How can I remove a specific item from an array?

Created 23.04.2011 22:17
Viewed 8.14M times
9384 votes

I have an array of numbers and I'm using the .push() method to add elements to it.

Is there a simple way to remove a specific element from an array?

I'm looking for the equivalent of something like:

array.remove(number);

I have to use core JavaScript. Frameworks are not allowed.

3
Comments
array.remove(index) or array.pull(index) would make a lot of sense. splice is very useful, but a remove() or pull() method would be welcome... Search the internet, you will find a lot of "What is the opposite of push() in JavaScript?" questions. Would be great if the answare could be as simples as plain english: Pull! by Gustavo Gonçalves, 24.10.2020 15:51
Opposite of push is pop by EnderShadow8, 27.03.2021 05:14
Answers 50
12
13597

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) {
  array.splice(index, 1);
}

// 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;
}
23.04.2011 22:23
Comments
Serious question: why doesn't JavaScript allow the simple and intuitive method of removing an element at an index? A simple, elegant, myArray.remove(index); seems to be the best solution and is implemented in many other languages (a lot of them older than JavaScript.) by default123, 10.09.2020 00:16
@Andrew sets and arrays are two completely different collection types. by Stefan Fabian, 01.12.2020 09:44
@StefanFabian Yes, they are, but in many many cases they will be interchangeable. by Andrew, 01.12.2020 21:08
You can simplify this solution by counting down instead of up: for ( var i = ary.length - 1; i >= 0; i-- ) { if ( ary[i] === value ) { ary.remove(i)} } by Luke Dupin, 08.12.2020 22:30
function remove(item,array) { var new_array = [] new_ array = array.filter((ar)=> ar != item) return new_array } by Rashid Iqbal, 07.01.2021 16:04
Remove all elements: const array = [2, 5, 9]; array.length = 0; console.log(array); by Rathish, 22.03.2021 07:42
Return statement is misleading. Array is already modified. It can confuse that origin array will be untouched. by mikep, 23.03.2021 12:08
Year 2279. The mankind, after colonizing Mars, starts exploring new frontiers. But Javascript, still didn't get a .remove() function... by Bob, 25.03.2021 14:02
Array.pull() is what humankind needs the most. by a2br, 08.04.2021 16:03
I'm a bit late to the party, but here's my two cents: @a2br: Array.unshift() is basically what pull() would be if it existed! @Bob: Personally, I think it's good that nothing similar to Array.remove() exists. We don't want JavaScript to end up like PHP, now do we? xD by OOPS Studio, 15.04.2021 09:12
@OOPSStudio Oh yeah, you're right! I didn't pick a correct name, I forgot push's destructive behavior. I hate destructive methods when removing stuff. by a2br, 15.04.2021 12:22
Show remaining 7 comments
1
1594

Edited on 2016 October

  • Do it simple, intuitive and explicit (Occam's razor)
  • Do it immutable (original array stay 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 using the filter polyfill from Mozilla.

Removing item (ECMA-262 Edition 5 code aka oldstyle 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 45 version, Firefox before 22 version, and Safari before 10 version. 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 47 version, Firefox before 43 version, Safari before 9 version, and Edge before 14 version so here is polyfill from Mozilla.

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

19.12.2013 19:54
Comments
what if content of array are objects and nested objects by Ranjeet Thorat, 11.01.2021 09:51
0
1448

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];
23.04.2011 22:20
1
528

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. 

23.04.2011 22:32
Comments
It's kinda funny that splice returns another array built out of the removed elements. I wrote something which assumed splice would return the newly modified list (like what immutable collections would do, for example). So, in this particular case of only one item in the list, and that item being removed, the returned list is exactly identical to the original one after splicing that one item. So, my app went into an infinite loop. by Teddy, 25.07.2020 10:55
0
321

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.

10.08.2013 19:21
0
212

There are two major approaches:

  1. splice(): anArray.splice(index, 1);

  2. delete: delete anArray[index];

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.

