How do I remove a property from a JavaScript object?

Question

Given an object:

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

How do I remove the property regex to end up with the following myObject?

let myObject = {
  "ircEvent": "PRIVMSG",
  "method": "newURI"
};

Hierophant

Answer 1

To remove a property from an object (mutating the object), you can do it by using the delete keyword, 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.

If you'd like a new object with all the keys of the original except some, you could use destructuring.

Demo

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

// assign the key regex to the variable _ indicating it will be unused
const { regex: _, ...newObj } = myObject;

console.log(newObj);   // has no 'regex' key
console.log(myObject); // remains unchanged

Wembley

This is a nice way to do it, but only good if you're actually going to use regex, otherwise eslint will complain about an unused variable. – Maguire

@Maguire you can rename the variable in your destructuring assignment to meet your argsIgnorePattern in the no-unused-vars rule. Easy problem to solve. – Bowel

The problem I've had with this approach is that if the destructing is inside a conditional, it makes ESlint go bonkers. – Injun

Article is gone - so sad when good content goes away, glad we have the wayback machine: web.archive.org/web/20210224201033/http://perfectionkills.com/… :) – Consociate

Understanding delete article: I found the URL perfectionkills.com/understanding-delete works (without the last slash) – Dormie

I used const {[dynamic]: regex, ...newObj} = myObject; for dynamic keys on the object. So cool! Thanks – Pomade

@nickf, "const {regex, ...newObj} = myObject" does the same, then what is the use of "regex: _"? – Zackzackariah

the destructuring syntax is very appealing, except now I have a bunch of unused consts in my code and my linter yells at me; can that be avoided somehow? – Lignocellulose

i test on my code the regex not deleted, its still appears – Elsyelton

ah object , how about array of object ? – Elsyelton

When using the destructuring method, you can make ESLint happy without having to reassign the dropped properties by setting the ignoreRestSiblings rule option to true: eslint.org/docs/latest/rules/no-unused-vars#ignorerestsiblings – Brothel

hi i have array '{62 => 105, 59 => 3.33}' i wnat to delete 59 and its value how i can do that – Flite

In my case I needed to specify "varsIgnorePattern": "^_" of the no-unused-vars rule in order to make ESLint happy. – Myron

Answer 2

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.

Jacobson

Answer 3

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);

Arrive

How do I destructure if the property name varies, i.e if I have it in a variable? – Extraterritorial

See this answer below; stackoverflow.com/a/52301527 – Arrive

Why is this preferable to delete()? "modern" isn't really a reason... – Psalterium

I'm not saying it is, I'm provinding an alternative. Although delete used to have some performance implications, which I think are already described in other answers on this page. – Arrive

@Psalterium One big reason why you would prefer this is because it does not mutate the original object. Very important with frameworks like React. I was actually looking through for an answer that does not mutate the original object. – Dolores

Can I put using object destructuring the props to exclude in an array of props instead of a comma separated list? – Cominform

Answer 4

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.

Watermelon

Answer 5

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

Strictly speaking, it's impossible to truly delete anything in JavaScript. The delete operator neither deletes objects nor frees memory. Rather, it sets its operand to undefined and manipulates the parent object so that the member is gone.

let parent = {
    member: { str: "Hello" }
};
let secondref = parent.member;

delete parent.member;
parent.member;        // undefined
secondref;            // { str: "Hello" }

The object is not deleted. Only the reference is. Memory is only freed by the garbage collector when all references to an object are removed.

Another important caveat is that the delete operator will not reorganize structures for you, which has results that can seem counterintuitive. Deleting an array index, for example, will leave a "hole" in it.

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

This is because arrays are objects. So indices are the same as keys.

let fauxarray = {0: 1, 1: 2, length: 2};
fauxarray.__proto__ = [].__proto__;
fauxarray.push(3);
fauxarray;                // [1, 2, 3]
Array.isArray(fauxarray); // false
Array.isArray([1, 2, 3]); // true

Different built-in functions in JavaScript handle arrays with holes in them differently.

