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Find all duplicates in array of objects [duplicate]

I need to check a JavaScript array to see if there are any duplicate values. What's the easiest way to do this? I just need to find what the duplicated values are - I don't actually need their indexes or how many times they are duplicated.
I know I can loop through the array and check all the other values for a match, but it seems like there should be an easier way.
Similar question:
Get all unique values in a JavaScript array (remove duplicates)

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)}`);
In case, if you are to return as a function for duplicates. This is for similar type of case.
Reference: https://stackoverflow.com/a/57532964/8119511

If you want to elimate the duplicates, try this great solution:
function eliminateDuplicates(arr) {
var i,
len = arr.length,
out = [],
obj = {};
for (i = 0; i < len; i++) {
obj[arr[i]] = 0;
}
for (i in obj) {
out.push(i);
}
return out;
}
console.log(eliminateDuplicates([1,6,7,3,6,8,1,3,4,5,1,7,2,6]))
Source:
http://dreaminginjavascript.wordpress.com/2008/08/22/eliminating-duplicates/

This is my answer from the duplicate thread (!):
When writing this entry 2014 - all examples were for-loops or jQuery. JavaScript has the perfect tools for this: sort, map and reduce.
Find duplicate items
var names = ['Mike', 'Matt', 'Nancy', 'Adam', 'Jenny', 'Nancy', 'Carl']
const uniq = names
.map((name) => {
return {
count: 1,
name: name
};
})
.reduce((result, b) => {
result[b.name] = (result[b.name] || 0) + b.count;
return result;
}, {});
const duplicates = Object.keys(uniq).filter((a) => uniq[a] > 1);
console.log(duplicates); // [ 'Nancy' ]
More functional syntax:
#Dmytro-Laptin pointed out some code that can be removed. This is a more compact version of the same code. Using some ES6 tricks and higher-order functions:
const names = ['Mike', 'Matt', 'Nancy', 'Adam', 'Jenny', 'Nancy', 'Carl'];
const count = names =>
names.reduce((result, value) => ({ ...result,
[value]: (result[value] || 0) + 1
}), {}); // don't forget to initialize the accumulator
const duplicates = dict =>
Object.keys(dict).filter((a) => dict[a] > 1);
console.log(count(names)); // { Mike: 1, Matt: 1, Nancy: 2, Adam: 1, Jenny: 1, Carl: 1 }
console.log(duplicates(count(names))); // [ 'Nancy' ]

UPDATED: Short one-liner to get the duplicates:
[1, 2, 2, 4, 3, 4].filter((e, i, a) => a.indexOf(e) !== i) // [2, 4]
To get the array without duplicates simply invert the condition:
[1, 2, 2, 4, 3, 4].filter((e, i, a) => a.indexOf(e) === i) // [1, 2, 3, 4]
Note that this answer’s main goal is to be short. If you need something performant for a big array, one possible solution is to sort your array first (if it is sortable) then do the following to get the same kind of results as above:
myHugeSortedArray.filter((e, i, a) => a[i-1] === e)
Here is an example for a 1 000 000 integers array:
const myHugeIntArrayWithDuplicates =
[...Array(1_000_000).keys()]
// adding two 0 and four 9 duplicates
.fill(0, 2, 4).fill(9, 10, 14)
console.time("time")
console.log(
myHugeIntArrayWithDuplicates
// a possible sorting method for integers
.sort((a, b) => a > b ? 1 : -1)
.filter((e, i, a) => a[i-1] === e)
)
console.timeEnd("time")
On my AMD Ryzen 7 5700G dev machine it outputs:
[ 0, 0, 9, 9, 9, 9 ]
time: 22.738ms
As pointed out in the comments both the short solution and the performant solution will return an array with several time the same duplicate if it occurs more than once in the original array:
[1, 1, 1, 2, 2, 2, 2].filter((e, i, a) => a.indexOf(e) !== i) // [1, 1, 2, 2, 2]
If unique duplicates are wanted then a function like
function duplicates(arr) {
return [...new Set(arr.filter((e, i, a) => a.indexOf(e) !== i))]
}
can be used so that duplicates([1, 1, 1, 2, 2, 2, 2]) returns [1, 2].
When all you need is to check that there are no duplicates as asked in this question you can use the every() method:
[1, 2, 3].every((e, i, a) => a.indexOf(e) === i) // true
[1, 2, 1].every((e, i, a) => a.indexOf(e) === i) // false
Note that every() doesn't work for IE 8 and below.

Find duplicate values in an array
This should be one of the shortest ways to actually find duplicate values in an array. As specifically asked for by the OP, this does not remove duplicates but finds them.
var input = [1, 2, 3, 1, 3, 1];
var duplicates = input.reduce(function(acc, el, i, arr) {
if (arr.indexOf(el) !== i && acc.indexOf(el) < 0) acc.push(el); return acc;
}, []);
document.write(duplicates); // = 1,3 (actual array == [1, 3])
This doesn't need sorting or any third party framework. It also doesn't need manual loops. It works with every value indexOf() (or to be clearer: the strict comparision operator) supports.
Because of reduce() and indexOf() it needs at least IE 9.

You can add this function, or tweak it and add it to Javascript's Array prototype:
Array.prototype.unique = function () {
var r = new Array();
o:for(var i = 0, n = this.length; i < n; i++)
{
for(var x = 0, y = r.length; x < y; x++)
{
if(r[x]==this[i])
{
alert('this is a DUPE!');
continue o;
}
}
r[r.length] = this[i];
}
return r;
}
var arr = [1,2,2,3,3,4,5,6,2,3,7,8,5,9];
var unique = arr.unique();
alert(unique);

UPDATED: The following uses an optimized combined strategy. It optimizes primitive lookups to benefit from hash O(1) lookup time (running unique on an array of primitives is O(n)). Object lookups are optimized by tagging objects with a unique id while iterating through so so identifying duplicate objects is also O(1) per item and O(n) for the whole list. The only exception is items that are frozen, but those are rare and a fallback is provided using an array and indexOf.
var unique = function(){
var hasOwn = {}.hasOwnProperty,
toString = {}.toString,
uids = {};
function uid(){
var key = Math.random().toString(36).slice(2);
return key in uids ? uid() : uids[key] = key;
}
function unique(array){
var strings = {}, numbers = {}, others = {},
tagged = [], failed = [],
count = 0, i = array.length,
item, type;
var id = uid();
while (i--) {
item = array[i];
type = typeof item;
if (item == null || type !== 'object' && type !== 'function') {
// primitive
switch (type) {
case 'string': strings[item] = true; break;
case 'number': numbers[item] = true; break;
default: others[item] = item; break;
}
} else {
// object
if (!hasOwn.call(item, id)) {
try {
item[id] = true;
tagged[count++] = item;
} catch (e){
if (failed.indexOf(item) === -1)
failed[failed.length] = item;
}
}
}
}
// remove the tags
while (count--)
delete tagged[count][id];
tagged = tagged.concat(failed);
count = tagged.length;
// append primitives to results
for (i in strings)
if (hasOwn.call(strings, i))
tagged[count++] = i;
for (i in numbers)
if (hasOwn.call(numbers, i))
tagged[count++] = +i;
for (i in others)
if (hasOwn.call(others, i))
tagged[count++] = others[i];
return tagged;
}
return unique;
}();
If you have ES6 Collections available, then there is a much simpler and significantly faster version. (shim for IE9+ and other browsers here: https://github.com/Benvie/ES6-Harmony-Collections-Shim)
function unique(array){
var seen = new Set;
return array.filter(function(item){
if (!seen.has(item)) {
seen.add(item);
return true;
}
});
}

var a = ["a","a","b","c","c"];
a.filter(function(value,index,self){ return (self.indexOf(value) !== index )})

This should get you what you want, Just the duplicates.
function find_duplicates(arr) {
var len=arr.length,
out=[],
counts={};
for (var i=0;i<len;i++) {
var item = arr[i];
counts[item] = counts[item] >= 1 ? counts[item] + 1 : 1;
if (counts[item] === 2) {
out.push(item);
}
}
return out;
}
find_duplicates(['one',2,3,4,4,4,5,6,7,7,7,'pig','one']); // -> ['one',4,7] in no particular order.

Find non-unique values from 3 arrays (or more):
ES2015
// 🚩🚩 🚩 🚩 🚩
var arr = [1,2,2,3,3,4,5,6,2,3,7,8,5,22],
arr2 = [1,2,511,12,50],
arr3 = [22,0],
merged,
nonUnique;
// Combine all the arrays to a single one
merged = arr.concat(arr2, arr3)
// create a new (dirty) Array with only the non-unique items
nonUnique = merged.filter((item,i) => merged.includes(item, i+1))
// Cleanup - remove duplicate & empty items items
nonUnique = [...new Set(nonUnique)]
console.log(nonUnique)
PRE-ES2015:
In the below example I chose to superimpose a unique method on top of the Array prototype, allowing access from everywhere and has more "declarative" syntax. I do not recommend this approach on large projects, since it might very well collide with another method with the same custom name.
Array.prototype.unique = function () {
var arr = this.sort(), i=arr.length; // input must be sorted for this to work
while(i--)
arr[i] === arr[i-1] && arr.splice(i,1) // remove duplicate item
return arr
}
Array.prototype.nonunique = function () {
var arr = this.sort(), i=arr.length, res = []; // input must be sorted for this to work
while(i--)
arr[i] === arr[i-1] && (res.indexOf(arr[i]) == -1) && res.push(arr[i])
return res
}
// 🚩🚩 🚩 🚩 🚩
var arr = [1,2,2,3,3,4,5,6,2,3,7,8,5,22],
arr2 = [1,2,511,12,50],
arr3 = [22,0],
// merge all arrays & call custom Array Prototype - "unique"
unique = arr.concat(arr2, arr3).unique(),
nonunique = arr.concat(arr2, arr3).nonunique()
console.log(unique) // [1,12,2,22,3,4,5,50,511,6,7,8]
console.log(nonunique) // [1,12,2,22,3,4,5,50,511,6,7,8]

using underscore.js
function hasDuplicate(arr){
return (arr.length != _.uniq(arr).length);
}

The simplest and quickest way is to use the Set object:
const numbers = [1, 2, 3, 2, 4, 5, 5, 6];
const set = new Set(numbers);
const duplicates = numbers.filter(item => {
if (set.has(item)) {
set.delete(item);
return false;
} else {
return true;
}
});
// OR more concisely
const duplicates = numbers.filter(item => !set.delete(item));
console.log(duplicates);
// [ 2, 5 ]

This is my proposal (ES6):
let a = [1, 2, 3, 4, 2, 2, 4, 1, 5, 6]
let b = [...new Set(a.sort().filter((o, i) => o !== undefined && a[i + 1] !== undefined && o === a[i + 1]))]
// b is now [1, 2, 4]

Here's the simplest solution I could think of:
const arr = [-1, 2, 2, 2, 0, 0, 0, 500, -1, 'a', 'a', 'a']
const filtered = arr.filter((el, index) => arr.indexOf(el) !== index)
// => filtered = [ 2, 2, 0, 0, -1, 'a', 'a' ]
const duplicates = [...new Set(filtered)]
console.log(duplicates)
// => [ 2, 0, -1, 'a' ]
That's it.
Note:
It works with any numbers including 0, strings and negative numbers e.g. -1 -
Related question: Get all unique values in a JavaScript array (remove duplicates)
The original array arr is preserved (filter returns the new array instead of modifying the original)
The filtered array contains all duplicates; it can also contain more than 1 same value (e.g. our filtered array here is [ 2, 2, 0, 0, -1, 'a', 'a' ])
If you want to get only values that are duplicated (you don't want to have multiple duplicates with the same value) you can use [...new Set(filtered)] (ES6 has an object Set which can store only unique values)
Hope this helps.