21.12.2012 11:32
1
180
Array.prototype.remove_by_value = function(val) {
 for (var i = 0; i < this.length; i++) {
  if (this[i] === val) {
   this.splice(i, 1);
   i--;
  }
 }
 return this;
}[
 // call like
 (1, 2, 3, 4)
].remove_by_value(3);

Array.prototype.remove_by_value = function(val) {
  for (var i = 0; i < this.length; i++) {
    if (this[i] === val) {
      this.splice(i, 1);
      i--;
    }
  }
  return this;
}

var rooms = ['hello', 'something']

rooms = rooms.remove_by_value('hello')

console.log(rooms)

23.04.2011 22:20
Comments
Many frown on modifying prototypes that don't belong to you. by snarf, 14.09.2020 22:34
0
132

There is no 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 nodejs for array with 1000 elements (average over 10000 runs):

indexof is approximately 10x 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
19.09.2013 01:53
1
102

This provides a predicate instead of a value.

NOTE: it will update the given array, and return the affected rows.

Usage

var removed = helper.removeOne(arr, row => row.id === 5 );

var removed = helper.remove(arr, row => row.name.startsWith('BMW'));

Definition

var helper = {
 // Remove and return the first occurrence

 removeOne: 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

 remove: 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;
 },
};
02.05.2014 12:00
Comments
put your code in code snippet so other users could see the result by Masoud Aghaei, 02.02.2021 19:55
0
96

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.

10.02.2014 22:06
0
91

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. Don't know why but I confirmed John's original implementation doesn't have this problem.

30.08.2013 19:07
1
89

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]
30.05.2014 09:57
Comments
though elegant and concise, OP clearly mentioned core JS only by EigenFool, 16.04.2020 08:56
2
89

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"]
04.10.2016 08:07
Comments
This answer is nice because it creates a copy of the original array, instead of modifying the original directly. by Claudio Holanda, 16.03.2017 15:27
Note: Array.prototype.filter is ECMAScript 5.1 (No IE8). for more specific solutions: stackoverflow.com/a/54390552/8958729 by Chang, 27.01.2019 19:45
0
79

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].

18.10.2012 10:13
0
76

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/)

04.05.2018 05:17
2
70

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)

02.04.2013 10:56
Comments
@RolandIllig Except the use of a for in-loop and the fact that the script could stopped earlier, by returning the result from the loop directly. The upvotes are reasonable ;) by yckart, 30.12.2014 16:13
I should also reiterate yckart's comment that for( i = 0; i < arr.length; i++ ) would be a better approach since it preserves the exact indices versus whatever order the browser decides to store the items (with for in). Doing so also lets you get the array index of a value if you need it. by Beejor, 21.08.2016 23:20
2
60

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)

07.10.2016 19:42
Comments
Why not just use filter then? array.filter((_, index) => index !== removedIndex);. by Sebastian Simon, 07.04.2020 18:30
@user4642212 you are right! also, I liked the underscore of Golang style by Abdennour TOUMI, 07.04.2020 20:37
2
59

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']

08.01.2019 14:34
Comments
This is the answer that worked for me when trying to update an array based on radio button toggling. by jdavid05, 03.04.2019 11:52
Beware, if "stringToRemoveFromArray" is not located your in array, this will remove last element of array. by Fusion, 06.04.2019 23:35
3
59

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']
25.08.2015 19:34
Comments
That's not core JS as the OP requested, is it? by some-non-descript-user, 24.09.2015 21:01
@some-non-descript-user You are right. But a lot of users like me come here looking for a general answer not just for the OP only. by Chun Yang, 01.10.2015 03:38
@ChunYang You are absolutely right. I am already using lodash, why not just use it if it saves time. by int-i, 08.06.2018 21:48
0
51

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

Enter image description here

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

Enter image description here

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

Enter image description here

09.12.2019 16:04
0
45

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.

10.05.2017 09:39
2
40

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.

16.01.2014 11:27
Comments
This is going to have terrible behavior if your array is really long and there are several instances of the element in it. The indexOf method of array will start at the beginning every time, so your cost is going to be O(n^2). by Zag, 24.05.2019 18:22
@Zag: It has a name: Shlemiel the Painter's Algorithm by Peter Mortensen, 01.09.2019 22:11
0
35

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]
13.03.2013 09:28
1
37

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;
});
09.04.2015 10:00
Comments
This is how I like to do it. Using an arrow function it can be a one-liner. I'm curious about performance. Also worth nothing that this replaces the array. Any code with a reference to the old array will not notice the change. by joeytwiddle, 29.07.2016 08:50
0
36

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.

