How do I remove a property from a JavaScript object?

5 173

874

Say I create an object as follows:

var 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?

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

johnstok

Posted 2008-10-16T10:57:45.830

Reputation: 40 399

5

Find here a benchmark which compares "delete" vs "undefined" vs "null" http://jsben.ch/#/BWMiw

– EscapeNetscape – 2017-03-06T11:31:56.850

Answers

7 159

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.

nickf

Posted 2008-10-16T10:57:45.830

Reputation: 367 419

I'm finding that I get an error when using Object.defineProperty( window, "b", ... ) on a var that was already declared on window, for example: "var b = 500;" even if I delete it, the error reads: "Uncaught TypeError: Cannot redefine property: b". (Latest version of Chrome.) – Resist Design – 2012-07-28T09:22:43.157

It's worth noting that you can use the 3rd example given here to delete items inside of a for-in loop. for(var k in obj) { delete obj[k]; } will unset every property of the object (not that you'd ever have a use for this, specifically -- but you could add a condition, like if (obj[k] === 2)). – Patrick M – 2012-10-26T22:03:04.000

11@ChristopherPfohl works for me. Like I said, it's actually quite in-depth, so it's a bit difficult to summarize. The basic response in the answer above is sufficient for almost all cases, the blog goes into some more of the edge cases and the reasons those cases exist. – nickf – 2012-12-20T01:57:12.327

1@GeorgeJempty You can use a variable for the property name itself in the delete expression: var x = 'x'; delete object[x]; – Jacob Relkin – 2013-07-09T20:07:16.583

90An upshot of one of the observations at the "understanding delete" link above, is that, since you cannot necessarily delete a variable, but only object properties, you therefore cannot delete an object property "by reference" -- var value=obj['prop']; delete value //doesn't work – George Jempty – 2010-04-16T16:24:33.760

I know this is old but I ran into a gotcha after using delete and then checking an object with jquery's .isEmptyObject the function will return false. – Cfreak – 2014-05-05T19:20:58.907

1@geoidesic it's not an array, it's an object. If you want to remove an item from an array, you should use .splice() – nickf – 2014-05-22T16:42:54.493

2

isn't delete forbidden when using JS Strict mode? doc on strict mode. How could I delete the property when using strict mode?

– code ninja – 2014-08-29T16:15:11.523

1

Question for my comment above: http://stackoverflow.com/questions/25572222/how-to-remove-a-property-from-a-javascript-object-when-using-strict-mode thanks

– code ninja – 2014-08-29T16:22:29.757

It's possible for deleting keys from an object to actually slow it down, if the Javascript engine is optimizing memory. Addy Osmani talks about it in a couple different places.

– iabw – 2014-09-03T13:08:05.333

If you delete a property from an object and wonder why it's still there, it's probably picking it up from the prototype (although browser implementation varies). – Matthew Strawbridge – 2016-06-02T11:09:06.767

1Is it possible to use associative array syntax with delete? Say i have the name of the property as a string i.e. 'regex'. – johnstok – 2008-10-16T11:04:14.787

31

Checked, it also works with "delete myJSONObject['regex'];" See: http://developer.mozilla.org/en/Core_JavaScript_1.5_Reference/Operators/Special_Operators/delete_Operator

– johnstok – 2008-10-16T11:06:42.533

Irritatingly MS fail to mention that in their own JScript documentation on MSDN. – AnthonyWJones – 2008-10-16T12:41:45.697

16So it doesn't actually delete it? It just becomes undefined, but the key still exists? Am I missing something? – Doug Molineux – 2011-08-10T02:21:35.107

126@Pete no, it does remove it. Given: var x = {a : 'A', b : 'B'}; Compare: delete x.a; typeof x.a; /* "undefined" */ x.hasOwnProperty('a'); /* false */ to x.b = undefined; typeof x.b; /* "undefined" */; x.hasOwnProperty('b'); /* true */ – nickf – 2011-08-10T08:29:19.273

would anyone please let me know How I can delete multiple properties from same array using single expression ? – Aakriti.G – 2018-01-22T09:37:14.903

To understand how delete works, I can recommend reading on its function equivalent, Reflect.deleteProperty -- both perform exactly the same operation, but the function may be a bit more understandable.

– amn – 2018-11-06T15:53:54.420

753

Operator delete is unexpectedly slow!

Look at the benchmark.

Delete is the only true way to remove object's properties without any leftovers, but it works ~ 100 times slower, compared to its "alternative", setting object[key] = undefined.

This alternative is not the correct answer to this question! But, if you use it with care, you can dramatically speed up some algorithms. If you are using delete in loops and you have problems with performance, read the verbose explanation.