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

• A naive for loop will yield 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, but will not remove it

Example:

let array = [1, 2, 3]; // [1,2,3]
delete array[1];       // [1, empty, 3]
array.map(x => 0);     // [0, empty, 0]

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. The opposite operation is push

Array#shift

Array#shift is similar to pop, except it removes the first element. The opposite operation is unshift.

Rundlet

Answer 6

Spread Syntax (ES6)

To complete Koen's answer, in case you want to remove a 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 a few common ways to remove a property from an object (or to address it as such).
Each one has its own pros and cons (check this performance comparison):

Delete Operator

It is 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];

Re-Assignment

In certain conditions, it can be handy. It is more than two times faster than a delete operator, but both key and value are not deleted and can be iterated (or skipped).

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 worst performance out of the above and is not recommended for use when you need to remove many properties at a time.

{ [key]: val, ...rest } = obj;

Eachern

This isn't removing the property, it's creating a shallow copy and not copying across the specified key and value. That's a very big difference. – Poetics

This is a mostly good answer but the last example is absurdism. It is analagous to someone asking how to get rid of an empy coca cola bottle located on their desk and my answer is to order a brand new desk and then move every single item from the old desk onto the new desk except for the empty bottle of coca-cola. you are creating an entirely new set of stuff in order to get rid of just subset of that stuff. – Desta

Answer 7

Another alternative is to use the Underscore.js library.

Note that _.pick() and _.omit() both return a copy of the object and don't directly modify the original object. Assigning the result to the original object should do the trick (not shown).

Reference: link _.pick(object, *keys)

Return a copy of the object, filtered to only have values for the whitelisted keys (or array of valid keys).

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

_.pick(myJSONObject, "ircEvent", "method");
=> {"ircEvent": "PRIVMSG", "method": "newURI"};

Reference: link _.omit(object, *keys)

Return a copy of the object, filtered to omit the blacklisted keys (or array of keys).

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

_.omit(myJSONObject, "regex");
=> {"ircEvent": "PRIVMSG", "method": "newURI"};

For arrays, _.filter() and _.reject() can be used in a similar manner.

Elegiac

_.omit(obj, 'key') is cool but I needed to omit a nested key so I used Object.assign(obj, { filters: { sort_by: '' }}) – Jellify

Answer 8

To clone an object without a property:

For example:

let object = { a: 1, b: 2, c: 3 };

And we need to delete a.

1. With an explicit prop key:

   const { a, ...rest } = object;
   object = rest;
   

2. With a variable prop key:

   const propKey = 'a';
   const { [propKey]: propValue, ...rest } = object;
   object = rest;
   

3. A cool arrow function 😎:

   const removeProperty = (propKey, { [propKey]: propValue, ...rest }) => rest;
   
   object = removeProperty('a', object);
   

4. For multiple properties

   const removeProperties = (object, ...keys) => (keys.length ? removeProperties(removeProperty(keys.pop(), object), ...keys) : object);
   

Usage

object = removeProperties(object, 'a', 'b') // result => { c: 3 }

Or

const propsToRemove = ['a', 'b']
object = removeProperties(object, ...propsToRemove) // result => { c: 3 }

Villenage

To avoid the following warning: Parameter 'propKey' implicitly has an 'any' type. ts(7006), add a type for propKey like so: const removeProperty = (propKey: string | number, { [propKey]: propValue, ...rest }) => rest – Rimester

Answer 9

The term you have used in your question title, Remove a property from a JavaScript object, can be interpreted in some different ways. The one is to remove it for whole the memory and the list of object keys or the other is just to remove it from your object. As it has been mentioned in some other answers, the delete keyword is the main part. Let's say you have your object like:

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

If you do:

console.log(Object.keys(myJSONObject));

the result would be:

["ircEvent", "method", "regex"]

You can delete that specific key from your object keys like:

delete myJSONObject["regex"];