30.10.2018 17:37
0
35

ES10 Update

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.

02.04.2020 10:15
0
31

You should never mutate your array. As this is against the functional programming pattern. You can create a new array without referencing the array 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.

26.12.2017 19:32
5
27

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.

25.04.2016 10:21
Comments
Note that .filter returns a new array, which is not exactly the same as removing the element from the same array. The benefit of this approach is that you can chain array methods together. eg: [1,2,3].filter(n => n%2).map(n => n*n) === [ 1, 9 ] by CodeOcelot, 06.05.2016 18:58
Great, if I have 600k elements in array and want to remove first 50k, can you imagine that slowness? This is not solution, there's need for function which just remove elements and returns nothing. by dev1223, 30.05.2016 15:11
@Seraph For that, you'd probably want to use splice or slice. by bjfletcher, 31.05.2016 22:04
@bjfletcher Thats even better, in process of removal, just allocate 50K elements and throw them somewhere. (with slice 550K elements, but without throwing them from the window). by dev1223, 31.05.2016 23:06
I'd prefer bjfletcher's answer, which could be as short as items= items.filter(x=>x!=3). Besides, the OP didn't state any requirement for large data set. by runsun, 27.08.2016 01:55
0
28

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);

Live Demo

12.08.2013 17:56
0
25
Array.prototype.removeItem = function(a) {
    for (i = 0; i < this.length; i++) {
        if (this[i] == a) {
            for (i2 = i; i2 < this.length - 1; i2++) {
                this[i2] = this[i2 + 1];
            }
            this.length = this.length - 1
            return;
        }
    }
}

var recentMovies = ['Iron Man', 'Batman', 'Superman', 'Spiderman'];
recentMovies.removeItem('Superman');
26.09.2013 00:12
0
26

You have 1 to 9 in the array, and you want remove 5. Use the below code:

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

var newNumberArray = numberArray.filter(m => {
  return m !== 5;
});

console.log("new Array, 5 removed", newNumberArray);


If you want to multiple values. Example:- 1,7,8

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

var newNumberArray = numberArray.filter(m => {
  return (m !== 1) && (m !== 7) && (m !== 8);
});

console.log("new Array, 1,7 and 8 removed", newNumberArray);


If you want to remove an array value in an array. Example: [3,4,5]

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

var newNumberArray = numberArray.filter(m => {
    return !removebleArray.includes(m);
});

console.log("new Array, [3,4,5] removed", newNumberArray);

Includes supported browser is link.

07.08.2018 09:37
1
22

enter image description here

2021 UPDATE

Your question is about how to remove a specific item from an array. By specific item you are referring to a number eg. remove number 5 from array. What I understand you are looking for something like:

// PSEUDOCODE, SCROLL FOR COPY-PASTE CODE
[1,2,3,4,5,6,8,5].remove(5) // result: [1,2,3,4,6,8]

As for 2021 the best way to achieve it is to use array filter function:

const input = [1,2,3,4,5,6,8,5];
const removeNumber = 5;
const result = input.filter(
    item => item != removeNumber
);

Above example uses array.prototype.filter function. It iterates over all array items, and returns only those satisfying arrow function. As a result, old array stays intact, while a new array called result contains all items that are not equal to five. You can test it yourself online.

You can visualize how array.prototype.filter like this:

enter image description here

Considerations

Code quality

Array.filter.prototype is far the most readable method to remove a number in this case. It leaves little place for mistakes and uses core JS functionality.

Why not array.prototype.map?

Array.prototype.map is sometimes consider as an alternative for array.prototype.filter for that use case. But it should not be used. The reason is that array.prototype.filter is conceptually used to filter items that satisfy arrow function (exactly what we need), while array.prototype.map is used to transform items. Since we don't change items while iterating over them, the proper function to use is array.prototype.filter.

Support

As of today (2.12.2020) 97,05% of Internet users browsers support array.prototype.filter. So generally speaking it is safe to use. However, IE6 - 8 does not support it. So if your use case requires support for these browsers there is a nice polyfill made by Chris Ferdinanti.

