Deleting array elements in JavaScript - delete vs splice

Question

What is the difference between using the delete operator on the array element as opposed to using the Array.splice method?

For example:

myArray = ['a', 'b', 'c', 'd'];

delete myArray[1];
//  or
myArray.splice (1, 1);

Why even have the splice method if I can delete array elements like I can with objects?

Answer 1

delete will delete the object property, but will not reindex the array or update its length. This makes it appears as if it is undefined:

> myArray = ['a', 'b', 'c', 'd']
  ["a", "b", "c", "d"]
> delete myArray[0]
  true
> myArray[0]
  undefined

Note that it is not in fact set to the value undefined, rather the property is removed from the array, making it appear undefined. The Chrome dev tools make this distinction clear by printing empty when logging the array.

> myArray[0]
  undefined
> myArray
  [empty, "b", "c", "d"]

myArray.splice(start, deleteCount) actually removes the element, reindexes the array, and changes its length.

> myArray = ['a', 'b', 'c', 'd']
  ["a", "b", "c", "d"]
> myArray.splice(0, 2)
  ["a", "b"]
> myArray
  ["c", "d"]

Answer 2

Array.remove() Method

John Resig, creator of jQuery created a very handy Array.remove method that I always use it in my projects.

// 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);
};

and here's some examples of how it could be used:

// Remove the second item from the array
array.remove(1);
// Remove the second-to-last item from the array
array.remove(-2);
// Remove the second and third items from the array
array.remove(1,2);
// Remove the last and second-to-last items from the array
array.remove(-2,-1);

John's website

Answer 3

Because delete only removes the object from the element in the array, the length of the array won't change. Splice removes the object and shortens the array.

The following code will display "a", "b", "undefined", "d"

myArray = ['a', 'b', 'c', 'd']; delete myArray[2];

for (var count = 0; count < myArray.length; count++) {
    alert(myArray[count]);
}

Whereas this will display "a", "b", "d"

myArray = ['a', 'b', 'c', 'd']; myArray.splice(2,1);

for (var count = 0; count < myArray.length; count++) {
    alert(myArray[count]);
}

Answer 4

I stumbled onto this question while trying to understand how to remove every occurrence of an element from an Array. Here's a comparison of splice and delete for removing every 'c' from the items Array.

var items = ['a', 'b', 'c', 'd', 'a', 'b', 'c', 'd'];

while (items.indexOf('c') !== -1) {
  items.splice(items.indexOf('c'), 1);
}

console.log(items); // ["a", "b", "d", "a", "b", "d"]

items = ['a', 'b', 'c', 'd', 'a', 'b', 'c', 'd'];

while (items.indexOf('c') !== -1) {
  delete items[items.indexOf('c')];
}

console.log(items); // ["a", "b", undefined, "d", "a", "b", undefined, "d"]
​

Answer 5

From Core JavaScript 1.5 Reference > Operators > Special Operators > delete Operator :

When you delete an array element, the array length is not affected. For example, if you delete a[3], a[4] is still a[4] and a[3] is undefined. This holds even if you delete the last element of the array (delete a[a.length-1]).

Answer 6

splice will work with numeric indices.

whereas delete can be used against other kind of indices..

example:

delete myArray['text1'];

Answer 7

As stated many times above, using splice() seems like a perfect fit. Documentation at Mozilla:

The splice() method changes the content of an array by removing existing elements and/or adding new elements.

var myFish = ['angel', 'clown', 'mandarin', 'sturgeon'];

myFish.splice(2, 0, 'drum'); 
// myFish is ["angel", "clown", "drum", "mandarin", "sturgeon"]

myFish.splice(2, 1); 
// myFish is ["angel", "clown", "mandarin", "sturgeon"]

Syntax

array.splice(start)
array.splice(start, deleteCount)
array.splice(start, deleteCount, item1, item2, ...)

Parameters

start

Index at which to start changing the array. If greater than the length of the array, actual starting index will be set to the length of the array. If negative, will begin that many elements from the end.

deleteCount

An integer indicating the number of old array elements to remove. If deleteCount is 0, no elements are removed. In this case, you should specify at least one new element. If deleteCount is greater than the number of elements left in the array starting at start, then all of the elements through the end of the array will be deleted.

If deleteCount is omitted, deleteCount will be equal to (arr.length - start).

item1, item2, ...