Then your objects key using Object.keys(myJSONObject) would be:

["ircEvent", "method"]

But the point is if you care about memory and you want to whole the object gets removed from the memory, it is recommended to set it to null before you delete the key:

myJSONObject["regex"] = null;
delete myJSONObject["regex"];

The other important point here is to be careful about your other references to the same object. For instance, if you create a variable like:

var regex = myJSONObject["regex"];

Or add it as a new pointer to another object like:

var myOtherObject = {};
myOtherObject["regex"] = myJSONObject["regex"];

Then even if you remove it from your object myJSONObject, that specific object won't get deleted from the memory, since the regex variable and myOtherObject["regex"] still have their values. Then how could we remove the object from the memory for sure?

The answer would be to delete all the references you have in your code, pointed to that very object and also not use var statements to create new references to that object. This last point regarding var statements, is one of the most crucial issues that we are usually faced with, because using var statements would prevent the created object from getting removed.

Which means in this case you won't be able to remove that object because you have created the regex variable via a var statement, and if you do:

delete regex; //False

The result would be false, which means that your delete statement haven't been executed as you expected. But if you had not created that variable before, and you only had myOtherObject["regex"] as your last existing reference, you could have done this just by removing it like:

myOtherObject["regex"] = null;
delete myOtherObject["regex"];

In other words, a JavaScript object is eligible to be killed as soon as there is no reference left in your code pointed to that object.

---

Update:

Thanks to @AgentME:

Setting a property to null before deleting it doesn't accomplish anything (unless the object has been sealed by Object.seal and the delete fails. That's not usually the case unless you specifically try).

To get more information on Object.seal: Object.seal()

Chassepot

"a JavaScript object gets killed as soon as there is no reference left in your code pointed to that object." Shouldn't this say "is eligible to be killed", as the garbage collector can procrastinate on killing the object? – Tangled

Answer 10

ECMAScript 2015 (or ES6) came with built-in Reflect object. It is possible to delete object property by calling Reflect.deleteProperty() function with target object and property key as parameters:

Reflect.deleteProperty(myJSONObject, 'regex');

which is equivalent to:

delete myJSONObject['regex'];

But if the property of the object is not configurable it cannot be deleted neither with deleteProperty function nor delete operator:

let obj = Object.freeze({ prop: "value" });
let success = Reflect.deleteProperty(obj, "prop");
console.log(success); // false
console.log(obj.prop); // value

Object.freeze() makes all properties of object not configurable (besides other things). deleteProperty function (as well as delete operator) returns false when tries to delete any of it's properties. If property is configurable it returns true, even if property does not exist.

The difference between delete and deleteProperty is when using strict mode:

"use strict";

let obj = Object.freeze({ prop: "value" });
Reflect.deleteProperty(obj, "prop"); // false
delete obj["prop"];
// TypeError: property "prop" is non-configurable and can't be deleted

Claudicant

Answer 11

Suppose you have an object that looks like this:

var Hogwarts = {
    staff : [
        'Argus Filch',
        'Filius Flitwick',
        'Gilderoy Lockhart',
        'Minerva McGonagall',
        'Poppy Pomfrey',
        ...
    ],
    students : [
        'Hannah Abbott',
        'Katie Bell',
        'Susan Bones',
        'Terry Boot',
        'Lavender Brown',
        ...
    ]
};

Deleting an object property

If you want to use the entire staff array, the proper way to do this, would be to do this:

delete Hogwarts.staff;

Alternatively, you could also do this:

delete Hogwarts['staff'];

Similarly, removing the entire students array would be done by calling delete Hogwarts.students; or delete Hogwarts['students'];.

Deleting an array index

Now, if you want to remove a single staff member or student, the procedure is a bit different, because both properties are arrays themselves.