Performance

Array.prototype.filter is great for most use cases. However if you are looking for some performance improvements for advanced data processing you can explore some other options like using pure for. Another great option is to rethink if really array you are processing has to be so big, it may be a sign that JavaScript should receive reduced array for processing from the data source.

02.12.2020 11:21
Comments
This is not "the best way to remove a specific item from an array". First off, .filter() removes ALL occurrences of removeNumber, not a specific entry. So in your example, if there were other elements of 5, they would also get removed, which is not what is wanted. Secondly, and closely tied to the first point, it's evaluating EVERY element in the array, so it's very inefficient if we know the index already. .filter() is evaluating every single element in the array with that condition. by Daniel Foust, 13.01.2021 01:55
0
20

Based on all the answers which were mainly correct and taking into account the best practices suggested (especially not using Array.prototype directly), I came up with the below code:

function arrayWithout(arr, values) {
  var isArray = function(canBeArray) {
    if (Array.isArray) {
      return Array.isArray(canBeArray);
    }
    return Object.prototype.toString.call(canBeArray) === '[object Array]';
  };

  var excludedValues = (isArray(values)) ? values : [].slice.call(arguments, 1);
  var arrCopy = arr.slice(0);

  for (var i = arrCopy.length - 1; i >= 0; i--) {
    if (excludedValues.indexOf(arrCopy[i]) > -1) {
      arrCopy.splice(i, 1);
    }
  }

  return arrCopy;
}

Reviewing the above function, despite the fact that it works fine, I realised there could be some performance improvement. Also using ES6 instead of ES5 is a much better approach. To that end, this is the improved code:

const arrayWithoutFastest = (() => {
  const isArray = canBeArray => ('isArray' in Array) 
    ? Array.isArray(canBeArray) 
    : Object.prototype.toString.call(canBeArray) === '[object Array]';

  let mapIncludes = (map, key) => map.has(key);
  let objectIncludes = (obj, key) => key in obj;
  let includes;

  function arrayWithoutFastest(arr, ...thisArgs) {
    let withoutValues = isArray(thisArgs[0]) ? thisArgs[0] : thisArgs;

    if (typeof Map !== 'undefined') {
      withoutValues = withoutValues.reduce((map, value) => map.set(value, value), new Map());
      includes = mapIncludes;
    } else {
      withoutValues = withoutValues.reduce((map, value) => { map[value] = value; return map; } , {}); 
      includes = objectIncludes;
    }

    const arrCopy = [];
    const length = arr.length;

    for (let i = 0; i < length; i++) {
      // If value is not in exclude list
      if (!includes(withoutValues, arr[i])) {
        arrCopy.push(arr[i]);
      }
    }

    return arrCopy;
  }

  return arrayWithoutFastest;  
})();

How to use:

const arr = [1,2,3,4,5,"name", false];

arrayWithoutFastest(arr, 1); // will return array [2,3,4,5,"name", false]
arrayWithoutFastest(arr, 'name'); // will return [2,3,4,5, false]
arrayWithoutFastest(arr, false); // will return [2,3,4,5]
arrayWithoutFastest(arr,[1,2]); // will return [3,4,5,"name", false];
arrayWithoutFastest(arr, {bar: "foo"}); // will return the same array (new copy)

I am currently writing a blog post in which I have benchmarked several solutions for Array without problem and compared the time it takes to run. I will update this answer with the link once I finish that post. Just to let you know, I have compared the above against lodash's without and in case the browser supports Map, it beats lodash! Notice that I am not using Array.prototype.indexOf or Array.prototype.includes as wrapping the exlcudeValues in a Map or Object makes querying faster!

28.01.2014 14:44
2
23

I know there are a lot of answers already, but many of them seem to over complicate the problem. Here is a simple, recursive way of removing all instances of a key - calls self until index isn't found. Yes, it only works in browsers with indexOf, but it's simple and can be easily polyfilled.