When should one use delete and when set value to undefined ?

An object may be seen as a set of key-value pairs. What I call a 'value' is a primitive or a reference to other object, connected to that 'key'.

Use delete, when you are passing the result object to the code on which you don't have control (or when you are not sure about your team or yourself).

It deletes the key from the hashmap.

 var obj = {
     field: 1     
 };
 delete obj.field;

Use setting to undefined, when you care about performance. It can give a serious boost to your code.

The key remains on its place in the hashmap, only the value is replaced with undefined. Understand, that for..in loop will still iterate over that key.

 var obj = {
     field: 1     
 };
 obj.field = undefined;

Using this method, not all ways of determining property existence will work as expected.

However, this code:

object.field === undefined

will behave equivalently for both methods.

Tests

To summarize, differences are all about ways of determining the property existence, and about for..in loop.

 console.log('* -> "Takes prototype inheritance into consideration, that means it lookups all over prototype chain too."');

 console.log(obj.field === undefined, 'obj.field === undefined', 'You get "undefined" value when querying for "field" in object-hashmap. *');

 console.log(obj["field"] === undefined, 'obj["field"] === undefined', 'Just another way to query (equivalent). *');

 console.log(typeof obj.field === "undefined", 'typeof obj.field === "undefined"', 'Get the value attached to "field" key, and check it\'s type is "undefined". *');

 console.log("field" in obj, '"field" in obj', 'This statement returns true if "field" key exists in the hashmap. False otherwise. *');

 console.log(obj.hasOwnProperty("field"), 'obj.hasOwnProperty("field")', 'This statement returns true if \'field\' key exists in the hashmap. The ONLY way NOT to lookup for property in the prototype chain!');
 //Object.keys().indexOf() is an overkill that runs much slower :)

 var counter = 0,
     key;
 for (key in obj) {
     counter++;
 }
 console.assert(counter === 0, 'counter === 0', '"field" is not iterated using "for .. in" loop. *');

Beware Of Memory Leaks!

While using obj[prop] = undefined is faster than doing delete obj[prop], another important consideration is that obj[prop] = undefined may not always be appropriate. delete obj[prop] removes prop from obj and erases it from memory whereas obj[prop] = undefined simply sets the value of prop to undefined which leaves prop still in memory. Therefore, in circumstances where there are many keys being created and deleted, using obj[prop] = undefined can force expensive memory reconciliation (causing the page to freeze up) and potentially an out-of-memory error. Examine the following code.

"use strict";
var theNodeList=[], i, current, numberOfNodes=65536, body=document.body, nodeRecords=[];
for (i = 0; i !== numberOfNodes; i++) {
    nodeRecords[i] = [];
    current = theNodeList[i] = document.createElement("div");
    current.textContent = i;
    document.body.appendChild( current );
}
var lastTime = -1;
requestAnimationFrame(function recordUpdates(){
    var currentTime = Math.round( performance.now()*1000 )
    for (i = 0; i !== numberOfNodes; i++) {
        if (lastTime !== -1) {
            // the previously collected data is no longer in use
            /*************************************************/
            /****/ nodeRecords[i][lastTime] = undefined; /****/
            /*************************************************/
        }
        nodeRecords[i][currentTime] = theNodeList[i].outerHTML;
    }
    lastTime = currentTime;
    requestAnimationFrame( recordUpdates );
});

In the code above, simply doing nodeRecords[i][lastTime] = undefined; will cause a massive memory leak because each animation frame. Each frame, all 65536 DOM elements will take up another 65536 individual slots, but the previous 65536 slots will only be set to undefined which leaves them hanging in the memory. Go ahead, try to run the above code in the console and see for yourself. After forcing an out-of-memory error, attempt to run it again except with the following version of the code that uses the delete operator instead.

"use strict";
var theNodeList=[], i, current, numberOfNodes=65536, body=document.body, nodeRecords=[];
for (i = 0; i !== numberOfNodes; i++) {
    nodeRecords[i] = [];
    current = theNodeList[i] = document.createElement("div");
    current.textContent = i;
    document.body.appendChild( current );
}
var lastTime = -1;
requestAnimationFrame(function recordUpdates(){
    var currentTime = Math.round( performance.now()*1000 )
    for (i = 0; i !== numberOfNodes; i++) {
        if (lastTime !== -1) {
            // the previously collected data is no longer in use
            /********************************************/
            /****/ delete nodeRecords[i][lastTime]; /****/
            /********************************************/
        }
        nodeRecords[i][currentTime] = theNodeList[i].outerHTML;
    }
    lastTime = currentTime;
    requestAnimationFrame( recordUpdates );
});

As seen in the above code snippet, there are some rare appropriate use cases for the delete operator. However, do not worry about this problem too much. This will only become a problem with long-lifespan objects that get new keys constantly added to them. In any other case (which is almost every case in real-world programming), it is most appropriate to use obj[prop] = undefined. The main purpose of this section is just to bring this to your attention so that in the rare chance that this does become a problem in your code, then you can more easily understand the problem and thus not have to waste hours dissecting your code to locate and understand this problem.

