How to Sort Strings in JavaScript

How to sort strings in JavaScript

Use String.prototype.localeCompare a per your example:

list.sort(function (a, b) {
    return ('' + a.attr).localeCompare(b.attr);
})

We force a.attr to be a string to avoid exceptions. localeCompare has been supported since Internet Explorer 6 and Firefox 1. You may also see the following code used that doesn't respect a locale:

if (item1.attr < item2.attr)
  return -1;
if ( item1.attr > item2.attr)
  return 1;
return 0;

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

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]));

Sort a string alphabetically using a function

You can use array sort function:

var sortAlphabets = function(text) {
    return text.split('').sort().join('');
};

STEPS

1. Convert string to array
2. Sort array
3. Convert back array to string

Demo

Sort array by firstname (alphabetically) in JavaScript

Suppose you have an array users. You may use users.sort and pass a function that takes two arguments and compare them (comparator)

It should return

• something negative if first argument is less than second (should be placed before the second in resulting array)
• something positive if first argument is greater (should be placed after second one)
• 0 if those two elements are equal.

In our case if two elements are a and b we want to compare a.firstname and b.firstname

Example:

users.sort(function(a, b){
    if(a.firstname < b.firstname) { return -1; }
    if(a.firstname > b.firstname) { return 1; }
    return 0;
})

This code is going to work with any type.

Note that in "real life"™ you often want to ignore case, correctly sort diacritics, weird symbols like ß, etc. when you compare strings, so you may want to use localeCompare. See other answers for clarity.

how to sort strings based on a substring JS

You can create a function to extract the last number id using a RegExp combined with String.prototype.match():

Regular expression: /\d+(?=\D*$)/

• \d matches a digit (equivalent to [0-9])
• + matches the previous token between one and unlimited times, as many times as possible, giving back as needed (greedy)
• Positive Lookahead (?=\D*$)
  Assert that the Regex below matches
  \D matches any character that's not a digit (equivalent to [^0-9])
• * matches the previous token between zero and unlimited times, as many times as possible, giving back as needed (greedy)
• $ asserts position at the end of the string, or before the line terminator right at the end of the string (if any)

And finally use Array.prototype.sort()

Code:

const arr = ["UserE | 2020", "UserA | 3", "UserB | 0", "UserC | 2", "UserD | 1"]
const getId = s => +s.match(/\d+(?=\D*$)/)[0] 

arr.sort((a, b) => getId(a) - getId(b))

console.log(arr)

Sort array of objects by string property value

It's easy enough to write your own comparison function:

function compare( a, b ) {
  if ( a.last_nom < b.last_nom ){
    return -1;
  }
  if ( a.last_nom > b.last_nom ){
    return 1;
  }
  return 0;
}

objs.sort( compare );

Or inline (c/o Marco Demaio):

objs.sort((a,b) => (a.last_nom > b.last_nom) ? 1 : ((b.last_nom > a.last_nom) ? -1 : 0))

Or simplified for numeric (c/o Andre Figueiredo):

objs.sort((a,b) => a.last_nom - b.last_nom); // b - a for reverse sort

How sort string with spaces in javascript?

The String.localeCompare method returns a number indicating whether a reference string comes before, or after, or is the same as the given string in sort order... which is the same as what Array.sort is supposed to return:

const array = [
  { attributes: { name: "abcd efg" } },
  { attributes: { name: "Übd cd" } },
  { attributes: { name: "Ku cdf" } },
  { attributes: { name: "ab" } }
];
array.sort((a, b) => a.attributes.name.toUpperCase().localeCompare(b.attributes.name.toUpperCase(), "de", { sensitivity: "base" }));
console.log(array);

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