Stand-alone function

function removeAll(array, key){
    var index = array.indexOf(key);

    if(index === -1) return;

    array.splice(index, 1);
    removeAll(array,key);
}

Prototype method

Array.prototype.removeAll = function(key){
    var index = this.indexOf(key);

    if(index === -1) return;

    this.splice(index, 1);
    this.removeAll(key);
}
03.02.2014 15:41
Comments
Just a note, 1 caveat with this method is the potential for stack overflows. Unless you're working with massive arrays, you shouldn't have an issue. by wharding28, 08.09.2015 05:57
But why a return in the middle? It is effectively a goto statement. by Peter Mortensen, 01.09.2019 22:14
0
17

I also ran into the situation where I had to remove an element from Array. .indexOf was not working in Internet Explorer, so I am sharing my working jQuery.inArray() solution:

var index = jQuery.inArray(val, arr);
if (index > -1) {
    arr.splice(index, 1);
    //console.log(arr);
}
08.10.2013 10:09
1
21

Immutable and one-liner way :

const newArr = targetArr.filter(e => e !== elementToDelete);
02.06.2020 16:18
Comments
An important clarification for new programmers: This does not remove the target item from the array. It creates an entirely new array that is a copy of the original array, except with the target item removed. by Daniel Waltrip, 07.07.2020 00:20
0
20

I have another good solution for removing from an array:

var words = ['spray', 'limit', 'elite', 'exuberant', 'destruction', 'present'];

const result = words.filter(word => word.length > 6);

console.log(result);
// expected output: Array ["exuberant", "destruction", "present"]

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/filter

05.06.2018 14:47
1
18

Remove by Index

A function that returns a copy of array without the element at index:

/**
* removeByIndex
* @param {Array} array
* @param {Number} index
*/
function removeByIndex(array, index){
      return array.filter(function(elem, _index){
          return index != _index;
    });
}
l = [1,3,4,5,6,7];
console.log(removeByIndex(l, 1));

$> [ 1, 4, 5, 6, 7 ]

Remove by Value

Function that return a copy of array without the Value.

/**
* removeByValue
* @param {Array} array
* @param {Number} value
*/
function removeByValue(array, value){
      return array.filter(function(elem, _index){
          return value != elem;
    });
}
l = [1,3,4,5,6,7];
console.log(removeByValue(l, 5));

$> [ 1, 3, 4, 6, 7]
12.05.2017 02:00
Comments
Are redundant constructions the norm around web developers? I have someone at work spraying stuff like this everywhere. Why not just return value != elem?! by Buffalo, 24.07.2017 11:59
0
17

Create new array:

var my_array = new Array();

Add elements to this array:

my_array.push("element1");

The function indexOf (returns index or -1 when not found):

var indexOf = function(needle)
{
    if (typeof Array.prototype.indexOf === 'function') // Newer browsers
    {
        indexOf = Array.prototype.indexOf;
    }
    else // Older browsers
    {
        indexOf = function(needle)
        {
            var index = -1;

            for (var i = 0; i < this.length; i++)
            {
                if (this[i] === needle)
                {
                    index = i;
                    break;
                }
            }
            return index;
        };
    }

    return indexOf.call(this, needle);
};

Check index of this element (tested with Firefox and Internet Explorer 8 (and later)):

var index = indexOf.call(my_array, "element1");

Remove 1 element located at index from the array

my_array.splice(index, 1);
07.08.2013 12:57
2
10

Use jQuery's InArray:

A = [1, 2, 3, 4, 5, 6];
A.splice($.inArray(3, A), 1);
//It will return A=[1, 2, 4, 5, 6]`   

Note: inArray will return -1, if the element was not found.

17.09.2014 10:14
Comments
but OP said: "good ol' fashioned JavaScript - no frameworks allowed" by CSᵠ, 12.12.2014 18:51
for Chrome 50.0, A.splice(-1, 1); will remove the last one in A. by Scott 混合理论, 17.06.2016 09:38
3
15

Oftentimes it's better to just create a new array with the filter function.

let array = [1,2,3,4];
array = array.filter(i => i !== 4); // [1,2,3]

This also improves readability IMHO. I'm not a fan of slice, although it know sometimes you should go for it.