The elements to add to the array, beginning at the start index. If you don't specify any elements, splice() will only remove elements from the array.

Return value

An array containing the deleted elements. If only one element is removed, an array of one element is returned. If no elements are removed, an empty array is returned.

[...]

Answer 8

It's probably also worth mentioning that splice only works on arrays. (Object properties can't be relied on to follow a consistent order.)

To remove the key-value pair from an object, delete is actually what you want:

delete myObj.propName;     // , or:
delete myObj["propName"];  // Equivalent.

Answer 9

delete Vs splice

when you delete an item from an array

var arr = [1,2,3,4]; delete arr[2]; //result [1, 2, 3:, 4]
console.log(arr)

when you splice

var arr = [1,2,3,4]; arr.splice(1,1); //result [1, 3, 4]
console.log(arr);

in case of delete the element is deleted but the index remains empty

while in case of splice element is deleted and the index of rest elements is reduced accordingly

Answer 10

delete acts like a non real world situation, it just removes the item, but the array length stays the same:

example from node terminal:

> var arr = ["a","b","c","d"];
> delete arr[2]
true
> arr
[ 'a', 'b', , 'd', 'e' ]

Here is a function to remove an item of an array by index, using slice(), it takes the arr as the first arg, and the index of the member you want to delete as the second argument. As you can see, it actually deletes the member of the array, and will reduce the array length by 1

function(arr,arrIndex){
    return arr.slice(0,arrIndex).concat(arr.slice(arrIndex + 1));
}

What the function above does is take all the members up to the index, and all the members after the index , and concatenates them together, and returns the result.

Here is an example using the function above as a node module, seeing the terminal will be useful:

> var arr = ["a","b","c","d"]
> arr
[ 'a', 'b', 'c', 'd' ]
> arr.length
4 
> var arrayRemoveIndex = require("./lib/array_remove_index");
> var newArray = arrayRemoveIndex(arr,arr.indexOf('c'))
> newArray
[ 'a', 'b', 'd' ] // c ya later
> newArray.length
3

please note that this will not work one array with dupes in it, because indexOf("c") will just get the first occurance, and only splice out and remove the first "c" it finds.

Answer 11

If you want to iterate a large array and selectively delete elements, it would be expensive to call splice() for every delete because splice() would have to re-index subsequent elements every time. Because arrays are associative in Javascript, it would be more efficient to delete the individual elements then re-index the array afterwards.

You can do it by building a new array. e.g

function reindexArray( array )
{
       var result = [];
        for( var key in array )
                result.push( array[key] );
        return result;
};

But I don't think you can modify the key values in the original array, which would be more efficient - it looks like you might have to create a new array.

Note that you don't need to check for the "undefined" entries as they don't actually exist and the for loop doesn't return them. It's an artifact of the array printing that displays them as undefined. They don't appear to exist in memory.

It would be nice if you could use something like slice() which would be quicker, but it does not re-index. Anyone know of a better way?

---

Actually, you can probably do it in place as follows which is probably more efficient, performance-wise:

reindexArray : function( array )
{
    var index = 0;                          // The index where the element should be
    for( var key in array )                 // Iterate the array
    {
        if( parseInt( key ) !== index )     // If the element is out of sequence
        {
            array[index] = array[key];      // Move it to the correct, earlier position in the array
            ++index;                        // Update the index
        }
    }

    array.splice( index );  // Remove any remaining elements (These will be duplicates of earlier items)
},

Answer 12

you can use something like 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';})); // [1,2,3,4,6]

Answer 13

Why not just filter? I think it is the most clear way to consider the arrays in js.

myArray = myArray.filter(function(item){
    return item.anProperty != whoShouldBeDeleted
});

Answer 14

The difference can be seen by logging the length of each array after the delete operator and splice() method are applied. For example:

delete operator

var trees = ['redwood', 'bay', 'cedar', 'oak', 'maple'];
delete trees[3];

console.log(trees); // ["redwood", "bay", "cedar", empty, "maple"]
console.log(trees.length); // 5

The delete operator removes the element from the array, but the "placeholder" of the element still exists. oak has been removed but it still takes space in the array. Because of this, the length of the array remains 5.

splice() method

var trees = ['redwood', 'bay', 'cedar', 'oak', 'maple'];
trees.splice(3,1);

console.log(trees); // ["redwood", "bay", "cedar", "maple"]
console.log(trees.length); // 4

The splice() method completely removes the target value and the "placeholder" as well. oak has been removed as well as the space it used to occupy in the array. The length of the array is now 4.

