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

Asked 2023-09-20 20:25:10 View 477,574

I have an array of numbers that I need to make sure are unique. I found the code snippet below on the internet and it works great until the array has a zero in it. I found this other script here on Stack Overflow that looks almost exactly like it, but it doesn't fail.

So for the sake of helping me learn, can someone help me determine where the prototype script is going wrong?

Array.prototype.getUnique = function() {
 var o = {}, a = [], i, e;
 for (i = 0; e = this[i]; i++) {o[e] = 1};
 for (e in o) {a.push (e)};
 return a;
}

More answers from duplicate question:

Similar question:

  • @hippietrail That older question is about finding and returning only the duplicates (I was confused too!). My question is more about why this function fails when an array has a zero in it. - anyone
  • For future readers, when start finding that you have to algorithmically modify the contents of your data structure all the time, (order them, remove repeating elements, etc.) or search for elements inside it at every iteration, it's safe to assume that you're using the wrong data structure in the first place and start using one that is more appropriate for the task at hand (in this case a hash set instead of array). - anyone
  • I copied the code from somewhere else, a loooong time ago... but it seems pretty straight-forward: o = object, a = array, i = index and e = umm, something :P - anyone
  • Possible duplicate of How to get unique values in an array - anyone
  • Just wanted to point out, a lot of people have suggested using JavaScript Set as a solution, proceed with caution because it is not supported in Internet Explorer. If you have to support IE, then use a polyfill. - anyone

Answers

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, array) {
  return array.indexOf(value) === index;
}

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

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

The native method filter will loop through the array and leave only those entries that pass the given callback function onlyUnique.

onlyUnique checks, if the given value is the first occurring. If not, it must be a duplicate and will not be copied.

This solution works without any extra library like jQuery or prototype.js.

It works for arrays with mixed value types too.

For old Browsers (<ie9), that do not support the native methods filter and indexOf you can find work arounds in the MDN documentation for filter and indexOf.

If you want to keep the last occurrence of a value, simply replace indexOf with lastIndexOf.

With ES6 this can be shorten to:

// usage example:
var myArray = ['a', 1, 'a', 2, '1'];
var unique = myArray.filter((value, index, array) => array.indexOf(value) === index);

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

Thanks to Camilo Martin for hint in comment.

ES6 has a native object Set to store unique values. To get an array with unique values you could now do this:

var myArray = ['a', 1, 'a', 2, '1'];

let unique = [...new Set(myArray)];

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

The constructor of Set takes an iterable object, like an Array, and the spread operator ... transform the set back into an Array. Thanks to Lukas Liese for hint in comment.

Answered   2023-09-20 20:25:10

  • This solution will run much slower, unfortunately. You're looping twice, once with filter and once with index of - anyone
  • In modern JS: .filter((v,i,a)=>a.indexOf(v)==i) (fat arrow notation). - anyone
  • let unique_values = [...new Set(random_array)]; developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/… - anyone
  • Lest any newcomer be scared away by the performance implications, consider "much slower" may very well be nanoseconds. If your array is relatively small (hundreds or less), using the nice concise one off version like this is good for reasons outside of performance, i.e. readability and maintainability. But the set version is quite nice and concise. - anyone
  • Comparing performance of .filter+indexOf and the Set-approach in Chrome 100 revealed that for an array with numbers and length 0 to 120 the filter-approach is even faster. With array length 200 the filter-approach takes 50% more time than with a Set (6 vs. 9 microseconds). For 5000 elements the filter operation takes over 3 milliseconds while Set still copes with my example in 173 microseconds. So "much slower", really depends on the use case. If someone is interested I can post the data as a separate Q&A. - anyone

Updated answer for ES6/ES2015: Using the Set and the spread operator (thanks le-m), the single line solution is:

let uniqueItems = [...new Set(items)]

Which returns

[4, 5, 6, 3, 2, 23, 1]