Here is mine simple and one line solution.
It searches not unique elements first, then makes found array unique with the use of Set.
So we have array of duplicates in the end.
var array = [1, 2, 2, 3, 3, 4, 5, 6, 2, 3, 7, 8, 5, 22, 1, 2, 511, 12, 50, 22];
console.log([...new Set(
array.filter((value, index, self) => self.indexOf(value) !== index))]
);

Shortest vanilla JS:
[1,1,2,2,2,3].filter((v,i,a) => a.indexOf(v) !== i) // [1, 2, 2]

one liner simple way
var arr = [9,1,2,4,3,4,9]
console.log(arr.filter((ele,indx)=>indx!==arr.indexOf(ele))) //get the duplicates
console.log(arr.filter((ele,indx)=>indx===arr.indexOf(ele))) //remove the duplicates

var a = [324,3,32,5,52,2100,1,20,2,3,3,2,2,2,1,1,1].sort();
a.filter(function(v,i,o){return i&&v!==o[i-1]?v:0;});
or when added to the prototyp.chain of Array
//copy and paste: without error handling
Array.prototype.unique =
function(){return this.sort().filter(function(v,i,o){return i&&v!==o[i-1]?v:0;});}
See here: https://gist.github.com/1305056

Fast and elegant way using es6 object destructuring and reduce
It runs in O(n) (1 iteration over the array) and doesn't repeat values that appear more than 2 times
const arr = ['hi', 'hi', 'hi', 'bye', 'bye', 'asd']
const {
dup
} = arr.reduce(
(acc, curr) => {
acc.items[curr] = acc.items[curr] ? acc.items[curr] += 1 : 1
if (acc.items[curr] === 2) acc.dup.push(curr)
return acc
}, {
items: {},
dup: []
},
)
console.log(dup)
// ['hi', 'bye']

You can use filter method and indexOf() to get all the duplicate values
function duplicate(arr) {
return duplicateArray = arr.filter((item, index) => arr.indexOf(item) !== index)
}
arr.indexOf(item) will always return the first index at which a given element can be
found

ES5 only (i.e., it needs a filter() polyfill for IE8 and below):
var arrayToFilter = [ 4, 5, 5, 5, 2, 1, 3, 1, 1, 2, 1, 3 ];
arrayToFilter.
sort().
filter( function(me,i,arr){
return (i===0) || ( me !== arr[i-1] );
});

Here is a very light and easy way:
var codes = dc_1.split(',');
var i = codes.length;
while (i--) {
if (codes.indexOf(codes[i]) != i) {
codes.splice(i,1);
}
}

ES6 offers the Set data structure which is basically an array that doesn't accept duplicates.
With the Set data structure, there's a very easy way to find duplicates in an array (using only one loop).
Here's my code
function findDuplicate(arr) {
var set = new Set();
var duplicates = new Set();
for (let i = 0; i< arr.length; i++) {
var size = set.size;
set.add(arr[i]);
if (set.size === size) {
duplicates.add(arr[i]);
}
}
return duplicates;
}

With ES6 (or using Babel or Typescipt) you can simply do:
var duplicates = myArray.filter(i => myArray.filter(ii => ii === i).length > 1);
https://es6console.com/j58euhbt/

Simple code with ES6 syntax (return sorted array of duplicates):
let duplicates = a => {d=[]; a.sort((a,b) => a-b).reduce((a,b)=>{a==b&&!d.includes(a)&&d.push(a); return b}); return d};
How to use:
duplicates([1,2,3,10,10,2,3,3,10]);

I have just figured out a simple way to achieve this using an Array filter
var list = [9, 9, 111, 2, 3, 4, 4, 5, 7];
// Filter 1: to find all duplicates elements
var duplicates = list.filter(function(value,index,self) {
return self.indexOf(value) !== self.lastIndexOf(value) && self.indexOf(value) === index;
});
console.log(duplicates);

This answer might also be helpful, it leverages js reduce operator/method to remove duplicates from array.
const result = [1, 2, 2, 3, 3, 3, 3].reduce((x, y) => x.includes(y) ? x : [...x, y], []);
console.log(result);

Higher ranked answers have a few inherent issues including the use of legacy javascript, incorrect ordering or with only support for 2 duplicated items.
Here's a modern solution which fixes those problems:
const arrayNonUniq = array => {
if (!Array.isArray(array)) {
throw new TypeError("An array must be provided!")
}
return array.filter((value, index) => array.indexOf(value) === index && array.lastIndexOf(value) !== index)
}
arrayNonUniq([1, 1, 2, 3, 3])
//=> [1, 3]
arrayNonUniq(["foo", "foo", "bar", "foo"])
//=> ['foo']
You can also use the npm package array-non-uniq.

The following function (a variation of the eliminateDuplicates function already mentioned) seems to do the trick, returning test2,1,7,5 for the input ["test", "test2", "test2", 1, 1, 1, 2, 3, 4, 5, 6, 7, 7, 10, 22, 43, 1, 5, 8]
Note that the problem is stranger in JavaScript than in most other languages, because a JavaScript array can hold just about anything. Note that solutions that use sorting might need to provide an appropriate sorting function--I haven't tried that route yet.
This particular implementation works for (at least) strings and numbers.
function findDuplicates(arr) {
var i,
len=arr.length,
out=[],
obj={};
for (i=0;i<len;i++) {
if (obj[arr[i]] != null) {
if (!obj[arr[i]]) {
out.push(arr[i]);
obj[arr[i]] = 1;
}
} else {
obj[arr[i]] = 0;
}
}
return out;
}

var arr = [2, 1, 2, 2, 4, 4, 2, 5];
function returnDuplicates(arr) {
return arr.reduce(function(dupes, val, i) {
if (arr.indexOf(val) !== i && dupes.indexOf(val) === -1) {
dupes.push(val);
}
return dupes;
}, []);
}
alert(returnDuplicates(arr));
This function avoids the sorting step and uses the reduce() method to push duplicates to a new array if it doesn't already exist in it.

Order an unordered array of numbers from 1-8, so that the end and starting integers are alternated eg [8,1,7,2,6,3,5,4,]

I'm a newbie to all of this and trying to improve myself by solving problems and challenges.
I came across a problem whereby I have an unordered array which contains 8 integers.
eg [2,3,1,4,6,5,8,7]
I need to sort it [1,2,3,4,5,6,7,8] and reorder the array so that the array starts with the end value and then the first value and so on eg [8,1,7,2,6,3,5,4,]
I worked out I could use map() to iterate across the array and then use push() with pop() and shift() however it leaves the last 2 numbers behind in the original array and I'm not sure why. I got around this by using a concat and a reverse but I still don't understand why pop and shift don't bring across all the elements.
Code below that doesn't pull all the elements:
const reorder = (array) => {
let store = []
array.sort((a, b) => a - b).map((item, i) => {
if (array) {
store.push(array.pop())
store.push(array.shift())
}
})
return store
}
reorder([2, 3, 1, 4, 6, 5, 8, 7]) // returns [8,1,7,2,6,3]
Code that works but I have to add a concat and a reverse:
const reorder = (array) => {
let store = []
array.sort((a, b) => a - b).map((item, i) => {
if (array) {
store.push(array.pop())
store.push(array.shift())
}
})
return store.concat(array.reverse())
}
reorder([2, 3, 1, 4, 6, 5, 8, 7]) //returns [8,1,7,2,6,3,5,4]
Thanks for any help

I would just bisect the array, sort them in opposite orders and then add each element from each array to a new array
Given that you want to then take the sorted bisected arrays and produce another single array, I'd then use Array.prototype.reduce:
const alternatingSort = function (array) {
array = array.sort();
const midpoint = Math.round(array.length / 2)
let arr1 = array.slice(0, midpoint);
let arr2 = array.slice(midpoint);
arr2 = arr2.sort(function (a, b) { return b - a });
return arr1.reduce(function (retVal, item, index) {
arr2[index] && retVal.push(arr2[index]);
retVal.push(item);
return retVal;
}, []);
}
console.log(alternatingSort([2, 3, 1, 4, 6, 5, 8, 7]));
console.log(alternatingSort([2, 3, 1, 4, 6, 5, 8])); // with odd number

As I've seen nobody explained why the original OP solution doesn't work, Here is why:
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/
Map calls a provided callback function once for each element in an array, in order, and constructs a new array from the results. callback is invoked only for indexes of the array which have assigned values (including undefined).
It is not called for missing elements of the array; that is:
1.Indexes that have never been set;
2.which have been deleted; or
3.which have never been assigned a value.
So what is happening in our code is that:
On the first iteration,
[(2), 3, 1, 4, 6, 5, 8, 7]
Map picks the first element(2) in the array, and delete the first and last characters in the array, so the array becomes
[3,(1), 4, 6, 5, 8]
Now, as map will not consider deleted elements, the second element(1) in the current array is called, also the first and last element in also removed:
[1, 4,(6), 5]
Now, map is trying to find the third element(6), and delete the first and last element:
[4,6]
Now, map is trying to find the fourth element, which is out of bound, so the map function will terminate.
So, you are strongly advised not to use Array.prototype.shift or Array.prototype.pop in Array.prototype.map.

You can do it following way:
const reorder = (array) => {
array.sort((a, b) => a - b);
const result = [];
const length = array.length;
for (let i = 0; i < length; i++) {
if (i % 2 === 0) {
result.push(array.pop());
} else {
result.push(array.shift());
}
}
return result;
}
const result = reorder([2, 3, 1, 4, 6, 5, 7]);
console.log(result);
Notice that I've intentionally made the array length to be an odd number. Some of the solutions here will break if the length is an odd number.

Personally I would sort, split in half and then just insert in. Not very fancy, but gets the job done.
function strangeWeave (arr) {
var sorted = arr.slice().sort()
var result = sorted.splice(0,Math.floor(sorted.length/2))
for (let i=0;sorted.length;i+=2) {
result.splice(i,0,sorted.pop())
}
return result
}
console.log(strangeWeave([1,2]))
console.log(strangeWeave([1,2,3]))
console.log(strangeWeave([1,2,3,4,5,6,7,8]))
console.log(strangeWeave([1,2,3,4,5,6,7,8,9]))

There is a much easier solution to sort two different arrays, one normal and one in reverse, then connect them together. Here is the code for that:
var myArray = [1, 3, 2, 4, 5, 7, 6, 8];
function getCopy(arr) {
var x = [];
for(var i = 0; i < arr.length; i++)
x.push(arr[i]);
return x;
}
function sortMyWay(arr) {
var sortedArr = [],
leftSide = getCopy(arr).sort()
.splice(0, Math.ceil(arr.length / 2)),
rightSide = getCopy(arr).sort().reverse()
.splice(0, Math.floor(arr.length / 2));
for(var i = 0; i < arr.length; i++)
i % 2
? sortedArr.push(leftSide[Math.floor(i / 2)])
: sortedArr.push(rightSide[Math.floor(i / 2)]);
console.log(sortedArr);
return sortedArr;
}
var sortedArr = sortMyWay(myArray);
Hope it helped!
Happy coding :)

How to remove an array element with JavaScript [duplicate]

How do I remove a specific value from an array? Something like:
array.remove(value);
Constraints: I have to use core JavaScript. Frameworks are not allowed.

