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How do I remove a property from a JavaScript object?


Question

Say I create an object as follows:

let myObject = {
    "ircEvent": "PRIVMSG",
    "method": "newURI",
    "regex": "^http://.*"
};

What is the best way to remove the property regex to end up with new myObject as follows?

let myObject = {
    "ircEvent": "PRIVMSG",
    "method": "newURI"
};
2020/01/20
1
6255
1/20/2020 4:22:08 AM

Accepted Answer

Like this:

delete myObject.regex;
// or,
delete myObject['regex'];
// or,
var prop = "regex";
delete myObject[prop];

Demo

var myObject = {
    "ircEvent": "PRIVMSG",
    "method": "newURI",
    "regex": "^http://.*"
};
delete myObject.regex;

console.log(myObject);

For anyone interested in reading more about it, Stack Overflow user kangax has written an incredibly in-depth blog post about the delete statement on their blog, Understanding delete. It is highly recommended.

2017/10/09
8441
10/9/2017 4:55:56 PM

Objects in JavaScript can be thought of as maps between keys and values. The delete operator is used to remove these keys, more commonly known as object properties, one at a time.

var obj = {
  myProperty: 1    
}
console.log(obj.hasOwnProperty('myProperty')) // true
delete obj.myProperty
console.log(obj.hasOwnProperty('myProperty')) // false

The delete operator does not directly free memory, and it differs from simply assigning the value of null or undefined to a property, in that the property itself is removed from the object. Note that if the value of a deleted property was a reference type (an object), and another part of your program still holds a reference to that object, then that object will, of course, not be garbage collected until all references to it have disappeared.

delete will only work on properties whose descriptor marks them as configurable.

2020/04/18

var myObject = {"ircEvent": "PRIVMSG", "method": "newURI", "regex": "^http://.*"};
    
delete myObject.regex;

console.log ( myObject.regex); // logs: undefined

This works in Firefox and Internet Explorer, and I think it works in all others.

2018/06/01

The delete operator is used to remove properties from objects.

const obj = { foo: "bar" }
delete obj.foo
obj.hasOwnProperty("foo") // false

Note that, for arrays, this is not the same as removing an element. To remove an element from an array, use Array#splice or Array#pop. For example:

arr // [0, 1, 2, 3, 4]
arr.splice(3,1); // 3
arr // [0, 1, 2, 4]

Details

delete in JavaScript has a different function to that of the keyword in C and C++: it does not directly free memory. Instead, its sole purpose is to remove properties from objects.

For arrays, deleting a property corresponding to an index, creates a sparse array (ie. an array with a "hole" in it). Most browsers represent these missing array indices as "empty".

var array = [0, 1, 2, 3]
delete array[2] // [0, 1, empty, 3]

Note that delete does not relocate array[3] into array[2].

Different built-in functions in JavaScript handle sparse arrays differently.

  • for...in will skip the empty index completely.

  • A traditional for loop will return undefined for the value at the index.

  • Any method using Symbol.iterator will return undefined for the value at the index.

  • forEach, map and reduce will simply skip the missing index.

So, the delete operator should not be used for the common use-case of removing elements from an array. Arrays have a dedicated methods for removing elements and reallocating memory: Array#splice() and Array#pop.

Array#splice(start[, deleteCount[, item1[, item2[, ...]]]])

Array#splice mutates the array, and returns any removed indices. deleteCount elements are removed from index start, and item1, item2... itemN are inserted into the array from index start. If deleteCount is omitted then elements from startIndex are removed to the end of the array.

let a = [0,1,2,3,4]
a.splice(2,2) // returns the removed elements [2,3]
// ...and `a` is now [0,1,4]

There is also a similarly named, but different, function on Array.prototype: Array#slice.

Array#slice([begin[, end]])

Array#slice is non-destructive, and returns a new array containing the indicated indices from start to end. If end is left unspecified, it defaults to the end of the array. If end is positive, it specifies the zero-based non-inclusive index to stop at. If end is negative it, it specifies the index to stop at by counting back from the end of the array (eg. -1 will omit the final index). If end <= start, the result is an empty array.

let a = [0,1,2,3,4]
let slices = [
    a.slice(0,2),
    a.slice(2,2),
    a.slice(2,3),
    a.slice(2,5) ]

//   a           [0,1,2,3,4]
//   slices[0]   [0 1]- - -   
//   slices[1]    - - - - -
//   slices[2]    - -[3]- -
//   slices[3]    - -[2 4 5]

Array#pop

Array#pop removes the last element from an array, and returns that element. This operation changes the length of the array.

2020/04/18

Old question, modern answer. Using object destructuring, an ECMAScript 6 feature, it's as simple as:

const { a, ...rest } = { a: 1, b: 2, c: 3 };

Or with the questions sample:

const myObject = {"ircEvent": "PRIVMSG", "method": "newURI", "regex": "^http://.*"};
const { regex, ...newObject } = myObject;
console.log(newObject);

You can see it in action in the Babel try-out editor.


Edit:

To reassign to the same variable, use a let:

let myObject = {"ircEvent": "PRIVMSG", "method": "newURI", "regex": "^http://.*"};
({ regex, ...myObject } = myObject);
console.log(myObject);
2016/12/01

Spread Syntax (ES6)

To whoever needs it...

To complete @Koen answer in this thread, in case you want to remove dynamic variable using the spread syntax, you can do it like so:

const key = 'a';
        
const { [key]: foo, ...rest } = { a: 1, b: 2, c: 3 };

console.log(foo);  // 1
console.log(rest); // { b: 2, c: 3 }

* foo will be a new variable with the value of a (which is 1).


EXTENDED ANSWER
There are few common ways to remove a property from an object.
Each one has it's own pros and cons (check this performance comparison):

Delete Operator
Readable and short, however, it might not be the best choice if you are operating on a large number of objects as its performance is not optimized.

delete obj[key];


Reassignment
More than 2X faster than delete, however the property is not deleted and can be iterated.

obj[key] = null;
obj[key] = false;
obj[key] = undefined;


Spread Operator
This ES6 operator allows us to return a brand new object, excluding any properties, without mutating the existing object. The downside is that it has the worse performance out of the above and not suggested to be used when you need to remove many properties at a time.

{ [key]: val, ...rest } = obj;
2019/12/18

Source: https://stackoverflow.com/questions/208105
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