If you know the index of your staff member, you could simply do this:

Hogwarts.staff.splice(3, 1);

If you do not know the index, you'll also have to do an index search:

Hogwarts.staff.splice(Hogwarts.staff.indexOf('Minerva McGonnagall') - 1, 1);

---

Note

While you technically can use delete for an array, using it would result in getting incorrect results when calling for example Hogwarts.staff.length later on. In other words, delete would remove the element, but it wouldn't update the value of length property. Using delete would also mess up your indexing.

So, when deleting values from an object, always first consider whether you're dealing with object properties or whether you're dealing with array values, and choose the appropriate strategy based on that.

If you want to experiment with this, you can use this Fiddle as a starting point.

Anglican

Answer 12

I personally use Underscore.js or Lodash for object and array manipulation:

myObject = _.omit(myObject, 'regex');

Nitramine

Answer 13

Using delete method is the best way to do that, as per MDN description, the delete operator removes a property from an object. So you can simply write:

delete myObject.regex;
// OR
delete myObject['regex'];

The delete operator removes a given property from an object. On successful deletion, it will return true, else false will be returned. However, it is important to consider the following scenarios:

• If the property which you are trying to delete does not exist, delete will not have any effect and will return true

• If a property with the same name exists on the object's prototype chain, then, after deletion, the object will use the property from the prototype chain (in other words, delete only has an effect on own properties).

• Any property declared with var cannot be deleted from the global scope or from a function's scope.

• As such, delete cannot delete any functions in the global scope (whether this is part of a function definition or a function (expression).
• Functions which are part of an object (apart from the
  global scope) can be deleted with delete.

• Any property declared with let or const cannot be deleted from the scope within which they were defined. Non-configurable properties cannot be removed. This includes properties of built-in objects like Math, Array, Object and properties that are created as non-configurable with methods like Object.defineProperty().

The following snippet gives another simple example:

var Employee = {
  age: 28,
  name: 'Alireza',
  designation: 'developer'
}

console.log(delete Employee.name);   // returns true
console.log(delete Employee.age);    // returns true

// When trying to delete a property that does 
// not exist, true is returned 
console.log(delete Employee.salary); // returns true

For more info about and seeing more examples visit the link below:

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/delete

Diacid

Answer 14

There are a few of ways of removing properties from an object:

1) Remove using the dot property accessor (mutable)

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

delete myObject.regex;
console.log(myObject);

2. Remove using square brackets property accessor (mutable)

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

delete myObject['regex'];
console.log(myObject);
// or
const name = 'ircEvent';
delete myObject[name];
console.log(myObject);

3) Removing without altering the original object is possible by using object destructuring and rest syntax (immutable)

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

const { regex, ...myObjectRest} = myObject;
console.log(myObjectRest); 

Jackelinejackelyn

Answer 15

Another solution, using Array#reduce.

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

myObject = Object.keys(myObject).reduce(function(obj, key) {
  if (key != "regex") {           //key you want to remove
    obj[key] = myObject[key];
  }
  return obj;
}, {});

console.log(myObject);

However, it will mutate the original object. If you want to create a new object without the specified key, just assign the reduce function to a new variable, e.g.:

(ES6)

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

const myNewObject = Object.keys(myObject).reduce((obj, key) => {
  key !== 'regex' ? obj[key] = myObject[key] : null;
  return obj;
}, {});

console.log(myNewObject);

Jahdol

Answer 16

There are a lot of good answers here but I just want to chime in that when using delete to remove a property in JavaScript, it is often wise to first check if that property exists to prevent errors.

E.g

var obj = {"property":"value", "property2":"value"};

if (obj && obj.hasOwnProperty("property2")) {
  delete obj.property2;
} else {
  //error handling
}

Due to the dynamic nature of JavaScript there are often cases where you simply don't know if the property exists or not. Checking if obj exists before the && also makes sure you don't throw an error due to calling the hasOwnProperty() function on an undefined object.

Sorry if this didn't add to your specific use case but I believe this to be a good design to adapt when managing objects and their properties.

Windywindzer

Answer 17

This post is very old and I find it very helpful so I decided to share the unset function I wrote in case someone else see this post and think why it's not so simple as it in PHP unset function.