Find the index of the array element you want to remove using indexOf, and then remove that index with splice.
The splice() method changes the contents of an array by removing
existing elements and/or adding new elements.
const array = [2, 5, 9];
console.log(array);
const index = array.indexOf(5);
if (index > -1) { // only splice array when item is found
array.splice(index, 1); // 2nd parameter means remove one item only
}
// array = [2, 9]
console.log(array);
The second parameter of splice is the number of elements to remove. Note that splice modifies the array in place and returns a new array containing the elements that have been removed.
For the reason of completeness, here are functions. The first function removes only a single occurrence (i.e. removing the first match of 5 from [2,5,9,1,5,8,5]), while the second function removes all occurrences:
function removeItemOnce(arr, value) {
var index = arr.indexOf(value);
if (index > -1) {
arr.splice(index, 1);
}
return arr;
}
function removeItemAll(arr, value) {
var i = 0;
while (i < arr.length) {
if (arr[i] === value) {
arr.splice(i, 1);
} else {
++i;
}
}
return arr;
}
// Usage
console.log(removeItemOnce([2,5,9,1,5,8,5], 5))
console.log(removeItemAll([2,5,9,1,5,8,5], 5))
In TypeScript, these functions can stay type-safe with a type parameter:
function removeItem<T>(arr: Array<T>, value: T): Array<T> {
const index = arr.indexOf(value);
if (index > -1) {
arr.splice(index, 1);
}
return arr;
}

Edited on 2016 October
Do it simple, intuitive and explicit (Occam's razor)
Do it immutable (original array stays unchanged)
Do it with standard JavaScript functions, if your browser doesn't support them - use polyfill
In this code example I use array.filter(...) function to remove unwanted items from an array. This function doesn't change the original array and creates a new one. If your browser doesn't support this function (e.g. Internet Explorer before version 9, or Firefox before version 1.5), consider polyfilling with core-js.
Removing item (ECMA-262 Edition 5 code AKA old style JavaScript)
var value = 3
var arr = [1, 2, 3, 4, 5, 3]
arr = arr.filter(function(item) {
return item !== value
})
console.log(arr)
// [ 1, 2, 4, 5 ]
Removing item (ECMAScript 6 code)
let value = 3
let arr = [1, 2, 3, 4, 5, 3]
arr = arr.filter(item => item !== value)
console.log(arr)
// [ 1, 2, 4, 5 ]
IMPORTANT ECMAScript 6 () => {} arrow function syntax is not supported in Internet Explorer at all, Chrome before version 45, Firefox before version 22, and Safari before version 10. To use ECMAScript 6 syntax in old browsers you can use BabelJS.
Removing multiple items (ECMAScript 7 code)
An additional advantage of this method is that you can remove multiple items
let forDeletion = [2, 3, 5]
let arr = [1, 2, 3, 4, 5, 3]
arr = arr.filter(item => !forDeletion.includes(item))
// !!! Read below about array.includes(...) support !!!
console.log(arr)
// [ 1, 4 ]
IMPORTANT array.includes(...) function is not supported in Internet Explorer at all, Chrome before version 47, Firefox before version 43, Safari before version 9, and Edge before version 14 but you can polyfill with core-js.
Removing multiple items (in the future, maybe)
If the "This-Binding Syntax" proposal is ever accepted, you'll be able to do this:
// array-lib.js
export function remove(...forDeletion) {
return this.filter(item => !forDeletion.includes(item))
}
// main.js
import { remove } from './array-lib.js'
let arr = [1, 2, 3, 4, 5, 3]
// :: This-Binding Syntax Proposal
// using "remove" function as "virtual method"
// without extending Array.prototype
arr = arr::remove(2, 3, 5)
console.log(arr)
// [ 1, 4 ]
Try it yourself in BabelJS :)
Reference
Array.prototype.includes
Functional composition

I don't know how you are expecting array.remove(int) to behave. There are three possibilities I can think of that you might want.
To remove an element of an array at an index i:
array.splice(i, 1);
If you want to remove every element with value number from the array:
for (var i = array.length - 1; i >= 0; i--) {
if (array[i] === number) {
array.splice(i, 1);
}
}
If you just want to make the element at index i no longer exist, but you don't want the indexes of the other elements to change:
delete array[i];

It depends on whether you want to keep an empty spot or not.
If you do want an empty slot:
array[index] = undefined;
If you don't want an empty slot:
//To keep the original:
//oldArray = [...array];
//This modifies the array.
array.splice(index, 1);
And if you need the value of that item, you can just store the returned array's element:
var value = array.splice(index, 1)[0];
If you want to remove at either end of the array, you can use array.pop() for the last one or array.shift() for the first one (both return the value of the item as well).
If you don't know the index of the item, you can use array.indexOf(item) to get it (in a if() to get one item or in a while() to get all of them). array.indexOf(item) returns either the index or -1 if not found.

A friend was having issues in Internet Explorer 8 and showed me what he did. I told him it was wrong, and he told me he got the answer here. The current top answer will not work in all browsers (Internet Explorer 8 for example), and it will only remove the first occurrence of the item.
Remove ALL instances from an array
function removeAllInstances(arr, item) {
for (var i = arr.length; i--;) {
if (arr[i] === item) arr.splice(i, 1);
}
}
It loops through the array backwards (since indices and length will change as items are removed) and removes the item if it's found. It works in all browsers.

There are two major approaches
splice(): anArray.splice(index, 1);
let fruits = ['Apple', 'Banana', 'Mango', 'Orange']
let removed = fruits.splice(2, 1);
// fruits is ['Apple', 'Banana', 'Orange']
// removed is ['Mango']
delete: delete anArray[index];
let fruits = ['Apple', 'Banana', 'Mango', 'Orange']
let removed = delete fruits(2);
// fruits is ['Apple', 'Banana', undefined, 'Orange']
// removed is true
Be careful when you use the delete for an array. It is good for deleting attributes of objects, but not so good for arrays. It is better to use splice for arrays.
Keep in mind that when you use delete for an array you could get wrong results for anArray.length. In other words, delete would remove the element, but it wouldn't update the value of the length property.
You can also expect to have holes in index numbers after using delete, e.g. you could end up with having indexes 1, 3, 4, 8, 9, and 11 and length as it was before using delete. In that case, all indexed for loops would crash, since indexes are no longer sequential.
If you are forced to use delete for some reason, then you should use for each loops when you need to loop through arrays. As the matter of fact, always avoid using indexed for loops, if possible. That way the code would be more robust and less prone to problems with indexes.

Array.prototype.removeByValue = function (val) {
for (var i = 0; i < this.length; i++) {
if (this[i] === val) {
this.splice(i, 1);
i--;
}
}
return this;
}
var fruits = ['apple', 'banana', 'carrot', 'orange'];
fruits.removeByValue('banana');
console.log(fruits);
// -> ['apple', 'carrot', 'orange']

There isn't any need to use indexOf or splice. However, it performs better if you only want to remove one occurrence of an element.
Find and move (move):
function move(arr, val) {
var j = 0;
for (var i = 0, l = arr.length; i < l; i++) {
if (arr[i] !== val) {
arr[j++] = arr[i];
}
}
arr.length = j;
}
Use indexOf and splice (indexof):
function indexof(arr, val) {
var i;
while ((i = arr.indexOf(val)) != -1) {
arr.splice(i, 1);
}
}
Use only splice (splice):
function splice(arr, val) {
for (var i = arr.length; i--;) {
if (arr[i] === val) {
arr.splice(i, 1);
}
}
}
Run-times on Node.js for an array with 1000 elements (averaged over 10,000 runs):
indexof is approximately 10 times slower than move. Even if improved by removing the call to indexOf in splice, it performs much worse than move.
Remove all occurrences:
move 0.0048 ms
indexof 0.0463 ms
splice 0.0359 ms
Remove first occurrence:
move_one 0.0041 ms
indexof_one 0.0021 ms

This provides a predicate instead of a value.
NOTE: it will update the given array, and return the affected rows.
Usage
var removed = helper.remove(arr, row => row.id === 5 );
var removed = helper.removeAll(arr, row => row.name.startsWith('BMW'));
Definition
var helper = {
// Remove and return the first occurrence
remove: function(array, predicate) {
for (var i = 0; i < array.length; i++) {
if (predicate(array[i])) {
return array.splice(i, 1);
}
}
},
// Remove and return all occurrences
removeAll: function(array, predicate) {
var removed = [];
for (var i = 0; i < array.length; ) {
if (predicate(array[i])) {
removed.push(array.splice(i, 1));
continue;
}
i++;
}
return removed;
},
};

You can do it easily with the filter method:
function remove(arrOriginal, elementToRemove){
return arrOriginal.filter(function(el){return el !== elementToRemove});
}
console.log(remove([1, 2, 1, 0, 3, 1, 4], 1));
This removes all elements from the array and also works faster than a combination of slice and indexOf.

Using filter is an elegant way to achieve this requirement.
filter will not mutate the original array.
const num = 3;
let arr = [1, 2, 3, 4];
const arr2 = arr.filter(x => x !== num);
console.log(arr); // [1, 2, 3, 4]
console.log(arr2); // [1, 2, 4]
You can use filter and then assign the result to the original array if you want to achieve a mutation removal behaviour.
const num = 3;
let arr = [1, 2, 3, 4];
arr = arr.filter(x => x !== num);
console.log(arr); // [1, 2, 4]
By the way, filter will remove all of the occurrences matched in the condition (not just the first occurrence) like you can see in the following example
const num = 3;
let arr = [1, 2, 3, 3, 3, 4];
arr = arr.filter(x => x !== num);
console.log(arr); // [1, 2, 4]
In case, you just want to remove the first occurrence, you can use the splice method
const num = 3;
let arr = [1, 2, 3, 3, 3, 4];
const idx = arr.indexOf(num);
arr.splice(idx, idx !== -1 ? 1 : 0);
console.log(arr); // [1, 2, 3, 3, 4]

John Resig posted a good implementation:
// Array Remove - By John Resig (MIT Licensed)
Array.prototype.remove = function(from, to) {
var rest = this.slice((to || from) + 1 || this.length);
this.length = from < 0 ? this.length + from : from;
return this.push.apply(this, rest);
};
If you don’t want to extend a global object, you can do something like the following, instead:
// Array Remove - By John Resig (MIT Licensed)
Array.remove = function(array, from, to) {
var rest = array.slice((to || from) + 1 || array.length);
array.length = from < 0 ? array.length + from : from;
return array.push.apply(array, rest);
};
But the main reason I am posting this is to warn users against the alternative implementation suggested in the comments on that page (Dec 14, 2007):
Array.prototype.remove = function(from, to) {
this.splice(from, (to=[0, from || 1, ++to - from][arguments.length]) < 0 ? this.length + to : to);
return this.length;
};
It seems to work well at first, but through a painful process I discovered it fails when trying to remove the second to last element in an array. For example, if you have a 10-element array and you try to remove the 9th element with this:
myArray.remove(8);
You end up with an 8-element array. I don't know why, but I confirmed John's original implementation doesn't have this problem.

You can use ES6. For example to delete the value '3' in this case:
var array=['1','2','3','4','5','6']
var newArray = array.filter((value)=>value!='3');
console.log(newArray);
Output :
["1", "2", "4", "5", "6"]

Underscore.js can be used to solve issues with multiple browsers. It uses in-build browser methods if present. If they are absent like in the case of older Internet Explorer versions it uses its own custom methods.
A simple example to remove elements from array (from the website):
_.without([1, 2, 1, 0, 3, 1, 4], 0, 1); // => [2, 3, 4]

If you want a new array with the deleted positions removed, you can always delete the specific element and filter out the array. It might need an extension of the array object for browsers that don't implement the filter method, but in the long term it's easier since all you do is this:
var my_array = [1, 2, 3, 4, 5, 6];
delete my_array[4];
console.log(my_array.filter(function(a){return typeof a !== 'undefined';}));
It should display [1, 2, 3, 4, 6].