Answered   2023-09-20 20:25:10

  • Notice, that inner array wouldn't work Array.from(new Set([[1,2],[1,2],[1,2,3]])) - anyone
  • Please note that if you use the Set and add objects instead of primitive values it will contain unique references to the objects. Thus the set s in let s = new Set([{Foo:"Bar"}, {Foo:"Bar"}]); will return this: Set { { Foo: 'Bar' }, { Foo: 'Bar' } } which is a Set with unique object references to objects that contain the same values. If you write let o = {Foo:"Bar"}; and then create a set with two references like so: let s2 = new Set([o,o]);, then s2 will be Set { { Foo: 'Bar' } } - anyone
  • In case anyone was wondering, this works well for strings too, e.g. [...new Set(["apple","apple","orange"])] results in ['apple', 'orange'] . Great! - anyone
  • In Typescript, use Array.from( new Set( items ) ) - anyone

I split all answers to 4 possible solutions:

  1. Use object { } to prevent duplicates
  2. Use helper array [ ]
  3. Use filter + indexOf
  4. Bonus! ES6 Sets method.

Here's sample codes found in answers:

Use object { } to prevent duplicates

function uniqueArray1( ar ) {
  var j = {};

  ar.forEach( function(v) {
    j[v+ '::' + typeof v] = v;
  });

  return Object.keys(j).map(function(v){
    return j[v];
  });
} 

Use helper array [ ]

function uniqueArray2(arr) {
    var a = [];
    for (var i=0, l=arr.length; i<l; i++)
        if (a.indexOf(arr[i]) === -1 && arr[i] !== '')
            a.push(arr[i]);
    return a;
}

Use filter + indexOf

function uniqueArray3(a) {
  function onlyUnique(value, index, self) { 
      return self.indexOf(value) === index;
  }

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

  return unique;
}

Use ES6 [...new Set(a)]

function uniqueArray4(a) {
  return [...new Set(a)];
}

And I wondered which one is faster. I've made sample Google Sheet to test functions. Note: ECMA 6 is not avaliable in Google Sheets, so I can't test it.

Here's the result of tests: enter image description here

I expected to see that code using object { } will win because it uses hash. So I'm glad that tests showed the best results for this algorithm in Chrome and IE. Thanks to @rab for the code.

Update 2020

Google Script enabled ES6 Engine. Now I tested the last code with Sets and it appeared faster than the object method.

Answered   2023-09-20 20:25:10

  • Makrov, so the uniqueItems = [...new Set(items)] appears to be the fastest and the most succinct of all the approaches? - anyone
  • Your solution only handles primitives, it won't work with objects, you'd need to JSON.stringify the v in the hash - anyone
  • you are GREAT for making this time test comparison, thanks, I tested now with an Array with hundreds of thousands of entries and indeed uniqueItems = [...new Set(items)] seems the fastest - anyone
  • "Object.keys(j).map ..." could be changed to use Object.values(j); - anyone

You can also use underscore.js.

console.log(_.uniq([1, 2, 1, 3, 1, 4]));
<script src="http://underscorejs.org/underscore-min.js"></script>

which will return:

[1, 2, 3, 4]

Answered   2023-09-20 20:25:10

  • Please do this folks. Don't jack something onto to the Array prototype. Please. - anyone
  • @JacobDalton Please don't do this. There's no need to add an extra library just for a small job that can be done with array = [...new Set(array)] - anyone
  • @JacobDalton why not? Is there a downside to "jacking something" onto the array? - anyone
  • If you want to find unique objects using object equality semantics, Lodash offers _.uniqWith(arrayOfObjects, _.isEqual). _.uniqWith([{a: 2}, {b: 3}, {a: 2}], _.isEqual) gives [{a: 2}, {b: 3}]. - anyone
  • This is a good solution overall, but surely the best solution if you already have a library to do that in your dependencies. This answer inspired me to look up something among my dependencies and, lo and behold, we already use lodash, which already has an uniq() funciton. So, even if it is not the same lib, thanks for the heads-up! - anyone

One Liner, Pure JavaScript

With ES6 syntax

list = list.filter((x, i, a) => a.indexOf(x) == i)

x --> item in array
i --> index of item
a --> array reference, (in this case "list")

enter image description here

With ES5 syntax

list = list.filter(function (x, i, a) { 
    return a.indexOf(x) == i; 
});

Browser Compatibility: IE9+

Answered   2023-09-20 20:25:10

Remove duplicates using Set.

