You could sort the array first and then all duplicates will be next to each other:
const numbers = [1, 2, 3, 1, 2, 3, 1, 2, 3]; const unique = [...numbers].sort().filter((v, i, arr) => i === 0 || arr[i - 1] !== v) console.log(unique) // [1, 2, 3]
why did you use an instance of the numbers array, not the number array itself? " [...numbers].sort() not numbers.sort() "
That is a good question, sort is a destructive function. This means running sort on numbers array will change original numbers array.
wow I forgot that
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You could sort the array first and then all duplicates will be next to each other:
why did you use an instance of the numbers array, not the number array itself? " [...numbers].sort() not numbers.sort() "
That is a good question, sort is a destructive function. This means running sort on numbers array will change original numbers array.
wow I forgot that