Here are a few ways to remove an item from an array using JavaScript.
All the method described do not mutate the original array, and instead create a new one.
If you know the index of an item
Suppose you have an array, and you want to remove an item in position i.
One method is to use slice():
const items = ['a', 'b', 'c', 'd', 'e', 'f']
const i = 3
const filteredItems = items.slice(0, i).concat(items.slice(i+1, items.length))
console.log(filteredItems)
slice() creates a new array with the indexes it receives. We simply create a new array, from start to the index we want to remove, and concatenate another array from the first position following the one we removed to the end of the array.
If you know the value
In this case, one good option is to use filter(), which offers a more declarative approach:
const items = ['a', 'b', 'c', 'd', 'e', 'f']
const valueToRemove = 'c'
const filteredItems = items.filter(item => item !== valueToRemove)
console.log(filteredItems)
This uses the ES6 arrow functions. You can use the traditional functions to support older browsers:
const items = ['a', 'b', 'c', 'd', 'e', 'f']
const valueToRemove = 'c'
const filteredItems = items.filter(function(item) {
return item !== valueToRemove
})
console.log(filteredItems)
or you can use Babel and transpile the ES6 code back to ES5 to make it more digestible to old browsers, yet write modern JavaScript in your code.
Removing multiple items
What if instead of a single item, you want to remove many items?
Let's find the simplest solution.
By index
You can just create a function and remove items in series:
const items = ['a', 'b', 'c', 'd', 'e', 'f']
const removeItem = (items, i) =>
items.slice(0, i-1).concat(items.slice(i, items.length))
let filteredItems = removeItem(items, 3)
filteredItems = removeItem(filteredItems, 5)
//["a", "b", "c", "d"]
console.log(filteredItems)
By value
You can search for inclusion inside the callback function:
const items = ['a', 'b', 'c', 'd', 'e', 'f']
const valuesToRemove = ['c', 'd']
const filteredItems = items.filter(item => !valuesToRemove.includes(item))
// ["a", "b", "e", "f"]
console.log(filteredItems)
Avoid mutating the original array
splice() (not to be confused with slice()) mutates the original array, and should be avoided.
(originally posted on my site https://flaviocopes.com/how-to-remove-item-from-array/)

Check out this code. It works in every major browser.
remove_item = function(arr, value) {
var b = '';
for (b in arr) {
if (arr[b] === value) {
arr.splice(b, 1);
break;
}
}
return arr;
};
var array = [1,3,5,6,5,9,5,3,55]
var res = remove_item(array,5);
console.log(res)

Removing a particular element/string from an array can be done in a one-liner:
theArray.splice(theArray.indexOf("stringToRemoveFromArray"), 1);
where:
theArray: the array you want to remove something particular from
stringToRemoveFromArray: the string you want to be removed and 1 is the number of elements you want to remove.
NOTE: If "stringToRemoveFromArray" is not located in the array, this will remove the last element of the array.
It's always good practice to check if the element exists in your array first, before removing it.
if (theArray.indexOf("stringToRemoveFromArray") >= 0){
theArray.splice(theArray.indexOf("stringToRemoveFromArray"), 1);
}
Depending if you have newer or older version of Ecmascript running on your client's computers:
var array=['1','2','3','4','5','6']
var newArray = array.filter((value)=>value!='3');
OR
var array = ['1','2','3','4','5','6'];
var newArray = array.filter(function(item){ return item !== '3' });
Where '3' is the value you want to be removed from the array.
The array would then become : ['1','2','4','5','6']

ES10
This post summarizes common approaches to element removal from an array as of ECMAScript 2019 (ES10).
1. General cases
1.1. Removing Array element by value using .splice()
| In-place: Yes |
| Removes duplicates: Yes(loop), No(indexOf) |
| By value / index: By index |
If you know the value you want to remove from an array you can use the splice method. First, you must identify the index of the target item. You then use the index as the start element and remove just one element.
// With a 'for' loop
const arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 0];
for( let i = 0; i < arr.length; i++){
if ( arr[i] === 5) {
arr.splice(i, 1);
}
} // => [1, 2, 3, 4, 6, 7, 8, 9, 0]
// With the .indexOf() method
const arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 0];
const i = arr.indexOf(5);
arr.splice(i, 1); // => [1, 2, 3, 4, 6, 7, 8, 9, 0]
1.2. Removing Array element using the .filter() method
| In-place: No |
| Removes duplicates: Yes |
| By value / index: By value |
The specific element can be filtered out from the array, by providing a filtering function. Such function is then called for every element in the array.
const value = 3
let arr = [1, 2, 3, 4, 5, 3]
arr = arr.filter(item => item !== value)
console.log(arr)
// [ 1, 2, 4, 5 ]
1.3. Removing Array element by extending Array.prototype
| In-place: Yes/No (Depends on implementation) |
| Removes duplicates: Yes/No (Depends on implementation) |
| By value / index: By index / By value (Depends on implementation) |
The prototype of Array can be extended with additional methods. Such methods will be then available to use on created arrays.
Note: Extending prototypes of objects from the standard library of JavaScript (like Array) is considered by some as an antipattern.
// In-place, removes all, by value implementation
Array.prototype.remove = function(item) {
for (let i = 0; i < this.length; i++) {
if (this[i] === item) {
this.splice(i, 1);
}
}
}
const arr1 = [1,2,3,1];
arr1.remove(1) // arr1 equals [2,3]
// Non-stationary, removes first, by value implementation
Array.prototype.remove = function(item) {
const arr = this.slice();
for (let i = 0; i < this.length; i++) {
if (arr[i] === item) {
arr.splice(i, 1);
return arr;
}
}
return arr;
}
let arr2 = [1,2,3,1];
arr2 = arr2.remove(1) // arr2 equals [2,3,1]
1.4. Removing Array element using the delete operator
| In-place: Yes |
| Removes duplicates: No |
| By value / index: By index |
Using the delete operator does not affect the length property. Nor does it affect the indexes of subsequent elements. The array becomes sparse, which is a fancy way of saying the deleted item is not removed but becomes undefined.
const arr = [1, 2, 3, 4, 5, 6];
delete arr[4]; // Delete element with index 4
console.log( arr ); // [1, 2, 3, 4, undefined, 6]
The delete operator is designed to remove properties from JavaScript objects, which arrays are objects.
1.5. Removing Array element using Object utilities (>= ES10)
| In-place: No |
| Removes duplicates: Yes |
| By value / index: By value |
ES10 introduced Object.fromEntries, which can be used to create the desired Array from any Array-like object and filter unwanted elements during the process.
const object = [1,2,3,4];
const valueToRemove = 3;
const arr = Object.values(Object.fromEntries(
Object.entries(object)
.filter(([ key, val ]) => val !== valueToRemove)
));
console.log(arr); // [1,2,4]
2. Special cases
2.1 Removing element if it's at the end of the Array
2.1.1. Changing Array length
| In-place: Yes |
| Removes duplicates: No |
| By value / index: N/A |
JavaScript Array elements can be removed from the end of an array by setting the length property to a value less than the current value. Any element whose index is greater than or equal to the new length will be removed.
const arr = [1, 2, 3, 4, 5, 6];
arr.length = 5; // Set length to remove element
console.log( arr ); // [1, 2, 3, 4, 5]
2.1.2. Using .pop() method
| In-place: Yes |
| Removes duplicates: No |
| By value / index: N/A |
The pop method removes the last element of the array, returns that element, and updates the length property. The pop method modifies the array on which it is invoked, This means unlike using delete the last element is removed completely and the array length reduced.
const arr = [1, 2, 3, 4, 5, 6];
arr.pop(); // returns 6
console.log( arr ); // [1, 2, 3, 4, 5]
2.2. Removing element if it's at the beginning of the Array
| In-place: Yes |
| Removes duplicates: No |
| By value / index: N/A |
The .shift() method works much like the pop method except it removes the first element of a JavaScript array instead of the last. When the element is removed the remaining elements are shifted down.
const arr = [1, 2, 3, 4];
arr.shift(); // returns 1
console.log( arr ); // [2, 3, 4]
2.3. Removing element if it's the only element in the Array
| In-place: Yes |
| Removes duplicates: N/A |
| By value / index: N/A |
The fastest technique is to set an array variable to an empty array.
let arr = [1];
arr = []; //empty array
Alternatively technique from 2.1.1 can be used by setting length to 0.

You can use lodash _.pull (mutate array), _.pullAt (mutate array) or _.without (does't mutate array),
var array1 = ['a', 'b', 'c', 'd']
_.pull(array1, 'c')
console.log(array1) // ['a', 'b', 'd']
var array2 = ['e', 'f', 'g', 'h']
_.pullAt(array2, 0)
console.log(array2) // ['f', 'g', 'h']
var array3 = ['i', 'j', 'k', 'l']
var newArray = _.without(array3, 'i') // ['j', 'k', 'l']
console.log(array3) // ['i', 'j', 'k', 'l']

ES6 & without mutation: (October 2016)
const removeByIndex = (list, index) =>
[
...list.slice(0, index),
...list.slice(index + 1)
];
output = removeByIndex([33,22,11,44],1) //=> [33,11,44]
console.log(output)

Performance
Today (2019-12-09) I conduct performance tests on macOS v10.13.6 (High Sierra) for chosen solutions. I show delete (A), but I do not use it in comparison with other methods, because it left empty space in the array.
The conclusions
the fastest solution is array.splice (C) (except Safari for small arrays where it has the second time)
for big arrays, array.slice+splice (H) is the fastest immutable solution for Firefox and Safari; Array.from (B) is fastest in Chrome
mutable solutions are usually 1.5x-6x faster than immutable
for small tables on Safari, surprisingly the mutable solution (C) is slower than the immutable solution (G)
Details
In tests, I remove the middle element from the array in different ways. The A, C solutions are in-place. The B, D, E, F, G, H solutions are immutable.
Results for an array with 10 elements
In Chrome the array.splice (C) is the fastest in-place solution. The array.filter (D) is the fastest immutable solution. The slowest is array.slice (F). You can perform the test on your machine here.
Results for an array with 1.000.000 elements
In Chrome the array.splice (C) is the fastest in-place solution (the delete (C) is similar fast - but it left an empty slot in the array (so it does not perform a 'full remove')). The array.slice-splice (H) is the fastest immutable solution. The slowest is array.filter (D and E). You can perform the test on your machine here.
var a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9];
var log = (letter,array) => console.log(letter, array.join `,`);
function A(array) {
var index = array.indexOf(5);
delete array[index];
log('A', array);
}
function B(array) {
var index = array.indexOf(5);
var arr = Array.from(array);
arr.splice(index, 1)
log('B', arr);
}
function C(array) {
var index = array.indexOf(5);
array.splice(index, 1);
log('C', array);
}
function D(array) {
var arr = array.filter(item => item !== 5)
log('D', arr);
}
function E(array) {
var index = array.indexOf(5);
var arr = array.filter((item, i) => i !== index)
log('E', arr);
}
function F(array) {
var index = array.indexOf(5);
var arr = array.slice(0, index).concat(array.slice(index + 1))
log('F', arr);
}
function G(array) {
var index = array.indexOf(5);
var arr = [...array.slice(0, index), ...array.slice(index + 1)]
log('G', arr);
}
function H(array) {
var index = array.indexOf(5);
var arr = array.slice(0);
arr.splice(index, 1);
log('H', arr);
}
A([...a]);
B([...a]);
C([...a]);
D([...a]);
E([...a]);
F([...a]);
G([...a]);
H([...a]);
This snippet only presents code used in performance tests - it does not perform tests itself.
Comparison for browsers: Chrome v78.0.0, Safari v13.0.4, and Firefox v71.0.0

OK, for example you have the array below:
var num = [1, 2, 3, 4, 5];
And we want to delete number 4. You can simply use the below code:
num.splice(num.indexOf(4), 1); // num will be [1, 2, 3, 5];
If you are reusing this function, you write a reusable function which will be attached to the native array function like below:
Array.prototype.remove = Array.prototype.remove || function(x) {
const i = this.indexOf(x);
if(i===-1)
return;
this.splice(i, 1); // num.remove(5) === [1, 2, 3];
}
But how about if you have the below array instead with a few [5]s in the array?
var num = [5, 6, 5, 4, 5, 1, 5];
We need a loop to check them all, but an easier and more efficient way is using built-in JavaScript functions, so we write a function which use a filter like below instead:
const _removeValue = (arr, x) => arr.filter(n => n!==x);
//_removeValue([1, 2, 3, 4, 5, 5, 6, 5], 5) // Return [1, 2, 3, 4, 6]
Also there are third-party libraries which do help you to do this, like Lodash or Underscore. For more information, look at lodash _.pull, _.pullAt or _.without.

