Related
After reading the MDN doc with examples of Destructuring assignment, and there is an example like:
const [a, b, ...{ pop, push }] = [1, 2];
console.log(a, b); // 1 2
console.log(pop, push); // [Function pop] [Function push]
and it says: This allows you to simultaneously unpack the properties and indices of arrays.
I took a second thought, isn't it will be fascinating if I could get the length of the array by doing this assignment at the same time getting the indices values out, so I tried this in my console:
const arr = [1, 2]
const [x, y, ...{length}] = arr
console.log(x) // 1
console.log(y) // 2
console.log(length) // 0 | ??? isn't it should give me 2?
// how about this...
const arr = [1, 2]
const {length} = arr
console.log(length) // 2 | ok well, it did work this way, but why not the destructuring assignment?
So, does anybody know why the length did not got assigned with value 2 in the first destructuring assignment? or did I found a bug in JavaScript?
What this does
[a, b, ...something] = someArray
is it takes the rest of the elements (index 2 and after) from someArray and puts them into a single variable as an array.
Doing
...{ pop, push }
takes two properties from that array: Array.prototype.pop and Array.prototype.push. But there aren't any more elements, because the original array had only 2 items, so doing
...{ length }
will give you 0.
If the original array did have more items, you'd see a length. For example:
const arr = [1, 2, 3, 4, 5]
const [x, y, ...{length}] = arr;
console.log(length);
The first two items are put into x and y. The last three items are taken as an array and then the length property is destructured from that array - so the length is 3.
Either add the number of destructured items previously to the length to get the actual length
const arr = [1, 2]
const [x, y, ...{length}] = arr
const actualLength = length + 2; // because 2 items were destructured; x and y
console.log(x) // 1
console.log(y) // 2
console.log(actualLength);
Or use object destructuring syntax instead
const arr = [1, 2]
const { 0: x, 1: y, length } = arr;
console.log(x) // 1
console.log(y) // 2
console.log(length);
Or, more understandable at a glance than either of the above - just do const { length } = arr on a separate line.
Let's see what happens when you only take one element off
const arr = [1, 2]
const [x, ...{length}] = arr
console.log("x", x) // 1
console.log("length", length) // 1
Notice how the length is one, because you start with 2 elements, then get rid of 1, and 2 - 1 is 1.
Why does this happen? ... means "rest", so it's the rest of the properties.
const arr = [1, 2]
const [x, ...rest] = arr
console.log("x", x) // 1
console.log("rest", rest) // [2]
When you use { } you get a property off of the rest, so it's the same as saying
const arr = [1, 2]
const [x, ...rest] = arr
console.log("x", x) // 1
const {length} = rest;
console.log("rest", rest) // [2]
console.log("length", length) // 1
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);
I tried to do a recursive function with clousure, but dosent work.
I tried to add each single number from array to finally return [1,2,3,4,5,6,7,8] but just return me [1,2,5] something weird is I put a console.log inside of else it's print me all numbers.
Could you explain me why happens this
var array = [1, 2, [3, 4], 5, [6, [7, 8]]]; // --> 1,2,3,4,5,6,7,8
function recursiveArrayPrint(array) {
let resultado = []
return function iterator() {
for (let i = 0; array.length > i; i++) {
if (Array.isArray(array[i])) {
recursiveArrayPrint(array[i])()
} else {
console.log(array[i])
resultado.push(array[i])
}
}
return resultado
}
}
const test = recursiveArrayPrint(array);
console.log('test: ',test())
the bug in your program
resultado is declared inside of recursiveArrayPrint
inside iterator you call recursiveArrayPrint, which creates a new resultado each time
only the outermost resultado is returned
function recursiveArrayPrint(array) {
let resultado = [] // #1
return function iterator() {
for (let i = 0; array.length > i; i++) {
if (Array.isArray(array[i])) {
recursiveArrayPrint(array[i])() // #2
} else {
console.log(array[i])
resultado.push(array[i])
}
}
return resultado // #3
}
}
how to fix
function recursiveArrayPrint(array) {
// no need for resultado
function* iterator() { // use generator
for (let i = 0; array.length > i; i++) {
if (Array.isArray(array[i])) {
yield *recursiveArrayPrint(array[i]) // yield
} else {
yield array[i] // yield, not console.log
// no need for resultado
}
}
// no need for resultado
}
return iterator()
}
const arr = [1, 2, [3, 4], 5, [6, [7, 8]]]
for (const x of recursiveArrayPrint(arr))
console.log(x)
1
2
3
4
5
6
7
8
we can do better
recursiveArrayPrint is not a good name. We've disentangled the console.log effect from the iteration and the caller is free to decide what to do with the output. We can just call it recursiveArray instead
Inner function iterator is a bit useless now. No need for this abstraction anymore
for loop with i is prone to off-by-one errors. We can write this in a clearer way using for..of loop
function* recursiveArray(array)
{ for (const x of array)
if (Array.isArray(x))
yield *recursiveArray(x)
else
yield x
}
const arr = [1, 2, [3, 4], 5, [6, [7, 8]]]
for (const x of recursiveArray(arr))
console.log(x)
If you are hoping to get the flattened array as a return value, JavaScript provides Array.from which turns any iterable into an array -
console.log(Array.from(recursiveArray(arr)))
[1,2,3,4,5,6,7,8]
not just for arrays
We can rename recursiveArray to traverse and it can operate as a generic function
Instead of assuming the first input is an array, we check isArray first and only then run the for..of loop. Otherwise simply yield the input
function* traverse(t)
{ if (Array.isArray(t))
for (const x of t)
yield *traverse(x)
else
yield t
}
const arr = [1, 2, [3, 4], 5, [6, [7, 8]]]
for (const x of traverse(arr))
console.log(x)
1
2
3
4
5
6
7
8
Writing robust functions will help you in the long run. traverse could easily be expanded to step through Objects now or other iterables. Now when non-array input is given, instead of an error, we get a sensible result -
for (const x of traverse(123))
console.log(x)
123
So what the recursiveArrayPrint function does, is that it creates a variable and returns a function that has closure over this variable. I think you get this part. The part that might confuse you, is that when you encounter an array, you call the recursiveArrayPrint function again, which will only create a new variable and return a new function that has closure over this variable, but this is in an entirely different scope.
