.

回答3:

var arr = ["a","b","c","A","B","Z"];

arr.sort((a,b)=>b.localeCompare(a))

console.log(arr)

来源：https://stackoverflow.com/questions/52030110/sorting-strings-in-descending-order-in-javascript-most-efficiently

标签

javascript

sorting