I'm pretty new to JavaScript and needed this functionality. I merely wrote this:
function removeFromArray(array, item, index) {
while((index = array.indexOf(item)) > -1) {
array.splice(index, 1);
}
}
Then when I want to use it:
//Set-up some dummy data
var dummyObj = {name:"meow"};
var dummyArray = [dummyObj, "item1", "item1", "item2"];
//Remove the dummy data
removeFromArray(dummyArray, dummyObj);
removeFromArray(dummyArray, "item2");
Output - As expected.
["item1", "item1"]
You may have different needs than I, so you can easily modify it to suit them. I hope this helps someone.

I want to answer based on ECMAScript 6. Assume you have an array like below:
let arr = [1,2,3,4];
If you want to delete at a special index like 2, write the below code:
arr.splice(2, 1); //=> arr became [1,2,4]
But if you want to delete a special item like 3 and you don't know its index, do like below:
arr = arr.filter(e => e !== 3); //=> arr became [1,2,4]
Hint: please use an arrow function for filter callback unless you will get an empty array.

If you have complex objects in the array you can use filters?
In situations where $.inArray or array.splice is not as easy to use. Especially if the objects are perhaps shallow in the array.
E.g. if you have an object with an Id field and you want the object removed from an array:
this.array = this.array.filter(function(element, i) {
return element.id !== idToRemove;
});

Update: This method is recommended only if you cannot use ECMAScript 2015 (formerly known as ES6). If you can use it, other answers here provide much neater implementations.
This gist here will solve your problem, and also deletes all occurrences of the argument instead of just 1 (or a specified value).
Array.prototype.destroy = function(obj){
// Return null if no objects were found and removed
var destroyed = null;
for(var i = 0; i < this.length; i++){
// Use while-loop to find adjacent equal objects
while(this[i] === obj){
// Remove this[i] and store it within destroyed
destroyed = this.splice(i, 1)[0];
}
}
return destroyed;
}
Usage:
var x = [1, 2, 3, 3, true, false, undefined, false];
x.destroy(3); // => 3
x.destroy(false); // => false
x; // => [1, 2, true, undefined]
x.destroy(true); // => true
x.destroy(undefined); // => undefined
x; // => [1, 2]
x.destroy(3); // => null
x; // => [1, 2]

You should never mutate your array as this is against the functional programming pattern. You can create a new array without referencing the one you want to change data of using the ECMAScript 6 method filter;
var myArray = [1, 2, 3, 4, 5, 6];
Suppose you want to remove 5 from the array, you can simply do it like this:
myArray = myArray.filter(value => value !== 5);
This will give you a new array without the value you wanted to remove. So the result will be:
[1, 2, 3, 4, 6]; // 5 has been removed from this array
For further understanding you can read the MDN documentation on Array.filter.

A more modern, ECMAScript 2015 (formerly known as Harmony or ES 6) approach. Given:
const items = [1, 2, 3, 4];
const index = 2;
Then:
items.filter((x, i) => i !== index);
Yielding:
[1, 2, 4]
You can use Babel and a polyfill service to ensure this is well supported across browsers.

You can do a backward loop to make sure not to screw up the indexes, if there are multiple instances of the element.
var myElement = "chocolate";
var myArray = ['chocolate', 'poptart', 'poptart', 'poptart', 'chocolate', 'poptart', 'poptart', 'chocolate'];
/* Important code */
for (var i = myArray.length - 1; i >= 0; i--) {
if (myArray[i] == myElement) myArray.splice(i, 1);
}
console.log(myArray);

counting number of occurrences in an array [duplicate]

I have an array:
[1, 2, 3, 5, 2, 8, 9, 2]
I would like to know how many 2s are in the array.
What is the most elegant way to do it in JavaScript without looping with for loop?

[this answer is a bit dated: read the edits, in the notion of 'equal' in javascript is ambiguous]
Say hello to your friends: map and filter and reduce and forEach and every etc.
(I only occasionally write for-loops in javascript, because of block-level scoping is missing, so you have to use a function as the body of the loop anyway if you need to capture or clone your iteration index or value. For-loops are more efficient generally, but sometimes you need a closure.)
The most readable way:
[....].filter(x => x==2).length
(We could have written .filter(function(x){return x==2}).length instead)
The following is more space-efficient (O(1) rather than O(N)), but I'm not sure how much of a benefit/penalty you might pay in terms of time (not more than a constant factor since you visit each element exactly once):
[....].reduce((total,x) => (x==2 ? total+1 : total), 0)
or as a commenter kindly pointed out:
[....].reduce((total,x) => total+(x==2), 0)
(If you need to optimize this particular piece of code, a for loop might be faster on some browsers... you can test things on jsperf.com.)
You can then be elegant and turn it into a prototype function:
[1, 2, 3, 5, 2, 8, 9, 2].count(2)
Like this:
Object.defineProperties(Array.prototype, {
count: {
value: function(value) {
return this.filter(x => x==value).length;
}
}
});
You can also stick the regular old for-loop technique (see other answers) inside the above property definition (again, that would likely be much faster).
2017 edit:
Whoops, this answer has gotten more popular than the correct answer. Actually, just use the accepted answer. While this answer may be cute, the js compilers probably don't (or can't due to spec) optimize such cases. So you should really write a simple for loop:
Object.defineProperties(Array.prototype, {
count: {
value: function(query) {
/*
Counts number of occurrences of query in array, an integer >= 0
Uses the javascript == notion of equality.
*/
var count = 0;
for(let i=0; i<this.length; i++)
if (this[i]==query)
count++;
return count;
}
}
});
You could define a version .countStrictEq(...) which used the === notion of equality. The notion of equality may be important to what you're doing! (for example [1,10,3,'10'].count(10)==2, because numbers like '4'==4 in javascript... hence calling it .countEq or .countNonstrict stresses it uses the == operator.)
Caveat:
Defining a common name on the prototype should be done with care. It is fine if you control your code, but bad if everyone wants to declare their own [].count function, especially if they behave differently. You may ask yourself "but .count(query) surely sounds quite perfect and canonical"... but consider perhaps you could do something like [].count(x=> someExpr of x). In that case you define functions like countIn(query, container) (under myModuleName.countIn), or something, or [].myModuleName_count().
Also consider using your own multiset data structure (e.g. like python's 'collections.Counter') to avoid having to do the counting in the first place. This works for exact matches of the form [].filter(x=> x==???).length (worst case O(N) down to O(1)), and modified will speed up queries of the form [].filter(filterFunction).length (roughly by a factor of #total/#duplicates).
class Multiset extends Map {
constructor(...args) {
super(...args);
}
add(elem) {
if (!this.has(elem))
this.set(elem, 1);
else
this.set(elem, this.get(elem)+1);
}
remove(elem) {
var count = this.has(elem) ? this.get(elem) : 0;
if (count>1) {
this.set(elem, count-1);
} else if (count==1) {
this.delete(elem);
} else if (count==0)
throw `tried to remove element ${elem} of type ${typeof elem} from Multiset, but does not exist in Multiset (count is 0 and cannot go negative)`;
// alternatively do nothing {}
}
}
Demo:
> counts = new Multiset([['a',1],['b',3]])
Map(2) {"a" => 1, "b" => 3}
> counts.add('c')
> counts
Map(3) {"a" => 1, "b" => 3, "c" => 1}
> counts.remove('a')
> counts
Map(2) {"b" => 3, "c" => 1}
> counts.remove('a')
Uncaught tried to remove element a of type string from Multiset, but does not exist in Multiset (count is 0 and cannot go negative)
sidenote: Though, if you still wanted the functional-programming way (or a throwaway one-liner without overriding Array.prototype), you could write it more tersely nowadays as [...].filter(x => x==2).length. If you care about performance, note that while this is asymptotically the same performance as the for-loop (O(N) time), it may require O(N) extra memory (instead of O(1) memory) because it will almost certainly generate an intermediate array and then count the elements of that intermediate array.

Modern JavaScript:
Note that you should always use triple equals === when doing comparison in JavaScript (JS). The triple equals make sure, that JS comparison behaves like double equals == in other languages (there is one exception, see below). The following solution shows how to solve this the functional way, which will ensure that you will never have out of bounds error:
// Let has local scope
let array = [1, 2, 3, 5, 2, 8, 9, 2]
// Functional filter with an Arrow function
// Filter all elements equal to 2 and return the length (count)
array.filter(x => x === 2).length // -> 3
The following anonymous Arrow function (lambda function) in JavaScript:
(x) => {
const k = 2
return k * x
}
may be simplified to this concise form for a single input:
x => 2 * x
where the return is implied.
Always use triple equals: === for comparison in JS, with the exception of when checking for nullability: if (something == null) {} as it includes a check for undefined, if you only use double equals as in this case.

Very simple:
var count = 0;
for(var i = 0; i < array.length; ++i){
if(array[i] == 2)
count++;
}

2017:
If someone is still interested in the question, my solution is the following:
const arrayToCount = [1, 2, 3, 5, 2, 8, 9, 2];
const result = arrayToCount.filter(i => i === 2).length;
console.log('number of the found elements: ' + result);

Here is an ES2017+ way to get the counts for all array items in O(N):
const arr = [1, 2, 3, 5, 2, 8, 9, 2];
const counts = {};
arr.forEach((el) => {
counts[el] = counts[el] ? (counts[el] + 1) : 1;
});
You can also optionally sort the output:
const countsSorted = Object.entries(counts).sort(([_, a], [__, b]) => a - b);
console.log(countsSorted) for your example array:
[
[ '2', 3 ],
[ '1', 1 ],
[ '3', 1 ],
[ '5', 1 ],
[ '8', 1 ],
[ '9', 1 ]
]

If you are using lodash or underscore the _.countBy method will provide an object of aggregate totals keyed by each value in the array. You can turn this into a one-liner if you only need to count one value:
_.countBy(['foo', 'foo', 'bar'])['foo']; // 2
This also works fine on arrays of numbers. The one-liner for your example would be:
_.countBy([1, 2, 3, 5, 2, 8, 9, 2])[2]; // 3

Weirdest way I can think of doing this is:
(a.length-(' '+a.join(' ')+' ').split(' '+n+' ').join(' ').match(/ /g).length)+1
Where:
a is the array
n is the number to count in the array
My suggestion, use a while or for loop ;-)

Not using a loop usually means handing the process over to some method that does use a loop.
Here is a way our loop hating coder can satisfy his loathing, at a price:
var a=[1, 2, 3, 5, 2, 8, 9, 2];
alert(String(a).replace(/[^2]+/g,'').length);
/* returned value: (Number)
3
*/
You can also repeatedly call indexOf, if it is available as an array method, and move the search pointer each time.
This does not create a new array, and the loop is faster than a forEach or filter.
It could make a difference if you have a million members to look at.
function countItems(arr, what){
var count= 0, i;
while((i= arr.indexOf(what, i))!= -1){
++count;
++i;
}
return count
}
countItems(a,2)
/* returned value: (Number)
3
*/