Do Not Always Set To undefined

One aspect of Javascript that is important to consider is polymorphism. Polymorphism is when assigning the same variable/slot-in-an-object different types as seen below.

var foo = "str";
foo = 100;          // variable foo is now labeled polymorphic by the browser
var bar = ["Some", "example"];
bar[2] = "text";    // bar is a monomorphic array here because all its entries have the
                    // same type: string primitive
bar[1] = undefined; // bar is now a polymorphic array

However, there are two major unfixable problems with polymorphic arrays:

  1. They are slow & memory inefficient. When accessing a specific index, instead of just getting the global type for the array, the browser instead has to get the type on a per-index basis whereby each index stores the additional metadata of its type.
  2. Once polymorphic, always polymorphic. When an array is made polymorphic, the polymorphism cannot be undone in Webkit browsers. So, even if you restore a polymorphic array to being non-polymorphic, it will still be stored by the browser as a polymorphic array.

One may liken polymorphism to a drug addiction. At first glance, it seems awesomely lucrative: nice pretty fluffy code. Then, the coder introduces their array to the drug of polymorphism. Instantly, the polymorphic array becomes less efficient, and it can never become as efficient as it was before since it is drugged. To correlate such circumstance to real life, someone on cocaine might not even be capable of operating a simple door handle, much less be able to calculate digits of PI. Likewise, an array on the drug of polymorphism cannot ever be as efficient as a monomorphic array.

But, how does a drug trip analogy relate to the delete operation? The answer inheres the last line of code in the snippet above. Thus let it be reexamined, this time with a twist.

var bar = ["Some", "example"];
bar[2] = "text";    // bar is not a polymorphic array here because all its entries have the
                    // same type: string primitive
bar[1] = "";        // bar is still a monomorphic array
bar[1] = undefined; // bar is now a polymorphic array

Observe. bar[1] = "" does not coerce polymorphism whereas bar[1] = undefined does. Therefore, one should always, whenever possible use the corresponding type for their objects so as to not accidentally cause polymorphism. One such person may use the following list as a general reference to get them going. However, please do not explicitly use the below ideas. Instead, use whatever works well for your code.

  • When using an array/variable typed to the boolean primitive, use either false or undefined as the empty value. While avoiding unnecessary polymorphism is good, rewriting all your code to explicitly forbid it will likely actually result in a decrease in performance. Use common judgement!
  • When using an array/variable typed to the number primitive, use 0 as the empty value. Note that internally, there are two types of numbers: fast integers (2147483647 to -2147483648 inclusive) and slow floating point doubles (anything other than that including NaN and Infinity). When an integer is demoted to a double, it cannot be promoted back to an integer.
  • When using an array/variable typed to the string primitive, use "" as the empty value.
  • When using a Symbol, wait, why are you using a Symbol?!?! Symbols are bad juju for performance. Everything programmed to use Symbols can be reprogrammed to not use Symbols, resulting in a faster code without Symbols. Symbols are really just super inefficient meta-sugar.
  • When using anything else, use null.

However, be mindful! Do not suddenly start doing this with all your preexisting code now as it would likely break such preexisting code and/or introduce strange bugs. Rather, such an efficient practice needs to be implemented from the start, and when converting preexisting code, it is recommended that you double, triple, quadruple check all the lines relating to that as trying to upgrade old code to this new practice can be as risky as it is rewarding.

Dan

Posted 2008-10-16T10:57:45.830

Reputation: 32 196

37a property is assigned to undefined still is a property of an object, so it will not be removed by GC, unless misread your last paragraph. – Lance – 2014-03-04T21:05:54.650

7I was wrong to touch the theme of GC here. Both methods have the same result for GC: they remove the value linked to the key. If that value was the last reference to some other object, that object would be cleaned up. – Dan – 2014-03-06T10:28:46.377

1Fun fact, you can cast an object back into fast properties mode after you delete from it. – Benjamin Gruenbaum – 2014-05-31T18:31:25.287

