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JavaScript array sorting shorthand

Submitted by: @import:30-seconds-of-code··
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shorthandjavascriptsortingarray

Problem

When sorting an array of primitive values (e.g. strings or numbers), you'll often see a lot of code that looks like this:
While this piece of code does the job, there is also a one-line alternative for it. The trick hinges on Array.prototype.sort() expecting either a positive or a negative value to perform a swap between two elements, thus allowing for more flexible values than 1 and -1. Subtracting the numeric values in an array is sufficient and can also be used to sort the array the other way around:
If you are working with string arrays, you should instead use String.prototype.localeCompare(), as it provides far greater flexibility, by accounting for specific locales and their unique needs:

Solution

const arr = [8, 2, 1, 4, 5, 0];
// Sort in ascending order
arr.sort((a, b) => {
  if (a > b) return 1;
  if (b > a) return -1
  return 0;
}); // [0, 1, 2, 4, 5, 8]


If you are working with string arrays, you should instead use String.prototype.localeCompare(), as it provides far greater flexibility, by accounting for specific locales and their unique needs:

Code Snippets

const arr = [8, 2, 1, 4, 5, 0];
// Sort in ascending order
arr.sort((a, b) => {
  if (a > b) return 1;
  if (b > a) return -1
  return 0;
}); // [0, 1, 2, 4, 5, 8]
const arr = [8, 2, 1, 4, 5, 0];
// Sort in ascending order
arr.sort((a, b) => a - b); // [0, 1, 2, 4, 5, 8]
// Sort in descending order
arr.sort((a, b) => b - a); // [8, 5, 4, 2, 1, 0]
const s = ['Hi', 'Hola', 'Hello'];
// Sort in ascending order
arr.sort((a, b) => a.localeCompare(b)); // ['Hello', 'Hi', 'Hola']
// Sort in descending order
arr.sort((a, b) => b.localeCompare(a)); // ['Hola', 'Hi', 'Hello']

Context

From 30-seconds-of-code: array-sorting-shorthand

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