Array with duplicates

const withDuplicates = [2, 2, 5, 5, 1, 1, 2, 2, 3, 3];

Get a new array without duplicates by using Set

const withoutDuplicates = Array.from(new Set(withDuplicates));

A shorter version

const withoutDuplicates = [...new Set(withDuplicates)];

Result: [2, 5, 1, 3]

Answered   2023-09-20 20:25:10

  • Completely duplicated answer for ones other ones given years before this one. - anyone
  • perhaps a new Set to remove duplicates - anyone

Many of the answers here may not be useful to beginners. If de-duping an array is difficult, will they really know about the prototype chain, or even jQuery?

In modern browsers, a clean and simple solution is to store data in a Set, which is designed to be a list of unique values.

const cars = ['Volvo', 'Jeep', 'Volvo', 'Lincoln', 'Lincoln', 'Ford'];
const uniqueCars = Array.from(new Set(cars));
console.log(uniqueCars);

The Array.from is useful to convert the Set back to an Array so that you have easy access to all of the awesome methods (features) that arrays have. There are also other ways of doing the same thing. But you may not need Array.from at all, as Sets have plenty of useful features like forEach.

If you need to support old Internet Explorer, and thus cannot use Set, then a simple technique is to copy items over to a new array while checking beforehand if they are already in the new array.

// Create a list of cars, with duplicates.
var cars = ['Volvo', 'Jeep', 'Volvo', 'Lincoln', 'Lincoln', 'Ford'];
// Create a list of unique cars, to put a car in if we haven't already.
var uniqueCars = [];

// Go through each car, one at a time.
cars.forEach(function (car) {
    // The code within the following block runs only if the
    // current car does NOT exist in the uniqueCars list
    // - a.k.a. prevent duplicates
    if (uniqueCars.indexOf(car) === -1) {
        // Since we now know we haven't seen this car before,
        // copy it to the end of the uniqueCars list.
        uniqueCars.push(car);
    }
});

To make this instantly reusable, let's put it in a function.

function deduplicate(data) {
    if (data.length > 0) {
        var result = [];

        data.forEach(function (elem) {
            if (result.indexOf(elem) === -1) {
                result.push(elem);
            }
        });

        return result;
    }
}

So to get rid of the duplicates, we would now do this.

var uniqueCars = deduplicate(cars);

The deduplicate(cars) part becomes the thing we named result when the function completes.

Just pass it the name of any array you like.

Answered   2023-09-20 20:25:10

  • How would this work if I wanted the new array to not be uniques, but be an array of values that were duplicated? So using the above example, the array I'm looking for is ["volvo","lincoln"] - anyone
  • @Jason I'd probably create a Map to store previously seen items and an array to store the duplicate items. Then loop through the cars array and check if the Map has the current item, if it does then push it to the duplicates array, if not then add it to the Map. I'd be happy to create a code example for you if you create a new question and we can continue the discussion there. - anyone
  • Just note that if the array is empty, the function does not return anything. - anyone
  • too bad you didn't show the set method as an extension of array.prototype - anyone
  • I deliberately did not show that. I consider it bad practice to modify Array.prototype and would strongly advise against doing so. Among other reasons, it can lead to future bugs if new methods are added to Array as part of the language. In fact, if your site is popular, it could even prevent new methods from being added to the language. That has actually happened. See: 2ality.com/2022/03/naming-conflicts.html - anyone

Using ES6 new Set

var array = [3,7,5,3,2,5,2,7];
var unique_array = [...new Set(array)];
console.log(unique_array);    // output = [3,7,5,2]