13.04.2019 08:51
Comments
@codepleb, can you elaborate on why you prefer filter over splice and why you think filter is more readable? by MHOOS, 20.06.2019 09:47
Albeit not recommended for lengthy arrays. by eightyfive, 20.06.2019 10:07
@MHOOS Slice has a lot of options and they are confusing IMHO. You can, if you want, pass a start and end variable and while the start index is included, the end index is not, etc. It's harder to read code playing with slice. If you don't use that too often, you often end up checking the docs during reviews to check if something is correct. Docs: developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/… by codepleb, 26.06.2019 16:12
0
12
[2,3,5].filter(i => ![5].includes(i))
24.04.2019 04:32
1
12

You can iterate over each array-item and splice it if it exist in your array.

function destroy(arr, val) {
    for (var i = 0; i < arr.length; i++) if (arr[i] === val) arr.splice(i, 1);
    return arr;
}
13.05.2013 12:22
Comments
destroy( [1,2,3,3,3,4,5], 3 ) returns [1,2,3,4,5]]. i should not be incremented when the array is spliced. by Renze de Waal, 23.01.2014 17:34
0
12

In CoffeeScript:

my_array.splice(idx, 1) for ele, idx in my_array when ele is this_value
30.01.2014 04:27
1
7

Use jQuery.grep():

var y = [1, 2, 3, 9, 4]
var removeItem = 9;

y = jQuery.grep(y, function(value) {
  return value != removeItem;
});
console.log(y)
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.2/jquery.min.js"></script>

28.06.2016 09:13
Comments
The OP specifically said no frameworks. Hence the downvote. by codingsplash, 06.08.2016 09:56
0
13

I think many of the JavaScript instructions are not well thought out for functional programming. Splice returns the deleted element where most of the time you need the reduced array. This is bad.

Imagine you are doing a recursive call and have to pass an array with one less item, probably without the current indexed item. Or imagine you are doing another recursive call and has to pass an array with an element pushed.

In neither of these cases you can do myRecursiveFunction(myArr.push(c)) or myRecursiveFunction(myArr.splice(i,1)). The first idiot will in fact pass the length of the array and the second idiot will pass the deleted element as a parameter.

So what I do in fact... For deleting an array element and passing the resulting to a function as a parameter at the same time I do as follows

myRecursiveFunction(myArr.slice(0,i).concat(a.slice(i+1)))

When it comes to push that's more silly... I do like,

myRecursiveFunction((myArr.push(c),myArr))

I believe in a proper functional language a method mutating the object it's called upon must return a reference to the very object as a result.

07.05.2016 20:10
2
13

2017-05-08

Most of the given answers work for strict comparison, meaning that both objects reference the exact same object in memory (or are primitive types), but often you want to remove a non-primitive object from an array that has a certain value. For instance, if you make a call to a server and want to check a retrieved object against a local object.

const a = {'field': 2} // Non-primitive object
const b = {'field': 2} // Non-primitive object with same value
const c = a            // Non-primitive object that reference the same object as "a"

assert(a !== b) // Don't reference the same item, but have same value
assert(a === c) // Do reference the same item, and have same value (naturally)

//Note: there are many alternative implementations for valuesAreEqual
function valuesAreEqual (x, y) {
   return  JSON.stringify(x) === JSON.stringify(y)
}


//filter will delete false values
//Thus, we want to return "false" if the item
// we want to delete is equal to the item in the array
function removeFromArray(arr, toDelete){
    return arr.filter(target => {return !valuesAreEqual(toDelete, target)})
}

const exampleArray = [a, b, b, c, a, {'field': 2}, {'field': 90}];
const resultArray = removeFromArray(exampleArray, a);

//resultArray = [{'field':90}]

There are alternative/faster implementations for valuesAreEqual, but this does the job. You can also use a custom comparator if you have a specific field to check (for example, some retrieved UUID vs a local UUID).

Also note that this is a functional operation, meaning that it does not mutate the original array.

08.05.2017 17:58
Comments
I like the idea, just think is a bit slow to do two stringify per element on the array. Anyway there are cases in which it will worth, thanks for sharing. by Adriano Spadoni, 25.07.2017 08:56
Thanks, I added an edit to clarify that valuesAreEqual can be substituted. I agree that the JSON approach is slow -- but it will always work. Should definitely use better comparison when possible. by Aidan Hoolachan, 07.10.2017 20:33
0
12

What a shame you have an array of integers, not an object where the keys are string equivalents of these integers.