It might make more sense to look at it this way.
var array = [1, 2, [3, 4], 5, [6, [7, 8]]]; // --> 1,2,3,4,5,6,7,8
function recursiveArrayPrint(array) {
let resultado = []
return function iterator() {
for (let i = 0; array.length > i; i++) {
if (Array.isArray(array[i])) {
// This is where a new closure is created, but with reference to another variable
let resultado1 = []
return function iterator1() {
for (let i = 0; array.length > i; i++) {
if (Array.isArray(array[i])) {
// This is where a new closure is created, but with reference to another variable
}
return resultado1
}
}
} else {
console.log(array[i])
resultado.push(array[i])
}
}
return resultado
}
}
const test = recursiveArrayPrint(array);
console.log('test: ',test())
I think what you intend to do is something more like this.
var array = [1, 2, [3, 4], 5, [6, [7, 8]]]; // --> 1,2,3,4,5,6,7,8
function recursiveArrayPrint() {
let resultado = []
return function iterator(array) {
for (let i = 0; array.length > i; i++) {
if (Array.isArray(array[i])) {
iterator(array[i])
} else {
console.log(array[i])
resultado.push(array[i])
}
}
return resultado
}
}
const test = recursiveArrayPrint();
console.log('test: ',test(array))
I think the main problem in your code is not just a difficulty in mastering the different elements of the JavaScript language or simply knowing how to play with the scope of variables for example.
No, it's just algorithmic, and that's arguably the hardest thing to do. We all fell into the trap of doing complicated things by trying in vain to make them simple.
It takes time to experience, to practice incessantly to accustom our brain to handling concepts in order to know how to put them together without fail.
Recursive solution:
const arr1 = [1, 2, [3, 4], 5, [6, [7, 8]]];
function recursiveflat(arr)
{
let res = []
for (let el of arr)
if (Array.isArray(el)) res.push(...recursiveflat(el) )
else res.push(el)
return res
}
console.log( JSON.stringify( recursiveflat( arr1 )))
.as-console-wrapper {max-height: 100%!important;top:0}
The idea of a closure as a recursive function seems incompatible to me, the only thing that can be done is to have a recursive function in a closure.
const arr1 = [1, 2, [3, 4], 5, [6, [7, 8]]]
, arr2 = ['a','b',[['c','d'],'e','f'],'g',['h',['i',['j'],'k'],'l'],'m']
function recursiveflatClosure(arr_in)
{
let resFLat = [] // global array is more efficient in this case
recursiveflat(arr_in) // builder
return ()=>resFLat // closure answer
function recursiveflat(arr)
{
for (let el of arr )
if (Array.isArray(el)) recursiveflat(el)
else resFLat.push(el)
}
}
let test1 = recursiveflatClosure( arr1 )
, test2 = recursiveflatClosure( arr2 )
console.log('test 1:', JSON.stringify( test1() ))
console.log('test 2:', JSON.stringify( test2() ))
.as-console-wrapper {max-height: 100%!important;top:0}
This is a somewhat strange requirement. It seems that the desired API is for a function that accepts a nested array and returns a function that returns the flattened version of it. It's unusual as generally we would not have any need for that intermediate function.
But we can update your code to get this to work. The biggest problem is that your internal recursive iterator function takes no argument. When you make a recursive call to it, there is no place to store the data you want to process. So let's make this accept an array argument:
function iterator(array) {
Now we note that this internal function is supposed to be recursive, but you're trying to call the outer function from inside. Let's call this instead:
if (Array.isArray(array[i])) {
iterator(array[i])
} else {
And now, we can't return that iterator function directly. We need to add a level of indirection, one that passes our original array into it.
function iterator(array) {
// ...
}
return function() {return iterator (array)}
Together, these make for what seems to be the smallest change to make your function work correctly:
const array = [1, 2, [3, 4], 5, [6, [7, 8]]]; // --> 1,2,3,4,5,6,7,8
function recursiveArrayPrint(array) {
let resultado = []
function iterator(array) {
for (let i = 0; array.length > i; i++) {
if (Array.isArray(array[i])) {
iterator(array[i])
} else {
console.log(array[i])
resultado.push(array[i])
}
}
return resultado
}
return function() {return iterator (array)}
}
const test = recursiveArrayPrint(array);
console.log('test: ',test())
.as-console-wrapper {max-height: 100% !important; top: 0}
However
This does not sound like a good idea. If you want a flattened version of an array, other answers and comments here suggest better ways, including better recursive solutions.
Moreover, recursion is a technique strongly associated with functional programming, and a fundamental tenet of functional programming is to avoid data mutation. But this solution is built around mutating your temporary array, resultado. I would look to the other answers for better ways of doing this.
A simple and clean flattening approach based on a reduce task and a recursively used reducer function ... no need of a closure either ...
function collectItemFlattened(list, item) {
return list.concat(
Array.isArray(item)
? item.reduce(collectItemFlattened, [])
: item
);
}
console.log(
[1, 2, [3, 4], 5, [6, [7, 8]]].reduce(collectItemFlattened, [])
);
.as-console-wrapper { min-height: 100%!important; top: 0; }
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);
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);