Using For Loop

var array = [3,7,5,3,2,5,2,7];

for(var i=0;i<array.length;i++) {
    for(var j=i+1;j<array.length;j++) {
        if(array[i]===array[j]) {
            array.splice(j,1);
        }
    }
}
console.log(array); // output = [3,7,5,2]

Answered   2023-09-20 20:25:10

I have since found a nice method that uses jQuery

arr = $.grep(arr, function(v, k){
    return $.inArray(v ,arr) === k;
});

Note: This code was pulled from Paul Irish's duck punching post - I forgot to give credit :P

Answered   2023-09-20 20:25:10

  • A concise solution, but calling inArray is way less efficient than calling hasOwnProperty. - anyone
  • This is also O(N^2), right? Whereas the dictionary or hasOwnProperty approach would likely be O(N*logN). - anyone

Magic

a.filter(e=>!(t[e]=e in t)) 

O(n) performance - we assume your array is in a and t={}. Explanation here (+Jeppe impr.)

let unique = (a,t={}) => a.filter(e=>!(t[e]=e in t));

// "stand-alone" version working with global t:
// a1.filter((t={},e=>!(t[e]=e in t)));

// Test data
let a1 = [5,6,0,4,9,2,3,5,0,3,4,1,5,4,9];
let a2 = [[2, 17], [2, 17], [2, 17], [1, 12], [5, 9], [1, 12], [6, 2], [1, 12]];
let a3 = ['Mike', 'Adam','Matt', 'Nancy', 'Adam', 'Jenny', 'Nancy', 'Carl'];

// Results
console.log(JSON.stringify( unique(a1) ))
console.log(JSON.stringify( unique(a2) ))
console.log(JSON.stringify( unique(a3) ))

Answered   2023-09-20 20:25:10

  • this look so super cool, that without a solid explanation i fell you're gonna mine bitcoins when i run this - anyone
  • what i meant is that you should expand your answer with some explanation and commented deconstruction of it. don't expect people will find useful answers like this. (though it really looks cool a probably works) - anyone
  • Not magic, but is much like the "Set"-answers, using O(1) key-lookups in the dictionary. Do you need to increment the counters though? How about "e=>!(t[e]=e in t)". Nice answer though. - anyone
  • @Jeppe when I run your improvement then I experience aha effect (before I don't know that I can use in operator outside the other construction than for loop :P) - Thank you - I appreciate it and will give +2 to your other good answers. - anyone
  • Well, the solution is so great indeed However it works well for tuples only The examples below work incorrect: unique(['2', 2]) // ['2']; unique([[1, 7], [1, '7'], ['1', 7], ['1', '7']]) // [1, 7] So be careful using this - anyone

The simplest, and fastest (in Chrome) way of doing this:

Array.prototype.unique = function() {
    var a = [];
    for (var i=0, l=this.length; i<l; i++)
        if (a.indexOf(this[i]) === -1)
            a.push(this[i]);
    return a;
}

Simply goes through every item in the array, tests if that item is already in the list, and if it's not, pushes to the array that gets returned.

According to JSBench, this function is the fastest of the ones I could find anywhere - feel free to add your own though.

The non-prototype version:

function uniques(arr) {
    var a = [];
    for (var i=0, l=arr.length; i<l; i++)
        if (a.indexOf(arr[i]) === -1 && arr[i] !== '')
            a.push(arr[i]);
    return a;
}

Sorting

When also needing to sort the array, the following is the fastest:

Array.prototype.sortUnique = function() {
    this.sort();
    var last_i;
    for (var i=0;i<this.length;i++)
        if ((last_i = this.lastIndexOf(this[i])) !== i)
            this.splice(i+1, last_i-i);
    return this;
}

or non-prototype:

function sortUnique(arr) {
    arr.sort();
    var last_i;
    for (var i=0;i<arr.length;i++)
        if ((last_i = arr.lastIndexOf(arr[i])) !== i)
            arr.splice(i+1, last_i-i);
    return arr;
}

This is also faster than the above method in most non-Chrome browsers.