10a property is assigned to undefined still is a property of an object, so it will not be removed by GC The GC doesn't manage anything about properties. It collects and removes values. As long as nothing references a value (an object, string, etc.) anymore, the GC does remove it from memory. – meandre – 2014-07-16T15:03:18.807

7

BTW, that is the dual problem to check if a property exist on Javascript object. Using in operator is the reliable but slow. Check if the property is not undefined "is not the correct answer" but it is a way faster. check

– rdllopes – 2014-11-03T17:32:30.673

All I can say is test any performance assertion about JS or browsers that A. isn't about obvious work avoidance and B. is more than 5 hours old. This will probably hold for a while though. JSON parsing something you JSON stringified and then regex replaced might be one avenue to explore for performance, key-clearing, and absolutely no messy leftovers, however. – Erik Reppen – 2016-02-03T02:07:52.467

The delete method isn't that slow..... – Charlie Fish – 2016-07-17T00:09:25.463

3

Is this answer still relevant? jsperf is currently down, but this benchmark seems to indicate that the speed difference is a mere 25%, which is nowhere close to the "~ 100 times slower" in this answer.

– Cerbrus – 2016-12-14T08:29:46.397

your "test" throws an error... Uncaught ReferenceError: obj is not defined – Mark Giblin – 2017-06-01T10:10:17.163

Hi Dan, I'm no expert so I'd like to ask what you mean by "unexpectedly slow"... First it appears to run millions of times each second if I understand and I don't understand if that "test" is reliable? What do we know about the way the Browser's interpreter (or whatnot) pre-optimizes the code? – wolfdawn – 2017-07-03T12:50:09.593

2Now, at the end of 2017, delete is still 98% slower than the other "alternatives" in your JSPerf test. However, when comparing things, one must use the same standards in all cases. It goes without saying that removing a property, irrespective if it's done using delete or any other function, will always be slower than assigning a value to that property, even 100 years from now. Just like removing a button from a webpage will always be slower than painting it in a transparent color - they're different operations, aka 'different standards' used for comparison... – Yin Cognyto – 2017-12-25T21:54:30.243

Take every jsperf benchmark with a huge grain of salt. Most of the time the benchmark is flawed, or the difference is so minimal it won't come close to making a noticeable difference. On top of that, it's a moving target because browsers are constantly optimizing slow APIs and code patterns, so what appears slower today could get super optimized tomorrow. Best to just write code that optimizes for maintainability/deletion than worrying about micro benchmarks – Hozefa – 2018-01-08T18:03:26.313

@Hozefa Yes take them with a grain of salt if they are written as terribly as this, but no do not take them with a grain of salt if they are written durably such as to inhibit dead code elimination. – Jack Giffin – 2018-04-18T00:53:54.837

Even though this answer has a lot of upvotes THIS answer to THIS question is a complete mistake in my opinion. Even though delete is slow, it is the correct answer to the question, because reasigning the property to be undefined wouldn't actually delete the property, therefore not answering the question – Alejandro Vales – 2018-05-29T13:22:58.947

209

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.

redsquare

Posted 2008-10-16T10:57:45.830

Reputation: 69 159

Is there any way to delete multiple properties from same array using one expression – Aakriti.G – 2018-01-22T09:38:01.997

When trying to delete multiple properties at once, would it not be better to just remap to a new object? – jasonscript – 2018-05-03T02:40:41.233

157

Update 2018-07-21: For a long time, I have felt embarrassed about this answer, so I think it's time that I touch it up a little bit. Just a little commentary, clarification, and formatting to help expedite the reading of the needlessly long and convoluted parts of this answer.


THE SHORT VERSION

The actual answer to the question

As others have said, you can use delete.

obj // {"foo": "bar"}
delete obj["foo"]
obj // {}
obj["foo"] // undefined

Array equivalent

Don't delete from an array. Use Array.prototype.splice instead.

arr // [1,2,3,4,5]
arr.splice(3,1); // 4
arr // [1,2,3,5]

THE LONG VERSION

JavaScript is an OOP Language, so everything is an object, including arrays. Thus, I feel it necessary to point out a particular caveat.

In arrays, unlike plain old objects, using delete leaves behind garbage in the form of null, creating a "hole" in the array.

var array = [1, 2, 3, 4];
delete array[2];
/* Expected result --> [1, 2, 4]
 * Actual result   --> [1, 2, null, 4]
 */

As you can see, delete doesn't always work as one might expect. The value is overwritten, but the memory is not reallocated. That is to say, array[4] isn't relocated to array[3]. Which is in contrast to Array.prototype.unshift, which inserts an element in the beginning of the array and shifts everything up (array[0] becomes array[1], etc.)