Answer 15

Others have already properly compared delete with splice.

Another interesting comparison is delete versus undefined: a deleted array item uses less memory than one that is just set to undefined;

For example, this code will not finish:

let y = 1;
let ary = [];
console.log("Fatal Error Coming Soon");
while (y < 4294967295)
{
    ary.push(y);
    ary[y] = undefined;
    y += 1;
}
console(ary.length);

It produces this error:

FATAL ERROR: CALL_AND_RETRY_LAST Allocation failed - JavaScript heap out of memory.

So, as you can see undefined actually takes up heap memory.

However, if you also delete the ary-item (instead of just setting it to undefined), the code will slowly finish:

let x = 1;
let ary = [];
console.log("This will take a while, but it will eventually finish successfully.");
while (x < 4294967295)
{
    ary.push(x);
    ary[x] = undefined;
    delete ary[x];
    x += 1;
}
console.log(`Success, array-length: ${ary.length}.`);

These are extreme examples, but they make a point about delete that I haven't seen anyone mention anywhere.

Answer 16

function remove_array_value(array, value) {
    var index = array.indexOf(value);
    if (index >= 0) {
        array.splice(index, 1);
        reindex_array(array);
    }
}
function reindex_array(array) {
   var result = [];
    for (var key in array) {
        result.push(array[key]);
    }
    return result;
}

example:

var example_arr = ['apple', 'banana', 'lemon'];   // length = 3
remove_array_value(example_arr, 'banana');

banana is deleted and array length = 2

Answer 17

They're different things that have different purposes.

splice is array-specific and, when used for deleting, removes entries from the array and moves all the previous entries up to fill the gap. (It can also be used to insert entries, or both at the same time.) splice will change the length of the array (assuming it's not a no-op call: theArray.splice(x, 0)).

delete is not array-specific; it's designed for use on objects: It removes a property (key/value pair) from the object you use it on. It only applies to arrays because standard (e.g., non-typed) arrays in JavaScript aren't really arrays at all*, they're objects with special handling for certain properties, such as those whose names are "array indexes" (which are defined as string names "...whose numeric value i is in the range +0 ≤ i < 2^32-1") and length. When you use delete to remove an array entry, all it does is remove the entry; it doesn't move other entries following it up to fill the gap, and so the array becomes "sparse" (has some entries missing entirely). It has no effect on length.

A couple of the current answers to this question incorrectly state that using delete "sets the entry to undefined". That's not correct. It removes the entry (property) entirely, leaving a gap.

Let's use some code to illustrate the differences:

console.log("Using `splice`:");
var a = ["a", "b", "c", "d", "e"];
console.log(a.length);            // 5
a.splice(0, 1);
console.log(a.length);            // 4
console.log(a[0]);                // "b"

console.log("Using `delete`");
var a = ["a", "b", "c", "d", "e"];
console.log(a.length);            // 5
delete a[0];
console.log(a.length);            // still 5
console.log(a[0]);                // undefined
console.log("0" in a);            // false
console.log(a.hasOwnProperty(0)); // false

console.log("Setting to `undefined`");
var a = ["a", "b", "c", "d", "e"];
console.log(a.length);            // 5
a[0] = undefined;
console.log(a.length);            // still 5
console.log(a[0]);                // undefined
console.log("0" in a);            // true
console.log(a.hasOwnProperty(0)); // true

---

* (that's a post on my anemic little blog)

Answer 18

Currently there are two ways to do this

1. using splice()

   arrayObject.splice(index, 1);

2. using delete

   delete arrayObject[index];

But I always suggest to use splice for array objects and delete for object attributes because delete does not update array length.

Answer 19

OK, imagine we have this array below:

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

Let's do delete first:

delete arr[1];

and this is the result:

[1, empty, 3, 4, 5];

empty! and let's get it:

arr[1]; //undefined

So means just the value deleted and it's undefined now, so length is the same, also it will return true...

Let's reset our array and do it with splice this time:

arr.splice(1, 1);

and this is the result this time:

[1, 3, 4, 5];

As you see the array length changed and arr[1] is 3 now...

Also this will return the deleted item in an Array which is [3] in this case...