The reason for writing this new unset function, is to keep the index of all other variables in this hash_map. Look at the following example, and see how the index of "test2" did not change after removing a value from the hash_map.

function unset(unsetKey, unsetArr, resort) {
  var tempArr = unsetArr;
  var unsetArr = {};
  delete tempArr[unsetKey];
  if (resort) {
    j = -1;
  }
  for (i in tempArr) {
    if (typeof(tempArr[i]) !== 'undefined') {
      if (resort) {
        j++;
      } else {
        j = i;
      }
      unsetArr[j] = tempArr[i];
    }
  }
  return unsetArr;
}

var unsetArr = ['test', 'deletedString', 'test2'];

console.log(unset('1', unsetArr, true)); // output Object {0: "test", 1: "test2"}
console.log(unset('1', unsetArr, false)); // output Object {0: "test", 2: "test2"}

Coffle

Answer 18

Using ramda#dissoc you will get a new object without the attribute regex:

const newObject = R.dissoc('regex', myObject);
// newObject !== myObject

You can also use other functions to achieve the same effect - omit, pick, ...

Daff

Answer 19

Try the following method. Assign the Object property value to undefined. Then stringify the object and parse.

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

myObject.regex = undefined;
myObject = JSON.parse(JSON.stringify(myObject));

console.log(myObject);

Whiteley

Answer 20

Using Lodash

import omit from 'lodash/omit';

const prevObject = {test: false, test2: true};
// Removes test2 key from previous object
const nextObject = omit(prevObject, 'test2');

Using Ramda

R.omit(['a', 'd'], {a: 1, b: 2, c: 3, d: 4}); //=> {b: 2, c: 3}

Astigmia

weird. your code _.omit(['a', 'd'], {a: 1, b: 2, c: 3, d: 4}); didn't work for me but instead _.omit({a: 1, b: 2, c: 3, d: 4}, ['a', 'd']); did work. – Horal

Answer 21

If you want to delete a property deeply nested in the object then you can use the following recursive function with path to the property as the second argument:

var deepObjectRemove = function(obj, path_to_key){
    if(path_to_key.length === 1){
        delete obj[path_to_key[0]];
        return true;
    }else{
        if(obj[path_to_key[0]])
            return deepObjectRemove(obj[path_to_key[0]], path_to_key.slice(1));
        else
            return false;
    }
};

Example:

var a = {
    level1:{
        level2:{
            level3: {
                level4: "yolo"
            }
        }
    }
};

deepObjectRemove(a, ["level1", "level2", "level3"]);
console.log(a);

//Prints {level1: {level2: {}}}

Rudich

Answer 22

Dan's assertion that 'delete' is very slow and the benchmark he posted were doubted. So I carried out the test myself in Chrome 59. It does seem that 'delete' is about 30 times slower:

var iterationsTotal = 10000000;  // 10 million
var o;
var t1 = Date.now(),t2;
for (let i=0; i<iterationsTotal; i++) {
   o = {a:1,b:2,c:3,d:4,e:5};
   delete o.a; delete o.b; delete o.c; delete o.d; delete o.e;
}
console.log ((t2=Date.now())-t1);  // 6135
for (let i=0; i<iterationsTotal; i++) {
   o = {a:1,b:2,c:3,d:4,e:5};
   o.a = o.b = o.c = o.d = o.e = undefined;
}
console.log (Date.now()-t2);  // 205

Note that I purposely carried out more than one 'delete' operations in one loop cycle to minimize the effect caused by the other operations.