I've looked through a lot of these answers and they all seem to use "brute force" as far as I can see. I haven't examined every single one, apologies if this is not so. For a smallish array this is fine, but what if you have 000s of integers in it?

Correct me if I'm wrong, but can't we assume that in a key => value map, of the kind which a JavaScript object is, that the key retrieval mechanism can be assumed to be highly engineered and optimised? (NB: if some super-expert tells me that this is not the case, I can suggest using ECMAScript 6's Map class instead, which certainly will be).

I'm just suggesting that, in certain circumstances, the best solution might be to convert your array to an object... the problem being, of course, that you might have repeating integer values. I suggest putting those in buckets as the "value" part of the key => value entries. (NB: if you are sure you don't have any repeating array elements this can be much simpler: values "same as" keys, and just go Object.values(...) to get back your modified array).

So you could do:

const arr = [ 1, 2, 55, 3, 2, 4, 55 ];
const f =    function( acc, val, currIndex ){
    // We have not seen this value before: make a bucket... NB: although val's typeof is 'number',
    // there is seamless equivalence between the object key (always string)
    // and this variable val.
    ! ( val in acc ) ? acc[ val ] = []: 0;
    // Drop another array index in the bucket
    acc[ val ].push( currIndex );
    return acc;
}
const myIntsMapObj = arr.reduce( f, {});

console.log( myIntsMapObj );

Output:

Object [ <1 empty slot>, Array1, Array[2], Array1, Array1, <5 empty slots>, 46 more… ]

It is then easy to delete all the numbers 55.

delete myIntsMapObj[ 55 ]; // Again, although keys are strings this works

You don't have to delete them all: index values are pushed into their buckets in order of appearance, so (for example):

myIntsMapObj[ 55 ].shift(); // And
myIntsMapObj[ 55 ].pop();

will delete the first and last occurrence respectively. You can count frequency of occurrence easily, replace all 55s with 3s by transferring the contents of one bucket to another, etc.

Retrieving a modified int array from your "bucket object" is slightly involved but not so much: each bucket contains the index (in the original array) of the value represented by the (string) key. Each of these bucket values is also unique (each is the unique index value in the original array): so you turn them into keys in a new object, with the (real) integer from the "integer string key" as value... then sort the keys and go Object.values( ... ).

This sounds very involved and time-consuming... but obviously everything depends on the circumstances and desired usage. My understanding is that all versions and contexts of JavaScript operate only in one thread, and the thread doesn't "let go", so there could be some horrible congestion with a "brute force" method: caused not so much by the indexOf ops, but multiple repeated slice/splice ops.

Addendum If you're sure this is too much engineering for your use case surely the simplest "brute force" approach is

const arr = [ 1, 2, 3, 66, 8, 2, 3, 2 ];
const newArray = arr.filter( number => number !== 3 );
console.log( newArray )

(Yes, other answers have spotted Array.prototype.filter...)

24.11.2017 19:17
0
12

Remove element at index i, without mutating the original array:

/**
* removeElement
* @param {Array} array
* @param {Number} index
*/
function removeElement(array, index) {
   return Array.from(array).splice(index, 1);
}

// Another way is
function removeElement(array, index) {
   return array.slice(0).splice(index, 1);
}
05.05.2017 01:32
2
12

Immutable way of removing an element from array using ES6 spread operator.

Let's say you want to remove 4.

let array = [1,2,3,4,5]
const index = array.indexOf(4)
let new_array = [...array.slice(0,index), ...array.slice(index+1, array.length)]
console.log(new_array)
=> [1, 2, 3, 5]
31.05.2020 11:58
Comments
An important clarification for new programmers: This does not delete the target item from the array. It creates an entirely new array that is a copy of the original array, except with the target item removed. The word "delete" implies that we are mutating something in place, not making a modified copy. by Daniel Waltrip, 07.07.2020 00:21
Yes, you are right. This is an immutable way of removing an element. Thanks for clearing this out. by Ahmad, 07.07.2020 17:26