Javascript 在Javascript中按降序对字符串进行排序（最有效）？
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Sorting strings in descending order in Javascript (Most efficiently)?
提问by Ole
W3CSchools has this example:
W3CSchools 有这个例子：
var fruits = ["Banana", "Orange", "Apple", "Mango"];
fruits.sort();
fruits.reverse();
Is this the most efficient way to sort strings in descending order in Javascript?
这是在 Javascript 中按降序对字符串进行排序的最有效方法吗？
Update
更新
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)?
答案之一是使用localeCompare. 只是好奇如果我们这样做reverse()，是否适用于所有语言环境（也许这是一个单独的问题 - 请在评论中告诉我）？
回答by colxi
If you consider
如果你考虑
obj.sort().reverse();
VS
VS
obj.sort((a, b) => (a > b ? -1 : 1))
VS
VS
obj.sort((a, b) => b.localeCompare(a) )
The performance winner is : obj.sort().reverse().
性能优胜者是：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).
使用 10.000 个元素的数组进行测试，
obj.sort().reverse()比obj.sort( function )（chrome 除外）和obj.sort( function )（使用localCompare）快。
Performance test here :
性能测试在这里：
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]));回答by Emeeus
Using just sortand reversea> Z, that is wrong if you want to order lower cases and upper cases strings:
如果您想对小写和大写字符串进行排序，则仅使用sort和reversea>Z是错误的：
var arr = ["a","b","c","A","B","Z"];
arr.sort().reverse();
console.log(arr)//<-- [ 'c', 'b', 'a', 'Z', 'B', 'A' ] wrong!!!English characters
英文字符
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)
使用locales 的特殊字符，在本例中为 es（西班牙语）
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:
在这种情况下，敏感性是基础：
Only strings that differ in base letters compare as unequal. Examples: a ≠ b, a = á, a = A.
只有基本字母不同的字符串比较为不相等。示例：a ≠ b，a = á，a = A。
回答by Mungai_Keren
var arr = ["a","b","c","A","B","Z"];
arr.sort((a,b)=>b.localeCompare(a))
console.log(arr)