Rattlepate

Answer 23

Object.assign() & Object.keys() & Array.map()

const obj = {
    "Filters":[
        {
            "FilterType":"between",
            "Field":"BasicInformationRow.A0",
            "MaxValue":"2017-10-01",
            "MinValue":"2017-09-01",
            "Value":"Filters value"
        }
    ]
};

let new_obj1 = Object.assign({}, obj.Filters[0]);
let new_obj2 = Object.assign({}, obj.Filters[0]);

/*

// old version

let shaped_obj1 = Object.keys(new_obj1).map(
    (key, index) => {
        switch (key) {
            case "MaxValue":
                delete new_obj1["MaxValue"];
                break;
            case "MinValue":
                delete new_obj1["MinValue"];
                break;
        }
        return new_obj1;
    }
)[0];


let shaped_obj2 = Object.keys(new_obj2).map(
    (key, index) => {
        if(key === "Value"){
            delete new_obj2["Value"];
        }
        return new_obj2;
    }
)[0];


*/


// new version!

let shaped_obj1 = Object.keys(new_obj1).forEach(
    (key, index) => {
        switch (key) {
            case "MaxValue":
                delete new_obj1["MaxValue"];
                break;
            case "MinValue":
                delete new_obj1["MinValue"];
                break;
            default:
                break;
        }
    }
);

let shaped_obj2 = Object.keys(new_obj2).forEach(
    (key, index) => {
        if(key === "Value"){
            delete new_obj2["Value"];
        }
    }
);

Jacobo

Answer 24

Property Removal in JavaScript

There are many different options presented on this page, not because most of the options are wrong—or because the answers are duplicates—but because the appropriate technique depends on the situation you're in and the goals of the tasks you and/or you team are trying to fulfill. To answer you question unequivocally, one needs to know:

1. The version of ECMAScript you're targeting
2. The range of object types you want to remove properties on and the type of property names you need to be able to omit (Strings only? Symbols? Weak references mapped from arbitrary objects? These have all been types of property pointers in JavaScript for years now)
3. The programming ethos/patterns you and your team use. Do you favor functional approaches and mutation is verboten on your team, or do you employ wild west mutative object-oriented techniques?
4. Are you looking to achieve this in pure JavaScript or are you willing & able to use a 3rd-party library?

Once those four queries have been answered, there are essentially four categories of "property removal" in JavaScript to chose from in order to meet your goals. They are:

Mutative object property deletion, unsafe

This category is for operating on object literals or object instances when you want to retain/continue to use the original reference and aren't using stateless functional principles in your code. An example piece of syntax in this category:

'use strict'
const iLikeMutatingStuffDontI = { myNameIs: 'KIDDDDD!', [Symbol.for('amICool')]: true }
delete iLikeMutatingStuffDontI[Symbol.for('amICool')] // true
Object.defineProperty({ myNameIs: 'KIDDDDD!', 'amICool', { value: true, configurable: false })
delete iLikeMutatingStuffDontI['amICool'] // throws

This category is the oldest, most straightforward & most widely supported category of property removal. It supports Symbol & array indexes in addition to strings and works in every version of JavaScript except for the very first release. However, it's mutative which violates some programming principles and has performance implications. It also can result in uncaught exceptions when used on non-configurable properties in strict mode.

Rest-based string property omission

This category is for operating on plain object or array instances in newer ECMAScript flavors when a non-mutative approach is desired and you don't need to account for Symbol keys:

const foo = { name: 'KIDDDDD!', [Symbol.for('isCool')]: true }
const { name, ...coolio } = foo // coolio doesn't have "name"
const { isCool, ...coolio2 } = foo // coolio2 has everything from `foo` because `isCool` doesn't account for Symbols :(

Mutative object property deletion, safe

This category is for operating on object literals or object instances when you want to retain/continue to use the original reference while guarding against exceptions being thrown on unconfigurable properties:

'use strict'
const iLikeMutatingStuffDontI = { myNameIs: 'KIDDDDD!', [Symbol.for('amICool')]: true }
Reflect.deleteProperty(iLikeMutatingStuffDontI, Symbol.for('amICool')) // true
Object.defineProperty({ myNameIs: 'KIDDDDD!', 'amICool', { value: true, configurable: false })
Reflect.deleteProperty(iLikeMutatingStuffDontI, 'amICool') // false

In addition, while mutating objects in-place isn't stateless, you can use the functional nature of Reflect.deleteProperty to do partial application and other functional techniques that aren't possible with delete statements.