Answered   2023-09-20 20:25:10

  • On Linux, Chrome 55.0.2883 prefers your arr.unique() and swilliams' arrclone2.sortFilter() is slowest (78% slower). However, Firefox 51.0.0 (with lots of addons) has swilliams as fastest (yet still slower by Ops/sec than any other Chrome result) with mottie's jQuery $.grep(arr, jqFilter) being slowest (46% slower). Your arr.uniq() was 30% slower. I ran each test twice and got consistent results. Rafael's arr.getUnique() got second place in both browsers. - anyone
  • jsPerf is buggy at the moment, so my edit to this test didn't commit everything, but it did result in adding two tests: Cocco's toUnique() beats Vamsi's ES6 list.filter() on both browsers, beating swilliams' sortFilter() for #1 on FF (sortFilter was 16% slower) and beating your sorted testing (which was slower by 2%) for #3 on Chrome. - anyone
  • Ah, I hadn't caught that those tests were trivially small and don't really matter. A comment to the accepted answer describes that problem and offers a correction in a revision to the test, in which Rafael's code is easily the fastest and Joetje50's arr.unique code is 98% slower. I've also made another revision as noted in this comment. - anyone
  • Well, actually the algorithm you implemented in unique function has O(n^2) complexity while the one in getUnique is O(n). The first one may be faster on small data sets, but how can you argue with the maths :) You can make sure the latter one is faster if you run it on an array of, say, 1e5 unique items - anyone
  • also used by lodash.uniq for input_array.length < 200, otherwise uses the [...new Set(input_array)] method. expressed as reducer: input_array.reduce((c, v) => {if (!c.includes(v)) c.push(v); return c;}, []) - anyone

We can do this using ES6 sets:

var duplicatesArray = [1, 2, 3, 4, 5, 1, 1, 1, 2, 3, 4];
var uniqueArray = [...new Set(duplicatesArray)];

console.log(uniqueArray); // [1,2,3,4,5]

Answered   2023-09-20 20:25:10

["Defects", "Total", "Days", "City", "Defects"].reduce(function(prev, cur) {
  return (prev.indexOf(cur) < 0) ? prev.concat([cur]) : prev;
 }, []);

[0,1,2,0,3,2,1,5].reduce(function(prev, cur) {
  return (prev.indexOf(cur) < 0) ? prev.concat([cur]) : prev;
 }, []);

Answered   2023-09-20 20:25:10

  • This worked very well with an array of objects! [].reduce((p, c) => (p.some((item) => c.id === item.id) ? p : p.concat(c)), []) - anyone

This has been answered a lot, but it didn't address my particular need.

Many answers are like this:

a.filter((item, pos, self) => self.indexOf(item) === pos);

But this doesn't work for arrays of complex objects.

Say we have an array like this:

const a = [
 { age: 4, name: 'fluffy' },
 { age: 5, name: 'spot' },
 { age: 2, name: 'fluffy' },
 { age: 3, name: 'toby' },
];

If we want the objects with unique names, we should use array.prototype.findIndex instead of array.prototype.indexOf:

a.filter((item, pos, self) => self.findIndex(v => v.name === item.name) === pos);

Answered   2023-09-20 20:25:10

  • Great solution, beware that a new array will return from a function. (it doesn't modify itself) - anyone
  • Works will with a complex array of objets - anyone
  • @EdgarQuintero only if the elements are actually the exact same object, so the array [ { a: 2 }, { a: 2 } ] won't work as many people might expect if you use the indexOf solution, but the findIndex solution might be useful - anyone

After looking into all the 90+ answers here, I saw there is room for one more:

Array.includes has a very handy second-parameter: "fromIndex", so by using it, every iteration of the filter callback method will search the array, starting from [current index] + 1 which guarantees not to include currently filtered item in the lookup and also saves time.

Note - this solution does not retain the order, as it removed duplicated items from left to right, but it wins the Set trick if the Array is a collection of Objects.