Honestly, aside from setting to null rather than undefined--which is legitimately weird--this behavior shouldn't be surprising, since delete is an unary operator, like typeof, that is hard-boiled into the language and is not supposed to care about the type of object it's being used on, whereas Array is a subclass of Object with methods specifically designed for working with arrays. So there's no good reason for delete to have a special case cooked in for re-shifting the array, as that would just slow things down with unnecessary work. In retrospect, my expectations were unrealistic.

Of course, it did surprise me. Because I wrote this to justify my crusade against "null garbage":

Ignoring the dangers and problems inherent in null, and the space wasted, this can be problematic if the array needs to be precise.

Which is a terrible justification for getting rid of the nulls--null is only dangerous if used improperly, and it has nothing to do with "precision". The real reason you shouldn't delete from an array is because leaving garbage-filled and messy data structures around is sloppy and bug-prone.

What follows is a contrived scenario that gets pretty long-winded, so you can skip to the section, The Solution, if you want. The only reason I leave this section in is because I think some people probably think it's funny, and I don't want to be "that guy" who posts a "funny" answer and then deletes all the "funny" from it later on.

...It's stupid, I know.

The contrived and long-winded PDP-11 scenario

For example, say you are creating a webapp that uses JSON-serialization to store an array used for 'tabs' in a string (in this case, localStorage). Let's also say that the code uses the numerical indices of the array's members to "title" them when drawing to the screen. Why are you doing this rather than just storing the "title" as well? Because... reasons.

Okay, let's just say that you're trying to save memory at the request of this one user who runs a PDP-11 minicomputer from the 1960's running UNIX, and wrote his own Elinks-based, JavaScript-compliant, line-printer-friendly browser because X11 is out of the question.

Increasingly stupid edge-case scenario aside, using delete on said array will result in null polluting the array, and probably causing bugs in the app later on. And if you check for null, it would straight up skip the numbers resulting in the tabs being rendered like [1] [2] [4] [5] ....

if (array[index] == null)
    continue;
else
    title = (index + 1).toString();
/* 0 -> "1"
 * 1 -> "2"
 * 2 -> (nothing)
 * 3 -> "4"
 */

Yeah, that's definitely not what you wanted.

Now, you could keep a second iterator, like j, to increment only when valid values are read from the array. But that wouldn't exactly solve the null issue, and you still have to please that troll PDP-11 user. Alas, his computer just doesn't have enough memory to hold that last integer (don't ask how he manages to handle a variable-width array...).

So, he sends you an email in anger:

Hey, your webapp broke my browser! I checked my localStorage database after your stupid code made my browser segfault, and this is what I found:

>"tabs:['Hello World', 'foo bar baz', null, null, null, null, null, null, null, null, null, null, null, null, null, null, null, null, ... ]"

After clearing my precious data, it segfaulted again, and I did a backtrace, and what do I find? WHAT DO I FIND!? YOU USE TOO MANY VARIABLES!

>var i = index;
>var j = 1;

Grr, I am angry now.
-Troll Davidson

About now, you're at your wit's end. This guy has been complaining non-stop about your app, and you want to tell him to shut up and go get a better computer.

The Solution: Array.prototype.splice

Luckily, arrays do have a specialized method for deleting indices and reallocating memory: Array.prototype.splice(). You could write something like this:

Array.prototype.remove = function(index){
  this.splice(index,1);
}
...
array = [1, 2, 3, 4];
array.remove(2);
// Result -> [1, 2, 4]

And just like that, you've pleased Mr. PDP-11. Hooray! (I'd still tell him off, though...)

Array.prototype.splice vs Array.prototype.slice

I feel it's important to point out the difference between these two similarly-named functions, as they are both very useful.

Array.prototype.splice(start, n)

.splice() mutates the array, and returns the removed indices. The array is sliced starting from the index, start, and n elements are sliced out. If n is unspecified, the entire array after start is sliced out (n = array.length - start).

let a = [5,4,3,2,1];
let chunk = a.splice(2,2);

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

chunk; // [3,2]
a;     // [5,4,1]

Array.prototype.slice(start, end)

.slice() is non-destructive and returns a new array containing the indicated indices from start to end. If end is left unspecified, the behavior is the same as .splice() (end = array.length). The behavior is a bit tricky since, for some reason, end indexes from 1 instead of 0. I don't know why it does this, but that's how it is. Also, if end <= start, the result is an empty array.