I'm a begin fan of js array's reduce function.
const myArray =[1, 2, 3, 5, 2, 8, 9, 2];
const count = myArray.reduce((count, num) => num === 2 ? count + 1 : count, 0)
In fact if you really want to get fancy you can create a count function on the Array prototype. Then you can reuse it.
Array.prototype.count = function(filterMethod) {
return this.reduce((count, item) => filterMethod(item)? count + 1 : count, 0);
}
Then do
const myArray =[1, 2, 3, 5, 2, 8, 9, 2]
const count = myArray.count(x => x==2)

Most of the posted solutions using array functions such as filter are incomplete because they aren't parameterized.
Here goes a solution with which the element to count can be set at run time.
function elementsCount(elementToFind, total, number){
return total += number==elementToFind;
}
var ar = [1, 2, 3, 5, 2, 8, 9, 2];
var elementToFind=2;
var result = ar.reduce(elementsCount.bind(this, elementToFind), 0);
The advantage of this approach is that could easily change the function to count for instance the number of elements greater than X.
You may also declare the reduce function inline
var ar = [1, 2, 3, 5, 2, 8, 9, 2];
var elementToFind=2;
var result = ar.reduce(function (elementToFind, total, number){
return total += number==elementToFind;
}.bind(this, elementToFind), 0);

Really, why would you need map or filter for this?
reduce was "born" for these kind of operations:
[1, 2, 3, 5, 2, 8, 9, 2].reduce( (count,2)=>count+(item==val), 0);
that's it! (if item==val in each iteration, then 1 will be added to the accumulator count, as true will resolve to 1).
As a function:
function countInArray(arr, val) {
return arr.reduce((count,item)=>count+(item==val),0)
}
Or, go ahead and extend your arrays:
Array.prototype.count = function(val) {
return this.reduce((count,item)=>count+(item==val),0)
}

It is better to wrap it into function:
let countNumber = (array,specificNumber) => {
return array.filter(n => n == specificNumber).length
}
countNumber([1,2,3,4,5],3) // returns 1

I use this:
function countElement(array, element) {
let tot = 0;
for(var el of array) {
if(el == element) {
tot++;
}
}
return tot;
}
var arr = ["a", "b", "a", "c", "d", "a", "e", "f", "a"];
console.log(countElement(arr, "a")); // 4

var arrayCount = [1,2,3,2,5,6,2,8];
var co = 0;
function findElement(){
arrayCount.find(function(value, index) {
if(value == 2)
co++;
});
console.log( 'found' + ' ' + co + ' element with value 2');
}
I would do something like that:
var arrayCount = [1,2,3,4,5,6,7,8];
function countarr(){
var dd = 0;
arrayCount.forEach( function(s){
dd++;
});
console.log(dd);
}

I believe what you are looking for is functional approach
const arr = ['a', 'a', 'b', 'g', 'a', 'e'];
const count = arr.filter(elem => elem === 'a').length;
console.log(count); // Prints 3
elem === 'a' is the condition, replace it with your own.

Array.prototype.count = function (v) {
var c = 0;
for (let i = 0; i < this.length; i++) {
if(this[i] === v){
c++;
}
}
return c;
}
var arr = [1, 2, 3, 5, 2, 8, 9, 2];
console.log(arr.count(2)); //3

Solution by recursion
function count(arr, value) {
if (arr.length === 1) {
return arr[0] === value ? 1 : 0;
} else {
return (arr.shift() === value ? 1 : 0) + count(arr, value);
}
}
count([1,2,2,3,4,5,2], 2); // 3

Create a new method for Array class in core level file and use it all over your project.
// say in app.js
Array.prototype.occurrence = function(val) {
return this.filter(e => e === val).length;
}
Use this anywhere in your project -
[1, 2, 4, 5, 2, 7, 2, 9].occurrence(2);
// above line returns 3

Here is a one liner in javascript.
Use map. Find the matching values (v === 2) in the array, returning an array of ones and zeros.
Use Reduce. Add all the values of the array for the total number found.
[1, 2, 3, 5, 2, 8, 9, 2]
.map(function(v) {
return v === 2 ? 1 : 0;
})
.reduce((a, b) => a + b, 0);
The result is 3.

Depending on how you want to run it:
const reduced = (array, val) => { // self explanatory
return array.filter((element) => element === val).length;
}
console.log(reduced([1, 2, 3, 5, 2, 8, 9, 2], 2));
// 3
const reducer = (array) => { // array to set > set.forEach > map.set
const count = new Map();
const values = new Set(array);
values.forEach((element)=> {
count.set(element, array.filter((arrayElement) => arrayElement === element).length);
});
return count;
}
console.log(reducer([1, 2, 3, 5, 2, 8, 9, 2]));
// Map(6) {1 => 1, 2 => 3, 3 => 1, 5 => 1, 8 => 1, …}

You can use built-in function Array.filter()
array.filter(x => x === element).length;
var arr = [1, 2, 3, 5, 2, 8, 9, 2];
// Count how many 2 there are in arr
var count = arr.filter(x => x === 2).length;
console.log(count);

One-liner function
const countBy = (a,f)=>a.reduce((p,v,i,x)=>p+!!f(v,i,x), 0)
countBy([1,2,3,4,5], v=>v%2===0) // 2

There are many ways to find out. I think the easiest way is to use the array filter method which is introduced in es6.
function itemCount(array, item) {
return array.filter(element => element === item).length
}
const myArray = [1,3,5,7,1,2,3,4,5,1,9,0,1]
const items = itemCount(myArray, 1)
console.log(items)

Something a little more generic and modern (in 2022):
import {pipe, count} from 'iter-ops';
const arr = [1, 2, 3, 5, 2, 8, 9, 2];
const n = pipe(arr, count(a => a === 2)).first; //=> 3
What's good about this:
It counts without creating a new array, so it is memory-efficient
It works the same for any Iterable and AsyncIterable

Another approach using RegExp
const list = [1, 2, 3, 5, 2, 8, 9, 2]
const d = 2;
const counter = (`${list.join()},`.match(new RegExp(`${d}\\,`, 'g')) || []).length
console.log(counter)
The Steps follows as below
Join the string using a comma Remember to append ',' after joining so as not to have incorrect values when value to be matched is at the end of the array
Match the number of occurrence of a combination between the digit and comma
Get length of matched items

I believe you can use the new Set array method of JavaScript to have unique values.
Example:
var arr = [1, 2, 3, 5, 2, 8, 9, 2]
var set = new Set(arr);
console.log(set);
// 1,2,3,5,8,9 . We get unique values as output.

You can use length property in JavaScript array:
var myarray = [];
var count = myarray.length;//return 0
myarray = [1,2];
count = myarray.length;//return 2

How can I remove a specific item from an array in JavaScript?

How do I remove a specific value from an array? Something like:
array.remove(value);
Constraints: I have to use core JavaScript. Frameworks are not allowed.

Find the index of the array element you want to remove using indexOf, and then remove that index with splice.
The splice() method changes the contents of an array by removing
existing elements and/or adding new elements.
const array = [2, 5, 9];
console.log(array);
const index = array.indexOf(5);
if (index > -1) { // only splice array when item is found
array.splice(index, 1); // 2nd parameter means remove one item only
}
// array = [2, 9]
console.log(array);
The second parameter of splice is the number of elements to remove. Note that splice modifies the array in place and returns a new array containing the elements that have been removed.
For the reason of completeness, here are functions. The first function removes only a single occurrence (i.e. removing the first match of 5 from [2,5,9,1,5,8,5]), while the second function removes all occurrences:
function removeItemOnce(arr, value) {
var index = arr.indexOf(value);
if (index > -1) {
arr.splice(index, 1);
}
return arr;
}
function removeItemAll(arr, value) {
var i = 0;
while (i < arr.length) {
if (arr[i] === value) {
arr.splice(i, 1);
} else {
++i;
}
}
return arr;
}
// Usage
console.log(removeItemOnce([2,5,9,1,5,8,5], 5))
console.log(removeItemAll([2,5,9,1,5,8,5], 5))
In TypeScript, these functions can stay type-safe with a type parameter:
function removeItem<T>(arr: Array<T>, value: T): Array<T> {
const index = arr.indexOf(value);
if (index > -1) {
arr.splice(index, 1);
}
return arr;
}

Edited on 2016 October
Do it simple, intuitive and explicit (Occam's razor)
Do it immutable (original array stays unchanged)
Do it with standard JavaScript functions, if your browser doesn't support them - use polyfill
In this code example I use array.filter(...) function to remove unwanted items from an array. This function doesn't change the original array and creates a new one. If your browser doesn't support this function (e.g. Internet Explorer before version 9, or Firefox before version 1.5), consider polyfilling with core-js.
Removing item (ECMA-262 Edition 5 code AKA old style JavaScript)
var value = 3
var arr = [1, 2, 3, 4, 5, 3]
arr = arr.filter(function(item) {
return item !== value
})
console.log(arr)
// [ 1, 2, 4, 5 ]
Removing item (ECMAScript 6 code)
let value = 3
let arr = [1, 2, 3, 4, 5, 3]
arr = arr.filter(item => item !== value)
console.log(arr)
// [ 1, 2, 4, 5 ]
IMPORTANT ECMAScript 6 () => {} arrow function syntax is not supported in Internet Explorer at all, Chrome before version 45, Firefox before version 22, and Safari before version 10. To use ECMAScript 6 syntax in old browsers you can use BabelJS.
Removing multiple items (ECMAScript 7 code)
An additional advantage of this method is that you can remove multiple items
let forDeletion = [2, 3, 5]
let arr = [1, 2, 3, 4, 5, 3]
arr = arr.filter(item => !forDeletion.includes(item))
// !!! Read below about array.includes(...) support !!!
console.log(arr)
// [ 1, 4 ]
IMPORTANT array.includes(...) function is not supported in Internet Explorer at all, Chrome before version 47, Firefox before version 43, Safari before version 9, and Edge before version 14 but you can polyfill with core-js.
Removing multiple items (in the future, maybe)
If the "This-Binding Syntax" proposal is ever accepted, you'll be able to do this:
// array-lib.js
export function remove(...forDeletion) {
return this.filter(item => !forDeletion.includes(item))
}
// main.js
import { remove } from './array-lib.js'
let arr = [1, 2, 3, 4, 5, 3]
// :: This-Binding Syntax Proposal
// using "remove" function as "virtual method"
// without extending Array.prototype
arr = arr::remove(2, 3, 5)
console.log(arr)
// [ 1, 4 ]
Try it yourself in BabelJS :)
Reference
Array.prototype.includes
Functional composition

I don't know how you are expecting array.remove(int) to behave. There are three possibilities I can think of that you might want.
To remove an element of an array at an index i:
array.splice(i, 1);
If you want to remove every element with value number from the array:
for (var i = array.length - 1; i >= 0; i--) {
if (array[i] === number) {
array.splice(i, 1);
}
}
If you just want to make the element at index i no longer exist, but you don't want the indexes of the other elements to change:
delete array[i];

It depends on whether you want to keep an empty spot or not.
If you do want an empty slot:
array[index] = undefined;
If you don't want an empty slot:
//To keep the original:
//oldArray = [...array];
//This modifies the array.
array.splice(index, 1);
And if you need the value of that item, you can just store the returned array's element:
var value = array.splice(index, 1)[0];
If you want to remove at either end of the array, you can use array.pop() for the last one or array.shift() for the first one (both return the value of the item as well).
If you don't know the index of the item, you can use array.indexOf(item) to get it (in a if() to get one item or in a while() to get all of them). array.indexOf(item) returns either the index or -1 if not found.