//                🚩              🚩 🚩
var list = [0,1,2,2,3,'a','b',4,5,2,'a']

console.log( 
  list.filter((v,i) => !list.includes(v,i+1))
)

// [0,1,3,"b",4,5,2,"a"]

Explanation:

For example, lets assume the filter function is currently iterating at index 2) and the value at that index happens to be 2. The section of the array that is then scanned for duplicates (includes method) is everything after index 2 (i+1):

           👇                    👇
[0, 1, 2,   2 ,3 ,'a', 'b', 4, 5, 2, 'a']
       👆   |---------------------------|

And since the currently filtered item's value 2 is included in the rest of the array, it will be filtered out, because of the leading exclamation mark which negates the filter rule.


If order is important, use this method:

//                🚩              🚩 🚩
var list = [0,1,2,2,3,'a','b',4,5,2,'a']

console.log( 
  // Initialize with empty array and fill with non-duplicates
  list.reduce((acc, v) => (!acc.includes(v) && acc.push(v), acc), [])
)

// [0,1,2,3,"a","b",4,5]

Answered   2023-09-20 20:25:10

  • Unfortunately, this keeps the LAST instance of each value, not the first. (which might be ok, but I think keeping the first is generally what's expected) - anyone
  • Although this is a nice optimization idea, I don't think people would want to use this since this's keeping the last item which basically mess up the array order. - anyone

You can simlply use the built-in functions Array.prototype.filter() and Array.prototype.indexOf()

array.filter((x, y) => array.indexOf(x) == y)

var arr = [1, 2, 3, 3, 4, 5, 5, 5, 6, 7, 8, 9, 6, 9];

var newarr = arr.filter((x, y) => arr.indexOf(x) == y);

console.log(newarr);

Answered   2023-09-20 20:25:10

This prototype getUnique is not totally correct, because if i have a Array like: ["1",1,2,3,4,1,"foo"] it will return ["1","2","3","4"] and "1" is string and 1 is a integer; they are different.

Here is a correct solution:

Array.prototype.unique = function(a){
    return function(){ return this.filter(a) }
}(function(a,b,c){ return c.indexOf(a,b+1) < 0 });

using:

var foo;
foo = ["1",1,2,3,4,1,"foo"];
foo.unique();

The above will produce ["1",2,3,4,1,"foo"].

Answered   2023-09-20 20:25:10

  • Note that $foo = 'bar' is the PHP way of declaring variables. It will work in javascript, but will create an implicit global, and generally shouldn't be done. - anyone
  • @CamiloMartin sorry but you're wrong, $foo is global because the example is not in a closure and he's missing the var keyword. Nothing to do with the dollar jsfiddle.net/robaldred/L2MRb - anyone
  • @Rob that's exactly what I'm saying, PHP people will think $foo is the way of declaring variables in javascript while actually var foo is. - anyone
[...new Set(duplicates)]

This is the simplest one and referenced from MDN Web Docs.

const numbers = [2,3,4,4,2,3,3,4,4,5,5,6,6,7,5,32,3,4,5]
console.log([...new Set(numbers)]) // [2, 3, 4, 5, 6, 7, 32]

Answered   2023-09-20 20:25:10

  • While this code may solve the question, including an explanation of how and why this solves the problem would really help to improve the quality of your post, and probably result in more up-votes. Remember that you are answering the question for readers in the future, not just the person asking now. Please edit your answer to add explanation, and give an indication of what limitations and assumptions apply. - anyone
  • Identical to previous answer which dates a year prior to this one - anyone

Without extending Array.prototype (it is said to be a bad practice) or using jquery/underscore, you can simply filter the array.

By keeping last occurrence:

    function arrayLastUnique(array) {
        return array.filter(function (a, b, c) {
            // keeps last occurrence
            return c.indexOf(a, b + 1) < 0;
        });
    },

or first occurrence:

    function arrayFirstUnique(array) {
        return array.filter(function (a, b, c) {
            // keeps first occurrence
            return c.indexOf(a) === b;
        });
    },

Well, it's only javascript ECMAScript 5+, which means only IE9+, but it's nice for a development in native HTML/JS (Windows Store App, Firefox OS, Sencha, Phonegap, Titanium, ...).