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

// a             [5,4,3,2,1]
// start          0 1 2 - -
// end, for...    - - - - -
//   chunks[0]  0 - - - - -   
//   chunks[1]    1 2 - - -
//   chunks[2]    1 2 3 - -
//   chunks[3]    1 2 3 4 5

chunks; // [ [], [], [3], [3,2,1] ]
a;      // [5,4,3,2,1]

That actually isn't what's happening, but it's easier to think of that way. According to MDN, here's what's actually happening:

// a             [5,4,3,2,1]
// start          0 1 2 - - -
// end, for...    - - - - - -
//   chunks[0]    0 - - - - -
//   chunks[1]    0 1 2 - - -
//   chunks[2]    0 1(2)3 - -
//   chunks[3]    0 1(2 3 4)5

The index specified by end is simply excluded from the slice. The parenthesized indices indicate what gets sliced. Either way, the behavior is not intuitive and it's bound to cause its fair share of off-by-one errors, so you might find it useful to make a wrapper function to more closely emulate the behavior of .splice():

function ez_slice(array, start = 0, n = null){
    if(!Array.isArray(array) || !is_number(start))
        return null;

    if(is_number(n))
        return array.slice(start, start + n);

    if(n === null)
        return array.slice(start);

    return null;
}

ez_slice([5,4,3,2,1], 2, 1) // [3]
ez_slice([5,4,3,2,1], 2)    // [3,2,1]

/* Fun fact: isNaN is unreliable.
 * [NaN, [], {}, 0, 1, Infinity, undefined, null, "Hi"].filter(isNaN)
 * [NaN, {}, undefined, "Hi"]
 *
 * What we want is...
 *
 * [NaN, [], {}, 0, 1, Infinity, undefined, null, "Hi"].filter(is_nan)
 * [NaN]
 */
function is_nan(num){
    return typeof num === "number"
        && num !== num;
}

function is_number(num){
    return !is_nan(num)
        && typeof num === "number"
        && isFinite(num);
}

Note that the wrapper function is designed to be very strict about types, and will return null if anything is off. That includes putting in a string like "3". It is left up to the programmer to be diligent about his types. This is to encourage good programming practice.

Update regarding is_array()

This is in regard to this (now-removed) snippet:

function is_array(array){
    return array !== null
        && typeof array === "object"
        && typeof array.length !== "undefined"
        && array.__proto__ === Array.prototype;
}

So as it turns out, there actually IS a built-in way to tell if an array is truly an array, and that is Array.isArray(), introduced in ECMAScript 5 (December 2009). I found this while looking to see if there was a question asking about telling arrays from objects, to see if there was either a better solution than mine, or to add mine if there were none. So, if you're using a version of JavaScript that is earlier than ECMA 5, there's your polyfill. However, I strongly recommend against using my is_array() function, as continuing to support old versions of JavaScript means continuing to support the old browsers that implement them, which means encouraging the use of insecure software and putting users at risk for malware. So please, use Array.isArray(). Use let and const. Use the new features that get added to the language. Don't use vendor prefixes. Delete that IE polyfill crap from your website. Delete that XHTML <!CDATA[[... crap, too--we moved to HTML5 back in 2014. The sooner everybody withdraws support for those old/esoteric browsers, the sooner the browser vendors will actually follow the web standard and embrace the new technology, and the sooner we can move on to a more secure web.

Braden Best

Posted 2008-10-16T10:57:45.830

Reputation: 5 257

16This delete keyword, however, is much more convenient, lol – Braden Best – 2012-09-18T00:59:47.370

11This approach doesn't modify the original object which might be still referenced elsewhere. This might or might not be a problem depending on how it's used but it's something to keep in mind. – Tamas Czinege – 2012-11-07T17:39:59.153

17

@B1KMusic Here's the way to delete an element from an Array: splice

– wulftone – 2013-01-25T20:20:43.963

2@wulftone nope, that splits the array and does nothing to delete a value. I really think the best way to delete from an array where specific values are needed to be deleted is to use delete and make a Garbage Collection function to clean it up. – Braden Best – 2013-01-26T20:37:09.073

@B1KMusic Oh I misunderstood. I thought by "delete" you meant "remove from the array." You actually mean to remove it from both the array and from memory at the same time, right? – wulftone – 2013-01-28T19:43:54.577

@wulftone umm, that's kinda what I assumed delete means in general. Yes, for some of my apps, I've come across weird glitches that were caused by using just delete array[index], and I had to write my own garbage collection method to keep the array clean so it wouldn't clutter the localStorage/skip a number. – Braden Best – 2013-01-28T22:13:38.270

@wulftone I actually did find a use for Array.prototype.splice(). I gave my answer a much needed update – Braden Best – 2013-01-28T22:48:24.013

4I don’t see splice in your edit, but remove ought to be Array.prototype.remove = function(index) { this.splice(index, 1); }; – Ry- – 2013-08-27T04:01:41.257

@minitech Yeah, that would be a much smarter way of doing it. At the time of the rewrite, I had only just begun to learn about splice(). – Braden Best – 2013-09-14T04:51:27.873

array.splice(index, 1).... – Derek 朕會功夫 – 2014-06-27T18:52:25.480

Nevermind that defining Array.prototype.remove() probably takes up much more than the approximately 8 bytes needed to define those two integers. – Patrick Roberts – 2016-01-27T04:03:50.557

@PatrickRoberts JavaScript's Number type is defined as a 64-bit double, so wouldn't it be 16 bytes?

