Count unique values in array javascript

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.

asked Feb 24, 2013 at 14:29

Ghoul FoolGhoul Fool

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3

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).

answered Jun 7, 2017 at 6:50

Web_DesignerWeb_Designer

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2

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.

answered Feb 24, 2013 at 14:34

kojirokojiro

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8

Why not something like:

var arr = ["jam", "beef", "cream", "jam"]
var uniqs = arr.reduce((acc, val) => {
  acc[val] = acc[val] === undefined ? 1 : acc[val] += 1;
  return acc;
}, {});
console.log(uniqs)

Pure Javascript, runs in O(n). Doesn't consume much space either unless your number of unique values equals number of elements (all the elements are unique).

answered Mar 7, 2018 at 16:09

Ash CatchemAsh Catchem

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1

Same as this solution, but less code.

let counts = {};
arr.forEach(el => counts[el] = 1  + (counts[el] || 0))

Stephen Rauch♦

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answered Aug 20, 2018 at 23:54

kryptoatomkryptoatom

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This expression gives you all the unique elements in the array without mutating it:

arr.filter(function(v,i) { return i==arr.lastIndexOf(v); })

You can chain it with this expression to build your string of results without sorting:

.forEach(function(v) {
     results+=v+" --> " + arr.filter(function(w){return w==v;}).length + " times\n";
});

In the first case the filter takes only includes the last of each specific element; in the second case the filter includes all the elements of that type, and .length gives the count.

answered Feb 24, 2013 at 14:32

PlynxPlynx

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3

This answer is for Beginners. Try this method you can solve this problem easily. You can find a full lesson for reduce, filter, map functions from This link.

const user = [1, 2, 2, 4, 8, 3, 3, 6, 5, 4, 8, 8];

const output = user.reduce(function (acc, curr) {
    if (acc[curr]) {
        acc[curr] = ++acc[curr];
    } else {
        acc[curr] = 1;
    }
    return acc;
}, {});

console.log(output);

answered Feb 3 at 16:28

Hishan_98Hishan_98

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function reomveDuplicates(array){
        var newarray = array.filter( (value, key)=>{
            return array.indexOf(value) == key
        });
        console.log("newarray", newarray);
    }
reomveDuplicates([1,2,5,2,1,8]);  

Using hash Map with the time complexity O(n)

function reomveDuplicates(array){

    var obj ={};
    let res=[];

    for( arg of array){
        obj[arg] = true;
    }

    console.log(Object.keys(obj));


    for(key in obj){
        res.push(Number(key));  // Only if you want in Number 
    }

    console.log(res);
}
reomveDuplicates([1,2,5,2,1,8]);  

answered Feb 22, 2021 at 15:09

2

In a modern, extensible and easy-to-read approach, here's one using iter-ops library:

import {pipe, distinct, count} from 'iter-ops';

const arr = ['jam', 'beef', 'cream', 'jam'];

const count = pipe(arr, distinct(), count()).first;

console.log(count); //=> 3

answered Nov 16, 2021 at 18:37

vitaly-tvitaly-t

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function check(arr) {
    var count = 0;

 for (var ele of arr) {
   
    if (typeof arr[ele] !== typeof (arr[ele+1])) {
      count++;
    } else {
      ("I don't know");
    }
  }
  return count;
}

answered Jan 8 at 9:03

1

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