Answered   2023-09-20 20:25:10

  • The fact that it's js 1.6 does not mean you can't use filter. At the MDN page they have an implementation for Internet Explorer, I mean, older browsers. Also: JS 1.6 refers only to Firefox's js engine, but the right thing to say it's that it is ECMAScript 5. - anyone

That's because 0 is a falsy value in JavaScript.

this[i] will be falsy if the value of the array is 0 or any other falsy value.

Answered   2023-09-20 20:25:10

  • Ahhhh, ok I see now... but would there be an easy fix to make it work? - anyone
Array.prototype.getUnique = function() {
    var o = {}, a = []
    for (var i = 0; i < this.length; i++) o[this[i]] = 1
    for (var e in o) a.push(e)
    return a
}

Answered   2023-09-20 20:25:10

  • I think this won't work if the array contains objects/arrays, and I'm not sure if it will preserve the type of scalars. - anyone
  • Yes, everything gets stringified. That could be fixed by storing the original value in o instead of just a 1, although equality comparison would still be stringwise (although, out of all the possible Javascript equalities, it doesn't seem too unreasonable). - anyone
  • The Array.prototype could be extended only with non enumerable methods .... Object.defineProperty(Array.prototype,"getUnique",{}) ... but the idea of using a helper object is very nice - anyone

Now using sets you can remove duplicates and convert them back to the array.

var names = ["Mike","Matt","Nancy", "Matt","Adam","Jenny","Nancy","Carl"];

console.log([...new Set(names)])

Another solution is to use sort & filter

var names = ["Mike","Matt","Nancy", "Matt","Adam","Jenny","Nancy","Carl"];
var namesSorted = names.sort();
const result = namesSorted.filter((e, i) => namesSorted[i] != namesSorted[i+1]);
console.log(result);

Answered   2023-09-20 20:25:10

If you're using Prototype framework there is no need to do 'for' loops, you can use http://prototypejs.org/doc/latest/language/Array/prototype/uniq/ like this:

var a = Array.uniq();  

Which will produce a duplicate array with no duplicates. I came across your question searching a method to count distinct array records so after uniq() I used size() and there was my simple result. p.s. Sorry if i mistyped something

edit: if you want to escape undefined records you may want to add compact() before, like this:

var a = Array.compact().uniq();  

Answered   2023-09-20 20:25:10

  • because i found a better answer, i think about topics are for all people not just for the one who asked - anyone
  • Thanks for time machine yet iirc around 15 years ago JS community had debate and result is - do not extend prototype cause of side effects and cause you pollute all JS arrays this way. - anyone

I had a slightly different problem where I needed to remove objects with duplicate id properties from an array. this worked.

let objArr = [{
  id: '123'
}, {
  id: '123'
}, {
  id: '456'
}];

objArr = objArr.reduce((acc, cur) => [
  ...acc.filter((obj) => obj.id !== cur.id), cur
], []);

console.log(objArr);

Answered   2023-09-20 20:25:10

If you're okay with extra dependencies, or you already have one of the libraries in your codebase, you can remove duplicates from an array in place using LoDash (or Underscore).

Usage

If you don't have it in your codebase already, install it using npm:

npm install lodash

Then use it as follows:

import _ from 'lodash';
let idArray = _.uniq ([
    1,
    2,
    3,
    3,
    3
]);
console.dir(idArray);

Out:

[ 1, 2, 3 ]

Answered   2023-09-20 20:25:10

  • You can also use lodash to remove objects with duplicate properties from an array: _.uniqWith(objectArray, _.isEqual). - anyone

Primitive values

With Set (Recommended)

var array = ["FreePhoenix888", "FreePhoenix888", "konard", "FreePhoenix888"];

let set = [...new Set(array)];

console.log(set); // ["FreePhoenix888", "konard"]