– Braden Best – 2016-01-27T04:06:23.250

@BradenBest okay sure, 16 bytes. I'm sure the function is still larger in memory. – Patrick Roberts – 2016-01-27T05:47:40.130

@PatrickRoberts It's a silly, contrived scenario anyway. A retroactive excuse to justify talking about arrays in a question about objects. – Braden Best – 2016-01-27T05:51:06.980

1This article is full of bull 1. It doesn't address the Question! 2. It is an exemplifying misuse of language and a of complain that "it doesn't work!" 3. Do not blame the JavaScript delete operator for Crockford's idiosyncratic error of putting null for an empty array index. He doesn't understand the meaning of null - he thinks it's a mistake. The mistake is his own and alone - there is no null in the deleted value of a given array index. There is no "hole" in the array - that's an empty index. Absolutely legitimate and expected. – Bekim Bacaj – 2016-11-16T07:34:00.387

Legitimate and expected, but not necessarily wanted, hence why I went over the splice function. And I'm not blaming anything on anything. I'm pointing out that the result of delete may not always be what you think it will, one of the many features of JavaScript that isn't quite logical.

– Braden Best – 2016-11-16T19:01:42.120

Not that leaving a null in the array is illogical, per se. However, in a dynamic language like js, where arrays are technically objects and can be easily resized and modified, you would think delete would splice it out or throw an error. Leaving in a sentinel value is something you would do in C. – Braden Best – 2016-11-16T19:14:10.830

And also, you're right that I didn't directly address the question. However, this is still useful information. Useful information that is not compatible with the comment format. Hence why it is posted as an answer. – Braden Best – 2016-11-16T19:16:36.393

On an array you can use array.splice() instead, that correctly controls the array. – E730 – 2017-10-25T13:31:50.423

@E730 you should probably read the full answer. Specifically, the implementation of Array.prototype.remove – Braden Best – 2017-10-26T00:32:25.177

@BradenBest, you're true, but the question is about removing object properties, not array entries. Arrays are actually made for adding and removing properties, and have an API for that. Objects don't have that, so using delete will be the best answer for this question. – E730 – 2017-10-26T08:07:45.800

@E730 old news. This has already been discussed in the comments. In fact, nearly a year ago. – Braden Best – 2017-10-26T08:09:20.473

Added a new section about slice vs splice, as well as a wrapper function for slice to make its behavior a little more intuitive. – Braden Best – 2018-07-19T16:52:15.880

...and now that section is longer than the original answer. – Braden Best – 2018-07-19T18:53:22.697

114

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);

Koen.

Posted 2008-10-16T10:57:45.830

Reputation: 14 952

78

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.

Thaddeus Albers

Posted 2008-10-16T10:57:45.830

Reputation: 2 660

4Keep in mind that if your object's keys are numbers, you may need to _.omit(collection, key.toString()) – Jordan Arseno – 2014-12-09T01:32:19.843

Why did you link to Wikipedia, and not to http://underscorejs.org/?

– E730 – 2017-10-26T08:09:30.273

@E730 Good question. I don't know why one of the editors changed it to point to wikipedia.org/wiki/Underscore.js. I changed it back to underscorejs.org. – Thaddeus Albers – 2018-02-03T19:54:28.870

Hmmmmm.... Underscore is ~100x slower than delete obj[prop] which is ~100x slower than obj[prop] = undefined. – Jack Giffin – 2018-04-14T23:00:57.963

40

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 gets 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 info on Object.seal: Object.seal()

Mehran Hatami

Posted 2008-10-16T10:57:45.830

Reputation: 9 378

You're wrong - only objects are passed by reference in JavaScript, so if myJSONObject.regex's value is a string and you assign it to some other object, the other object has a copy of this value. – Michał Perłakowski – 2016-01-10T16:58:19.133

You are right and this is a quote: "to be careful about your other references to the same object." – Mehran Hatami – 2016-01-10T17:34:48.110

31

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

madox2

Posted 2008-10-16T10:57:45.830

Reputation: 22 407

31

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.