Syntax-based string property omission

This category is for operating on plain object or array instances in newer ECMAScript flavors when a non-mutative approach is desired and you don't need to account for Symbol keys:

const foo = { name: 'KIDDDDD!', [Symbol.for('isCool')]: true }
const { name, ...coolio } = foo // coolio doesn't have "name"
const { isCool, ...coolio2 } = foo // coolio2 has everything from `foo` because `isCool` doesn't account for Symbols :(

Library-based property omission

This category is generally allows for greater functional flexibility, including accounting for Symbols & omitting more than one property in one statement:

const o = require("lodash.omit")
const foo = { [Symbol.for('a')]: 'abc', b: 'b', c: 'c' }
const bar = o(foo, 'a') // "'a' undefined"
const baz = o(foo, [ Symbol.for('a'), 'b' ]) // Symbol supported, more than one prop at a time, "Symbol.for('a') undefined"

Quadruple

Answer 25

Here's an ES6 way to remove the entry easily:

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

const removeItem = 'regex';

const { [removeItem]: remove, ...rest } = myObject;

console.log(remove); // "^http://.*"
console.log(rest); // Object { ircEvent: "PRIVMSG", method: "newURI" }

Besought

Answer 26

@johnstock, we can also use JavaScript's prototyping concept to add method to objects to delete any passed key available in calling object.

Above answers are appreciated.

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

// 1st and direct way 
delete myObject.regex; // delete myObject["regex"]
console.log(myObject); // { ircEvent: 'PRIVMSG', method: 'newURI' }

// 2 way -  by using the concept of JavaScript's prototyping concept
Object.prototype.removeFromObjectByKey = function(key) {
  // If key exists, remove it and return true
  if (this[key] !== undefined) {
    delete this[key]
    return true;
  }
  // Else return false
  return false;
}

var isRemoved = myObject.removeFromObjectByKey('method')
console.log(myObject) // { ircEvent: 'PRIVMSG' }

// More examples
var obj = {
  a: 45,
  b: 56,
  c: 67
}
console.log(obj) // { a: 45, b: 56, c: 67 }

// Remove key 'a' from obj
isRemoved = obj.removeFromObjectByKey('a')
console.log(isRemoved); //true
console.log(obj); // { b: 56, c: 67 }

// Remove key 'd' from obj which doesn't exist
var isRemoved = obj.removeFromObjectByKey('d')
console.log(isRemoved); // false
console.log(obj); // { b: 56, c: 67 }

Mulholland

Answer 27

You can use a filter like below

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

// Way 1

let filter1 = {}
  Object.keys({...myObject}).filter(d => {
  if(d !== 'regex'){
    filter1[d] = myObject[d];
  }
})

console.log(filter1)

// Way 2

let filter2 = Object.fromEntries(Object.entries({...myObject}).filter(d =>
d[0] !== 'regex'
))

console.log(filter2)

Corneous

but The filter() method creates an array filled with all array elements that pass a test so the sole purpose of the filter is reducing and creating a new array – Corneous

Yes it does but that's only useful if you are using the returned value of filter by resigning it to the original variable or another one, and also it only works if it's given a return value in the function passed in as a parameter to it, but none if those things are done here so it's only purpose is iterating the array elements, which is what forEach is for – Thymol

You could instead do let filter = Object.fromEntries(Object.entries(myObject).filter(d => d !== 'regex' )) – Thymol

Answer 28

I have used Lodash "unset" to make it happen for a nested object also... only this needs to write small logic to get the path of the property key which is expected by the omit method.