Without Set

function filterUniqueObjects(value, index, array) {
  return array.indexOf(value) === index;
}

// usage example:
var array = ["FreePhoenix888", "FreePhoenix888", "konard", "FreePhoenix888"];
var arrayOfUniqueItems = array.filter(filterUniqueObjects);

console.log(arrayOfUniqueItems); // ["FreePhoenix888", "konard"]

Objects

This example shows how you can filter not just an array of primitive values but an array of objects. I have added comments to make it easier to understand what you can change there depending on your requirements.

let array = [
  { name: '@deep-foundation/core', version: '0.0.2' },
  { name: '@deep-foundation/capacitor-device', version: '10.0.1' },
  { name: '@deep-foundation/capacitor-device', version: '10.0.2' },
];

// Of course you can inline this function as filter argument uniqueArray.filter((item, index, self) => self.findIndex(innerItem => innerItem.name === item.name) === index);
function filterUniqueObjects(value, index, self) {
  return (
    self.findIndex(
      // Modify this function as you desire. You may want to calculate uniqueness depending only on specific fields, not all
      (obj) => obj.name === value.name
    ) === index
  );
};

let uniqueArray = array
  .reverse() // If you want latest duplicates to remain
  .filter(filterUniqueObjects)
  .reverse(); // To get back to original order after first reverse

console.log(uniqueArray)

Answered   2023-09-20 20:25:10

I'm not sure why Gabriel Silveira wrote the function that way but a simpler form that works for me just as well and without the minification is:

Array.prototype.unique = function() {
  return this.filter(function(value, index, array) {
    return array.indexOf(value, index + 1) < 0;
  });
};

or in CoffeeScript:

Array.prototype.unique = ->
  this.filter( (value, index, array) ->
    array.indexOf(value, index + 1) < 0
  )

Answered   2023-09-20 20:25:10

Finding unique Array values in simple method

function arrUnique(a){
  var t = [];
  for(var x = 0; x < a.length; x++){
    if(t.indexOf(a[x]) == -1)t.push(a[x]);
  }
  return t;
}
arrUnique([1,4,2,7,1,5,9,2,4,7,2]) // [1, 4, 2, 7, 5, 9]

Answered   2023-09-20 20:25:10

  • How can this answer be correct? Expected result for unique array is [5,9] as per the input given [1,4,2,7,1,5,9,2,4,7,2] - anyone
  • @PoojaThapa this will extract the unique values from input array. - anyone

It appears we have lost Rafael's answer, which stood as the accepted answer for a few years. This was (at least in 2017) the best-performing solution if you don't have a mixed-type array:

Array.prototype.getUnique = function(){
    var u = {}, a = [];
    for (var i = 0, l = this.length; i < l; ++i) {
        if (u.hasOwnProperty(this[i])) {
            continue;
        }
        a.push(this[i]);
        u[this[i]] = 1;
    }
return a;
}

If you do have a mixed-type array, you can serialize the hash key:

Array.prototype.getUnique = function() {
    var hash = {}, result = [], key; 
    for ( var i = 0, l = this.length; i < l; ++i ) {
        key = JSON.stringify(this[i]);
        if ( !hash.hasOwnProperty(key) ) {
            hash[key] = true;
            result.push(this[i]);
        }
    }
    return result;
}

Answered   2023-09-20 20:25:10

strange this hasn't been suggested before.. to remove duplicates by object key (id below) in an array you can do something like this:

const uniqArray = array.filter((obj, idx, arr) => (
  arr.findIndex((o) => o.id === obj.id) === idx
)) 

Answered   2023-09-20 20:25:10

  • Don't both filter() and findIndex() have to iterate through the array? That would make this a double-loop and therefore twice as expensive to run as any other answer here. - anyone
  • @AdamKatz yes it will iterate over the array n+1 times. Please be aware that the other answers here using a combination of map, filter, indexOf, reduce etc. also must do this, it's sort of inherent in the problem. To avoid, you could use new Set(), or a lookup object similar to the answer by Grozz. - anyone