A friend was having issues in Internet Explorer 8 and showed me what he did. I told him it was wrong, and he told me he got the answer here. The current top answer will not work in all browsers (Internet Explorer 8 for example), and it will only remove the first occurrence of the item.
Remove ALL instances from an array
function removeAllInstances(arr, item) {
for (var i = arr.length; i--;) {
if (arr[i] === item) arr.splice(i, 1);
}
}
It loops through the array backwards (since indices and length will change as items are removed) and removes the item if it's found. It works in all browsers.

There are two major approaches
splice(): anArray.splice(index, 1);
let fruits = ['Apple', 'Banana', 'Mango', 'Orange']
let removed = fruits.splice(2, 1);
// fruits is ['Apple', 'Banana', 'Orange']
// removed is ['Mango']
delete: delete anArray[index];
let fruits = ['Apple', 'Banana', 'Mango', 'Orange']
let removed = delete fruits(2);
// fruits is ['Apple', 'Banana', undefined, 'Orange']
// removed is true
Be careful when you use the delete for an array. It is good for deleting attributes of objects, but not so good for arrays. It is better to use splice for arrays.
Keep in mind that when you use delete for an array you could get wrong results for anArray.length. In other words, delete would remove the element, but it wouldn't update the value of the length property.
You can also expect to have holes in index numbers after using delete, e.g. you could end up with having indexes 1, 3, 4, 8, 9, and 11 and length as it was before using delete. In that case, all indexed for loops would crash, since indexes are no longer sequential.
If you are forced to use delete for some reason, then you should use for each loops when you need to loop through arrays. As the matter of fact, always avoid using indexed for loops, if possible. That way the code would be more robust and less prone to problems with indexes.

Array.prototype.removeByValue = function (val) {
for (var i = 0; i < this.length; i++) {
if (this[i] === val) {
this.splice(i, 1);
i--;
}
}
return this;
}
var fruits = ['apple', 'banana', 'carrot', 'orange'];
fruits.removeByValue('banana');
console.log(fruits);
// -> ['apple', 'carrot', 'orange']

There isn't any need to use indexOf or splice. However, it performs better if you only want to remove one occurrence of an element.
Find and move (move):
function move(arr, val) {
var j = 0;
for (var i = 0, l = arr.length; i < l; i++) {
if (arr[i] !== val) {
arr[j++] = arr[i];
}
}
arr.length = j;
}
Use indexOf and splice (indexof):
function indexof(arr, val) {
var i;
while ((i = arr.indexOf(val)) != -1) {
arr.splice(i, 1);
}
}
Use only splice (splice):
function splice(arr, val) {
for (var i = arr.length; i--;) {
if (arr[i] === val) {
arr.splice(i, 1);
}
}
}
Run-times on Node.js for an array with 1000 elements (averaged over 10,000 runs):
indexof is approximately 10 times slower than move. Even if improved by removing the call to indexOf in splice, it performs much worse than move.
Remove all occurrences:
move 0.0048 ms
indexof 0.0463 ms
splice 0.0359 ms
Remove first occurrence:
move_one 0.0041 ms
indexof_one 0.0021 ms

This provides a predicate instead of a value.
NOTE: it will update the given array, and return the affected rows.
Usage
var removed = helper.remove(arr, row => row.id === 5 );
var removed = helper.removeAll(arr, row => row.name.startsWith('BMW'));
Definition
var helper = {
// Remove and return the first occurrence
remove: function(array, predicate) {
for (var i = 0; i < array.length; i++) {
if (predicate(array[i])) {
return array.splice(i, 1);
}
}
},
// Remove and return all occurrences
removeAll: function(array, predicate) {
var removed = [];
for (var i = 0; i < array.length; ) {
if (predicate(array[i])) {
removed.push(array.splice(i, 1));
continue;
}
i++;
}
return removed;
},
};

You can do it easily with the filter method:
function remove(arrOriginal, elementToRemove){
return arrOriginal.filter(function(el){return el !== elementToRemove});
}
console.log(remove([1, 2, 1, 0, 3, 1, 4], 1));
This removes all elements from the array and also works faster than a combination of slice and indexOf.

Using filter is an elegant way to achieve this requirement.
filter will not mutate the original array.
const num = 3;
let arr = [1, 2, 3, 4];
const arr2 = arr.filter(x => x !== num);
console.log(arr); // [1, 2, 3, 4]
console.log(arr2); // [1, 2, 4]
You can use filter and then assign the result to the original array if you want to achieve a mutation removal behaviour.
const num = 3;
let arr = [1, 2, 3, 4];
arr = arr.filter(x => x !== num);
console.log(arr); // [1, 2, 4]
By the way, filter will remove all of the occurrences matched in the condition (not just the first occurrence) like you can see in the following example
const num = 3;
let arr = [1, 2, 3, 3, 3, 4];
arr = arr.filter(x => x !== num);
console.log(arr); // [1, 2, 4]
In case, you just want to remove the first occurrence, you can use the splice method
const num = 3;
let arr = [1, 2, 3, 3, 3, 4];
const idx = arr.indexOf(num);
arr.splice(idx, idx !== -1 ? 1 : 0);
console.log(arr); // [1, 2, 3, 3, 4]

John Resig posted a good implementation:
// Array Remove - By John Resig (MIT Licensed)
Array.prototype.remove = function(from, to) {
var rest = this.slice((to || from) + 1 || this.length);
this.length = from < 0 ? this.length + from : from;
return this.push.apply(this, rest);
};
If you don’t want to extend a global object, you can do something like the following, instead:
// Array Remove - By John Resig (MIT Licensed)
Array.remove = function(array, from, to) {
var rest = array.slice((to || from) + 1 || array.length);
array.length = from < 0 ? array.length + from : from;
return array.push.apply(array, rest);
};
But the main reason I am posting this is to warn users against the alternative implementation suggested in the comments on that page (Dec 14, 2007):
Array.prototype.remove = function(from, to) {
this.splice(from, (to=[0, from || 1, ++to - from][arguments.length]) < 0 ? this.length + to : to);
return this.length;
};
It seems to work well at first, but through a painful process I discovered it fails when trying to remove the second to last element in an array. For example, if you have a 10-element array and you try to remove the 9th element with this:
myArray.remove(8);
You end up with an 8-element array. I don't know why, but I confirmed John's original implementation doesn't have this problem.

You can use ES6. For example to delete the value '3' in this case:
var array=['1','2','3','4','5','6']
var newArray = array.filter((value)=>value!='3');
console.log(newArray);
Output :
["1", "2", "4", "5", "6"]

Underscore.js can be used to solve issues with multiple browsers. It uses in-build browser methods if present. If they are absent like in the case of older Internet Explorer versions it uses its own custom methods.
A simple example to remove elements from array (from the website):
_.without([1, 2, 1, 0, 3, 1, 4], 0, 1); // => [2, 3, 4]

If you want a new array with the deleted positions removed, you can always delete the specific element and filter out the array. It might need an extension of the array object for browsers that don't implement the filter method, but in the long term it's easier since all you do is this:
var my_array = [1, 2, 3, 4, 5, 6];
delete my_array[4];
console.log(my_array.filter(function(a){return typeof a !== 'undefined';}));
It should display [1, 2, 3, 4, 6].

Here are a few ways to remove an item from an array using JavaScript.
All the method described do not mutate the original array, and instead create a new one.
If you know the index of an item
Suppose you have an array, and you want to remove an item in position i.
One method is to use slice():
const items = ['a', 'b', 'c', 'd', 'e', 'f']
const i = 3
const filteredItems = items.slice(0, i).concat(items.slice(i+1, items.length))
console.log(filteredItems)
slice() creates a new array with the indexes it receives. We simply create a new array, from start to the index we want to remove, and concatenate another array from the first position following the one we removed to the end of the array.
If you know the value
In this case, one good option is to use filter(), which offers a more declarative approach:
const items = ['a', 'b', 'c', 'd', 'e', 'f']
const valueToRemove = 'c'
const filteredItems = items.filter(item => item !== valueToRemove)
console.log(filteredItems)
This uses the ES6 arrow functions. You can use the traditional functions to support older browsers:
const items = ['a', 'b', 'c', 'd', 'e', 'f']
const valueToRemove = 'c'
const filteredItems = items.filter(function(item) {
return item !== valueToRemove
})
console.log(filteredItems)
or you can use Babel and transpile the ES6 code back to ES5 to make it more digestible to old browsers, yet write modern JavaScript in your code.
Removing multiple items
What if instead of a single item, you want to remove many items?
Let's find the simplest solution.
By index
You can just create a function and remove items in series:
const items = ['a', 'b', 'c', 'd', 'e', 'f']
const removeItem = (items, i) =>
items.slice(0, i-1).concat(items.slice(i, items.length))
let filteredItems = removeItem(items, 3)
filteredItems = removeItem(filteredItems, 5)
//["a", "b", "c", "d"]
console.log(filteredItems)
By value
You can search for inclusion inside the callback function:
const items = ['a', 'b', 'c', 'd', 'e', 'f']
const valuesToRemove = ['c', 'd']
const filteredItems = items.filter(item => !valuesToRemove.includes(item))
// ["a", "b", "e", "f"]
console.log(filteredItems)
Avoid mutating the original array
splice() (not to be confused with slice()) mutates the original array, and should be avoided.
(originally posted on my site https://flaviocopes.com/how-to-remove-item-from-array/)

Check out this code. It works in every major browser.
remove_item = function(arr, value) {
var b = '';
for (b in arr) {
if (arr[b] === value) {
arr.splice(b, 1);
break;
}
}
return arr;
};
var array = [1,3,5,6,5,9,5,3,55]
var res = remove_item(array,5);
console.log(res)

Removing a particular element/string from an array can be done in a one-liner:
theArray.splice(theArray.indexOf("stringToRemoveFromArray"), 1);
where:
theArray: the array you want to remove something particular from
stringToRemoveFromArray: the string you want to be removed and 1 is the number of elements you want to remove.
NOTE: If "stringToRemoveFromArray" is not located in the array, this will remove the last element of the array.
It's always good practice to check if the element exists in your array first, before removing it.
if (theArray.indexOf("stringToRemoveFromArray") >= 0){
theArray.splice(theArray.indexOf("stringToRemoveFromArray"), 1);
}
Depending if you have newer or older version of Ecmascript running on your client's computers:
var array=['1','2','3','4','5','6']
var newArray = array.filter((value)=>value!='3');
OR
var array = ['1','2','3','4','5','6'];
var newArray = array.filter(function(item){ return item !== '3' });
Where '3' is the value you want to be removed from the array.
The array would then become : ['1','2','4','5','6']

