Sorting strings in descending order in Javascript (Most efficiently)? - Stack Overflow

admin管理员组

文章数量:1134236

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)?

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)?

• javascript
• sorting

Share Improve this question Follow edited Aug 27, 2018 at 15:20 Ole asked Aug 26, 2018 at 20:40 OleOle 46.8k6868 gold badges236236 silver badges439439 bronze badges 9

• 1 By what measure of efficiency? – David Thomas Commented Aug 26, 2018 at 20:42
• Possible duplicate stackoverflow.com/questions/1063007/… – Asons Commented Aug 26, 2018 at 20:49
• 1 .sort() and .reverse() is already the most efficient way. – Sebastian Simon Commented Aug 26, 2018 at 20:52
• 1 .sort((a, b) => -(a>b)||+(a<b)) – Bergi Commented Aug 26, 2018 at 20:53
• 2 reverse() doesn't care about the locales, it only modifies the indexes of the array in reverse order – colxi Commented Aug 26, 2018 at 21:25

 |  Show 4 more comments

5 Answers 5

Sorted by: Reset to default 234

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 than obj.sort( function ) (except on chrome), and obj.sort( function ) (using localCompare).

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]));

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.

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.

本文标签： Sorting strings in descending order in Javascript (Most efficiently)Stack Overflow