Get All Non-Unique Values (I.E.: Duplicate/More Than One Occurrence) in an Array

Get all non-unique values (i.e.: duplicate/more than one occurrence) in an array

You could sort the array and then run through it and then see if the next (or previous) index is the same as the current. Assuming your sort algorithm is good, this should be less than O(n²):

const findDuplicates = (arr) => {

  let sorted_arr = arr.slice().sort(); // You can define the comparing function here. 

  // JS by default uses a crappy string compare.

  // (we use slice to clone the array so the

  // original array won't be modified)

  let results = [];

  for (let i = 0; i < sorted_arr.length - 1; i++) {

    if (sorted_arr[i + 1] == sorted_arr[i]) {

      results.push(sorted_arr[i]);

    }

  }

  return results;

}

let duplicatedArray = [9, 9, 111, 2, 3, 4, 4, 5, 7];

console.log(`The duplicates in ${duplicatedArray} are ${findDuplicates(duplicatedArray)}`);

Get all non-unique values (i.e.: duplicate/more than one occurrence) in an array

You could sort the array and then run through it and then see if the next (or previous) index is the same as the current. Assuming your sort algorithm is good, this should be less than O(n²):

const findDuplicates = (arr) => {

  let sorted_arr = arr.slice().sort(); // You can define the comparing function here. 

  // JS by default uses a crappy string compare.

  // (we use slice to clone the array so the

  // original array won't be modified)

  let results = [];

  for (let i = 0; i < sorted_arr.length - 1; i++) {

    if (sorted_arr[i + 1] == sorted_arr[i]) {

      results.push(sorted_arr[i]);

    }

  }

  return results;

}

let duplicatedArray = [9, 9, 111, 2, 3, 4, 4, 5, 7];

console.log(`The duplicates in ${duplicatedArray} are ${findDuplicates(duplicatedArray)}`);

Sorting an array, having all non-unique values nested

Have a look at this one-liner. Guess this is what you need.

Note that this isn't the most efficient way to to this - but guess it's beautyfull!

const array = [1,2,4,591,392,391,2,5,10,2,1,1,1,20,20];
const newArray = [...new Set(array)].sort((a, b) => a-b).map(e => new Array(array.filter( i => i === e).length).fill(e)).map(e => e.length === 1 ? e[0] : e);
console.log(newArray);

Get all unique values in a JavaScript array (remove duplicates)

With JavaScript 1.6 / ECMAScript 5 you can use the native filter method of an Array in the following way to get an array with unique values:

function onlyUnique(value, index, self) {
  return self.indexOf(value) === index;
}

// usage example:
var a = ['a', 1, 'a', 2, '1'];
var unique = a.filter(onlyUnique);

console.log(unique); // ['a', 1, 2, '1']

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