ES10
This post summarizes common approaches to element removal from an array as of ECMAScript 2019 (ES10).
1. General cases
1.1. Removing Array element by value using .splice()
| In-place: Yes |
| Removes duplicates: Yes(loop), No(indexOf) |
| By value / index: By index |
If you know the value you want to remove from an array you can use the splice method. First, you must identify the index of the target item. You then use the index as the start element and remove just one element.
// With a 'for' loop
const arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 0];
for( let i = 0; i < arr.length; i++){
if ( arr[i] === 5) {
arr.splice(i, 1);
}
} // => [1, 2, 3, 4, 6, 7, 8, 9, 0]
// With the .indexOf() method
const arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 0];
const i = arr.indexOf(5);
arr.splice(i, 1); // => [1, 2, 3, 4, 6, 7, 8, 9, 0]
1.2. Removing Array element using the .filter() method
| In-place: No |
| Removes duplicates: Yes |
| By value / index: By value |
The specific element can be filtered out from the array, by providing a filtering function. Such function is then called for every element in the array.
const value = 3
let arr = [1, 2, 3, 4, 5, 3]
arr = arr.filter(item => item !== value)
console.log(arr)
// [ 1, 2, 4, 5 ]
1.3. Removing Array element by extending Array.prototype
| In-place: Yes/No (Depends on implementation) |
| Removes duplicates: Yes/No (Depends on implementation) |
| By value / index: By index / By value (Depends on implementation) |
The prototype of Array can be extended with additional methods. Such methods will be then available to use on created arrays.
Note: Extending prototypes of objects from the standard library of JavaScript (like Array) is considered by some as an antipattern.
// In-place, removes all, by value implementation
Array.prototype.remove = function(item) {
for (let i = 0; i < this.length; i++) {
if (this[i] === item) {
this.splice(i, 1);
}
}
}
const arr1 = [1,2,3,1];
arr1.remove(1) // arr1 equals [2,3]
// Non-stationary, removes first, by value implementation
Array.prototype.remove = function(item) {
const arr = this.slice();
for (let i = 0; i < this.length; i++) {
if (arr[i] === item) {
arr.splice(i, 1);
return arr;
}
}
return arr;
}
let arr2 = [1,2,3,1];
arr2 = arr2.remove(1) // arr2 equals [2,3,1]
1.4. Removing Array element using the delete operator
| In-place: Yes |
| Removes duplicates: No |
| By value / index: By index |
Using the delete operator does not affect the length property. Nor does it affect the indexes of subsequent elements. The array becomes sparse, which is a fancy way of saying the deleted item is not removed but becomes undefined.
const arr = [1, 2, 3, 4, 5, 6];
delete arr[4]; // Delete element with index 4
console.log( arr ); // [1, 2, 3, 4, undefined, 6]
The delete operator is designed to remove properties from JavaScript objects, which arrays are objects.
1.5. Removing Array element using Object utilities (>= ES10)
| In-place: No |
| Removes duplicates: Yes |
| By value / index: By value |
ES10 introduced Object.fromEntries, which can be used to create the desired Array from any Array-like object and filter unwanted elements during the process.
const object = [1,2,3,4];
const valueToRemove = 3;
const arr = Object.values(Object.fromEntries(
Object.entries(object)
.filter(([ key, val ]) => val !== valueToRemove)
));
console.log(arr); // [1,2,4]
2. Special cases
2.1 Removing element if it's at the end of the Array
2.1.1. Changing Array length
| In-place: Yes |
| Removes duplicates: No |
| By value / index: N/A |
JavaScript Array elements can be removed from the end of an array by setting the length property to a value less than the current value. Any element whose index is greater than or equal to the new length will be removed.
const arr = [1, 2, 3, 4, 5, 6];
arr.length = 5; // Set length to remove element
console.log( arr ); // [1, 2, 3, 4, 5]
2.1.2. Using .pop() method
| In-place: Yes |
| Removes duplicates: No |
| By value / index: N/A |
The pop method removes the last element of the array, returns that element, and updates the length property. The pop method modifies the array on which it is invoked, This means unlike using delete the last element is removed completely and the array length reduced.
const arr = [1, 2, 3, 4, 5, 6];
arr.pop(); // returns 6
console.log( arr ); // [1, 2, 3, 4, 5]
2.2. Removing element if it's at the beginning of the Array
| In-place: Yes |
| Removes duplicates: No |
| By value / index: N/A |
The .shift() method works much like the pop method except it removes the first element of a JavaScript array instead of the last. When the element is removed the remaining elements are shifted down.
const arr = [1, 2, 3, 4];
arr.shift(); // returns 1
console.log( arr ); // [2, 3, 4]
2.3. Removing element if it's the only element in the Array
| In-place: Yes |
| Removes duplicates: N/A |
| By value / index: N/A |
The fastest technique is to set an array variable to an empty array.
let arr = [1];
arr = []; //empty array
Alternatively technique from 2.1.1 can be used by setting length to 0.

You can use lodash _.pull (mutate array), _.pullAt (mutate array) or _.without (does't mutate array),
var array1 = ['a', 'b', 'c', 'd']
_.pull(array1, 'c')
console.log(array1) // ['a', 'b', 'd']
var array2 = ['e', 'f', 'g', 'h']
_.pullAt(array2, 0)
console.log(array2) // ['f', 'g', 'h']
var array3 = ['i', 'j', 'k', 'l']
var newArray = _.without(array3, 'i') // ['j', 'k', 'l']
console.log(array3) // ['i', 'j', 'k', 'l']

ES6 & without mutation: (October 2016)
const removeByIndex = (list, index) =>
[
...list.slice(0, index),
...list.slice(index + 1)
];
output = removeByIndex([33,22,11,44],1) //=> [33,11,44]
console.log(output)

Performance
Today (2019-12-09) I conduct performance tests on macOS v10.13.6 (High Sierra) for chosen solutions. I show delete (A), but I do not use it in comparison with other methods, because it left empty space in the array.
The conclusions
the fastest solution is array.splice (C) (except Safari for small arrays where it has the second time)
for big arrays, array.slice+splice (H) is the fastest immutable solution for Firefox and Safari; Array.from (B) is fastest in Chrome
mutable solutions are usually 1.5x-6x faster than immutable
for small tables on Safari, surprisingly the mutable solution (C) is slower than the immutable solution (G)
Details
In tests, I remove the middle element from the array in different ways. The A, C solutions are in-place. The B, D, E, F, G, H solutions are immutable.
Results for an array with 10 elements
In Chrome the array.splice (C) is the fastest in-place solution. The array.filter (D) is the fastest immutable solution. The slowest is array.slice (F). You can perform the test on your machine here.
Results for an array with 1.000.000 elements
In Chrome the array.splice (C) is the fastest in-place solution (the delete (C) is similar fast - but it left an empty slot in the array (so it does not perform a 'full remove')). The array.slice-splice (H) is the fastest immutable solution. The slowest is array.filter (D and E). You can perform the test on your machine here.
var a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9];
var log = (letter,array) => console.log(letter, array.join `,`);
function A(array) {
var index = array.indexOf(5);
delete array[index];
log('A', array);
}
function B(array) {
var index = array.indexOf(5);
var arr = Array.from(array);
arr.splice(index, 1)
log('B', arr);
}
function C(array) {
var index = array.indexOf(5);
array.splice(index, 1);
log('C', array);
}
function D(array) {
var arr = array.filter(item => item !== 5)
log('D', arr);
}
function E(array) {
var index = array.indexOf(5);
var arr = array.filter((item, i) => i !== index)
log('E', arr);
}
function F(array) {
var index = array.indexOf(5);
var arr = array.slice(0, index).concat(array.slice(index + 1))
log('F', arr);
}
function G(array) {
var index = array.indexOf(5);
var arr = [...array.slice(0, index), ...array.slice(index + 1)]
log('G', arr);
}
function H(array) {
var index = array.indexOf(5);
var arr = array.slice(0);
arr.splice(index, 1);
log('H', arr);
}
A([...a]);
B([...a]);
C([...a]);
D([...a]);
E([...a]);
F([...a]);
G([...a]);
H([...a]);
This snippet only presents code used in performance tests - it does not perform tests itself.
Comparison for browsers: Chrome v78.0.0, Safari v13.0.4, and Firefox v71.0.0

OK, for example you have the array below:
var num = [1, 2, 3, 4, 5];
And we want to delete number 4. You can simply use the below code:
num.splice(num.indexOf(4), 1); // num will be [1, 2, 3, 5];
If you are reusing this function, you write a reusable function which will be attached to the native array function like below:
Array.prototype.remove = Array.prototype.remove || function(x) {
const i = this.indexOf(x);
if(i===-1)
return;
this.splice(i, 1); // num.remove(5) === [1, 2, 3];
}
But how about if you have the below array instead with a few [5]s in the array?
var num = [5, 6, 5, 4, 5, 1, 5];
We need a loop to check them all, but an easier and more efficient way is using built-in JavaScript functions, so we write a function which use a filter like below instead:
const _removeValue = (arr, x) => arr.filter(n => n!==x);
//_removeValue([1, 2, 3, 4, 5, 5, 6, 5], 5) // Return [1, 2, 3, 4, 6]
Also there are third-party libraries which do help you to do this, like Lodash or Underscore. For more information, look at lodash _.pull, _.pullAt or _.without.

I'm pretty new to JavaScript and needed this functionality. I merely wrote this:
function removeFromArray(array, item, index) {
while((index = array.indexOf(item)) > -1) {
array.splice(index, 1);
}
}
Then when I want to use it:
//Set-up some dummy data
var dummyObj = {name:"meow"};
var dummyArray = [dummyObj, "item1", "item1", "item2"];
//Remove the dummy data
removeFromArray(dummyArray, dummyObj);
removeFromArray(dummyArray, "item2");
Output - As expected.
["item1", "item1"]
You may have different needs than I, so you can easily modify it to suit them. I hope this helps someone.

I want to answer based on ECMAScript 6. Assume you have an array like below:
let arr = [1,2,3,4];
If you want to delete at a special index like 2, write the below code:
arr.splice(2, 1); //=> arr became [1,2,4]
But if you want to delete a special item like 3 and you don't know its index, do like below:
arr = arr.filter(e => e !== 3); //=> arr became [1,2,4]
Hint: please use an arrow function for filter callback unless you will get an empty array.

If you have complex objects in the array you can use filters?
In situations where $.inArray or array.splice is not as easy to use. Especially if the objects are perhaps shallow in the array.
E.g. if you have an object with an Id field and you want the object removed from an array:
this.array = this.array.filter(function(element, i) {
return element.id !== idToRemove;
});

Update: This method is recommended only if you cannot use ECMAScript 2015 (formerly known as ES6). If you can use it, other answers here provide much neater implementations.
This gist here will solve your problem, and also deletes all occurrences of the argument instead of just 1 (or a specified value).
Array.prototype.destroy = function(obj){
// Return null if no objects were found and removed
var destroyed = null;
for(var i = 0; i < this.length; i++){
// Use while-loop to find adjacent equal objects
while(this[i] === obj){
// Remove this[i] and store it within destroyed
destroyed = this.splice(i, 1)[0];
}
}
return destroyed;
}
Usage:
var x = [1, 2, 3, 3, true, false, undefined, false];
x.destroy(3); // => 3
x.destroy(false); // => false
x; // => [1, 2, true, undefined]
x.destroy(true); // => true
x.destroy(undefined); // => undefined
x; // => [1, 2]
x.destroy(3); // => null
x; // => [1, 2]

You should never mutate your array as this is against the functional programming pattern. You can create a new array without referencing the one you want to change data of using the ECMAScript 6 method filter;
var myArray = [1, 2, 3, 4, 5, 6];
Suppose you want to remove 5 from the array, you can simply do it like this:
myArray = myArray.filter(value => value !== 5);
This will give you a new array without the value you wanted to remove. So the result will be:
[1, 2, 3, 4, 6]; // 5 has been removed from this array
For further understanding you can read the MDN documentation on Array.filter.

A more modern, ECMAScript 2015 (formerly known as Harmony or ES 6) approach. Given:
const items = [1, 2, 3, 4];
const index = 2;
Then:
items.filter((x, i) => i !== index);
Yielding:
[1, 2, 4]
You can use Babel and a polyfill service to ensure this is well supported across browsers.

You can do a backward loop to make sure not to screw up the indexes, if there are multiple instances of the element.
var myElement = "chocolate";
var myArray = ['chocolate', 'poptart', 'poptart', 'poptart', 'chocolate', 'poptart', 'poptart', 'chocolate'];
/* Important code */
for (var i = myArray.length - 1; i >= 0; i--) {
if (myArray[i] == myElement) myArray.splice(i, 1);
}
console.log(myArray);