Get all unique values in a JavaScript array (remove duplicates)
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;
}
M开发者_如何学运维ore answers from duplicate question:
- Remove duplicate values from JS array
Similar question:
- Get all non-unique values (i.e.: duplicate/more than one occurrence) in an array
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']
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((v, i, a) => a.indexOf(v) === i);
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.
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]
I split all answers to 4 possible solutions:
- Use object
{ }
to prevent duplicates - Use helper array
[ ]
- Use
filter + indexOf
- 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:
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.
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]
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")
With ES5 syntax
list = list.filter(function (x, i, a) {
return a.indexOf(x) == i;
});
Browser Compatibility: IE9+
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]
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.
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]
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
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.
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) ))
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]
["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;
}, []);
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"]
.
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.
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