Answer 20

If you have small array you can use filter:

myArray = ['a', 'b', 'c', 'd'];
myArray = myArray.filter(x => x !== 'b');

Answer 21

Performance

There are already many nice answer about functional differences - so here I want to focus on performance. Today (2020.06.25) I perform tests for Chrome 83.0, Safari 13.1 and Firefox 77.0 for solutions mention in question and additionally from chosen answers

Conclusions

• the splice (B) solution is fast for small and big arrays
• the delete (A) solution is fastest for big and medium fast for small arrays
• the filter (E) solution is fastest on Chrome and Firefox for small arrays (but slowest on Safari, and slow for big arrays)
• solution D is quite slow
• solution C not works for big arrays in Chrome and Safari

  function C(arr, idx) {
    var rest = arr.slice(idx + 1 || arr.length);
    arr.length = idx < 0 ? arr.length + idx : idx;
    arr.push.apply(arr, rest);
    return arr;
  }
  
  
  // Crash test
  
  let arr = [...'abcdefghij'.repeat(100000)]; // 1M elements
  
  try {
   C(arr,1)
  } catch(e) {console.error(e.message)}

Details

I perform following tests for solutions A B C D E (my)

• for small array (4 elements) - you can run test HERE
• for big array (1M elements) - you can run test HERE

function A(arr, idx) {
  delete arr[idx];
  return arr;
}

function B(arr, idx) {
  arr.splice(idx,1);
  return arr;
}

function C(arr, idx) {
  var rest = arr.slice(idx + 1 || arr.length);
  arr.length = idx < 0 ? arr.length + idx : idx;
  arr.push.apply(arr, rest);
  return arr;
}

function D(arr,idx){
    return arr.slice(0,idx).concat(arr.slice(idx + 1));
}

function E(arr,idx) {
  return arr.filter((a,i) => i !== idx);
}

myArray = ['a', 'b', 'c', 'd'];

[A,B,C,D,E].map(f => console.log(`${f.name} ${JSON.stringify(f([...myArray],1))}`));

This snippet only presents used solutions

Example results for Chrome

Answer 22

Easiest way is probably

var myArray = ['a', 'b', 'c', 'd'];
delete myArray[1]; // ['a', undefined, 'c', 'd']. Then use lodash compact method to remove false, null, 0, "", undefined and NaN
myArray = _.compact(myArray); ['a', 'c', 'd'];

Hope this helps. Reference: https://lodash.com/docs#compact

Answer 23

For those who wants to use Lodash can use: myArray = _.without(myArray, itemToRemove)

Or as I use in Angular2

import { without } from 'lodash';
...
myArray = without(myArray, itemToRemove);
...

Answer 24

delete: delete will delete the object property, but will not reindex the array or update its length. This makes it appears as if it is undefined:

splice: actually removes the element, reindexes the array, and changes its length.

Delete element from last

arrName.pop();

Delete element from first

arrName.shift();

Delete from middle

arrName.splice(starting index,number of element you wnt to delete);

Ex: arrName.splice(1,1);

Delete one element from last

arrName.splice(-1);

Delete by using array index number

 delete arrName[1];

Answer 25

If the desired element to delete is in the middle (say we want to delete 'c', which its index is 1), you can use:

var arr = ['a','b','c'];
var indexToDelete = 1;
var newArray = arr.slice(0,indexToDelete).combine(arr.slice(indexToDelete+1, arr.length))

Answer 26

IndexOf accepts also a reference type. Suppose the following scenario:

var arr = [{item: 1}, {item: 2}, {item: 3}];
var found = find(2, 3); //pseudo code: will return [{item: 2}, {item:3}]
var l = found.length;

while(l--) {
   var index = arr.indexOf(found[l])
      arr.splice(index, 1);
   }
   
console.log(arr.length); //1

Differently:

var item2 = findUnique(2); //will return {item: 2}
var l = arr.length;
var found = false;
  while(!found && l--) {
  found = arr[l] === item2;
}

console.log(l, arr[l]);// l is index, arr[l] is the item you look for

Answer 27

function deleteFromArray(array, indexToDelete){
  var remain = new Array();
  for(var i in array){
    if(array[i] == indexToDelete){
      continue;
    }
    remain.push(array[i]);
  }
  return remain;
}

myArray = ['a', 'b', 'c', 'd'];
deleteFromArray(myArray , 0);

// result : myArray = ['b', 'c', 'd'];