1. Method which returns the property path as an array

var a = {"bool":{"must":[{"range":{"price_index.final_price":{"gt":"450", "lt":"500"}}}, {"bool":{"should":[{"term":{"color_value.keyword":"Black"}}]}}]}};

function getPathOfKey(object,key,currentPath, t){
    var currentPath = currentPath || [];

    for(var i in object){
        if(i == key){
            t = currentPath;
        }
        else if(typeof object[i] == "object"){
            currentPath.push(i)
            return getPathOfKey(object[i], key,currentPath)
        }
    }
    t.push(key);
    return t;
}
document.getElementById("output").innerHTML =JSON.stringify(getPathOfKey(a,"price_index.final_price"))

<div id="output">

</div>

1. Then just using Lodash unset method remove property from object.

var unset = require('lodash.unset');
unset(a, getPathOfKey(a, "price_index.final_price"));

Cydnus

Answer 29

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


obj = Object.fromEntries(
    Object.entries(myObject).filter(function (m){
        return m[0] != "regex"/*or whatever key to delete*/
    }
))

console.log(obj)

You can also just treat the object like a2d array using Object.entries, and use splice to remove an element as you would in a normal array, or simply filter through the object, as one would an array, and assign the reconstructed object back to the original variable

Stockbreeder

Answer 30

If you don't want to modify the original object.

Remove a property without mutating the object

If mutability is a concern, you can create a completely new object by copying all the properties from the old, except the one you want to remove.

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

let prop = 'regex';
const updatedObject = Object.keys(myObject).reduce((object, key) => {
  if (key !== prop) {
    object[key] = myObject[key]
  }
  return object
}, {})

console.log(updatedObject);

Boigie

Answer 31

The straight way

As of 2023, you can do it in just one line:

delete myObject.regex;

OR

delete myObject["regex"];

(For instance if you have an object obj with a property called key-one, you cannot do obj.key-one, so here comes obj["key-one"]).

Courson

Answer 32

We can easily remove any property of an object by using the delete operator.

const object = {
  prop1: 10,
  prop2: 20,
  prop3: 30,
  "test prop": "This is test props"
}
console.log(object); // will print all 4 props

delete object.prop1;
delete object["test prop"];

console.log(object); // it will print only prop2 and prop3

See docs : MDN

Ochrea

Answer 33

Two ways to delete an object

1. using for ... in

    function deleteUser(key) {
   
        const newUsers = {};
        for (const uid in users) {
            if (uid !== key) {
                newUsers[uid] = users[uid];
            }
   
        return newUsers
    }
   

or

delete users[key]

Rainband

So you are recommending that to remove one property, the whole object should be copied into a new object without the targeted property?!? – Flooded

Is that valid code? A } seems to be missing. You can edit your answer - but without "Edit:", "Update:", or similar - the answer should appear as if it was written today. – Outshout

Answer 34

We can remove using

1. using delete object.property
2. using delete object['property']
3. using rest, remove multiple property

let myObject = {
  "ircEvent": "PRIVMSG",
  "method": "newURI",
  "regex": "^http://.*",
  "regex1": "^http://.*",
  "regex2": "^http://.*",
  "regex3": "^http://.*",
  "regex4": "^http://.*"
};

delete myObject.regex; // using delete object.property

// Or 

delete myObject['regex1']; // using delete object['property']

const { regex2, regex3, regex4, ...newMyObject } = myObject;

console.log(newMyObject);

Colossians

Answer 35

You can delete property from object using Delete property[key]

Fishbowl

Answer 36

In JavaScript, there are 2 common ways to remove properties from an object.

The first mutable approach is to use the delete object.property operator.

The second approach, which is immutable since it doesn't modify the original object, is to invoke the object destructuring and spread syntax: const {property, ...rest} = object

Lepine

Answer 37

Short answer

var obj = {
  data: 1,
  anotherData: 'sample'    
}
delete obj.data //this removes data from the obj

You are left with

var obj = {
  anotherData: 'sample'    
}

Extortioner

Answer 38

This is how I do.

const obj = {
  name: 'John',
  age: 30,
  gender: 'male'
};

delete obj.age;

console.log(obj); // { name: 'John', gender: 'male' }

Brundage