Count unique elements in array without sorting

Question

In JavaScript the following will find the number of elements in the array. Assuming there to be a minimum of one element in the array

arr = ["jam", "beef", "cream", "jam"] arr.sort(); var count = 1; var results = ""; for (var i = 0; i < arr.length; i++) {     if (arr[i] == arr[i+1])     {       count +=1;     }     else     {         results += arr[i] + " --> " + count + " times\n" ;         count=1;     } } 

Is it possible to do this without using sort() or without mutating the array in any way? I would imagine that the array would have to be re-created and then sort could be done on the newly created array, but I want to know what's the best way without sorting. And yes, I'm an artist, not a programmer, your honour.

Answer 1

A quick way to do this is to copy the unique elements into an Object.

var counts = {}; for (var i = 0; i < arr.length; i++) {     counts[arr[i]] = 1 + (counts[arr[i]] || 0); } 

When this loop is complete the counts object will have the count of each distinct element of the array.

Answer 2

The fast way to do this is with a new Set() object.

Sets are awesome and we should use them more often. They are fast, and supported by Chrome, Firefox, Microsoft Edge, and node.js.
— What is faster Set or Object? by Andrei Kashcha

The items in a Set will always be unique, as it only keeps one copy of each value you put in. Here's a function that uses this property:

function countUnique(iterable) {    return new Set(iterable).size;  }    console.log(countUnique('banana')); //=> 3  console.log(countUnique([5,6,5,6])); //=> 2  console.log(countUnique([window, document, window])); //=> 2

This can be used to count the items in any iterable (including an Array, String, TypedArray, and arguments object).