John Slegers

Posted 2008-10-16T10:57:45.830

Reputation: 27 166

24

The delete operator is the best way to do so.

A live example to show:

var foo = {bar: 'bar'};
delete foo.bar;
console.log('bar' in foo); // Logs false, because bar was deleted from foo.

Tarun Nagpal

Posted 2008-10-16T10:57:45.830

Reputation: 397

It might be worth noting that even though using the delete operator is healthy in regard of garbage collection, it can be unexpectedly slow, in no small part for the same reason.

– John Weisz – 2015-05-06T17:51:33.153

21

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

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

emil

Posted 2008-10-16T10:57:45.830

Reputation: 1 878

21

Spread Syntax (ES6)

To whoever need it...

To complete @Koen answer in this thread, 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(rest); // { b: 2, c: 3 }

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


UPDATE:
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;

Lior Elrom

Posted 2008-10-16T10:57:45.830

Reputation: 7 088

18

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);

kind user

Posted 2008-10-16T10:57:45.830

Reputation: 18 837

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"}

talsibony

Posted 2008-10-16T10:57:45.830

Reputation: 4 078

17

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 from 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: 'abc',
      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 example, visit the link below:

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

Alireza

Posted 2008-10-16T10:57:45.830

Reputation: 43 514

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.

Willem

Posted 2008-10-16T10:57:45.830

Reputation: 612

2delete foo.bar works even if bar doesn't exist, so your test is a bit too much, IMHO. – PhiLho – 2015-06-01T15:14:02.890

@PhiLho that depends on where you are running JavaScript. In Node.js I believe this causes your server to crash. – Willem – 2015-06-02T20:07:29.763

2delete foo.bar; only throws an exception if foo is falsy, or if you're in strict mode and foo is an object with an unconfigurable bar property. – Macil – 2015-06-02T21:28:10.020

I don't remember the exact issue I've had with this but I think the problem may appear when foo itself doesn't exist and you try to delete it's property. – Willem – 2015-06-02T22:20:12.013

Yes, you have to test if foo exists, otherwise foo.bar will throw an exception, but you don't need to check the existence for bar before deleting it. That's the "too much" part of my comment. :-) – PhiLho – 2015-06-03T09:16:35.987

Could be that I was checking for the property itself to determine the state of the machine or something. E.g. not necessary for OP's question. – Willem – 2015-06-15T10:06:52.703

12

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, ...

zatziky

Posted 2008-10-16T10:57:45.830

Reputation: 8 400

11

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: {}}}

ayushgp

Posted 2008-10-16T10:57:45.830

Reputation: 2 157

11

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);

Mohammed Safeer

Posted 2008-10-16T10:57:45.830

Reputation: 9 698

6

You can simply delete any property of an object using the delete keyword.

For example:

var obj = {key1:"val1",key2:"val2",key3:"val3"}

To remove any property, say key1, use the delete keyword like this:

delete obj.key1

Or you can also use array-like notation:

delete obj[key1]

Ref: MDN.

Kalpesh Patel

Posted 2008-10-16T10:57:45.830

Reputation: 2 229

6

Using ES6:

(Destructuring + Spread operator)

    const myObject = {
      regex: "^http://.*",
      b: 2,
      c: 3
    };
    const { regex, ...noRegex } = myObject;
    console.log(noRegex); // => { b: 2, c: 3 }

Srinivas

Posted 2008-10-16T10:57:45.830

Reputation: 493

5

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 purposedly carried out more than one 'delete' operations in one loop cycle to minimize the effect caused by the other operations.

Chong Lip Phang

Posted 2008-10-16T10:57:45.830

Reputation: 2 723

5

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"];
        }
    }
);

xgqfrms-gildata

Posted 2008-10-16T10:57:45.830

Reputation: 355

4

Try this

delete myObject['key'];

codemirror

Posted 2008-10-16T10:57:45.830

Reputation: 915

4

Hello You Can try this simple an sort

var obj = [];

obj.key1 = {name: "John", room: 1234};
obj.key2 = {name: "Jim", room: 1234};

delete(obj.key1);

Dhaval Gohel

Posted 2008-10-16T10:57:45.830

Reputation: 482

4

Consider creating a new object without the "regex" property because the original object could always be referenced by other parts of your program. Thus you should avoid manipulating it.

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

const { regex, ...newMyObject } = myObject;

console.log(newMyObject);

ideaboxer

Posted 2008-10-16T10:57:45.830

Reputation: 389

4

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"

james_womack

Posted 2008-10-16T10:57:45.830

Reputation: 7 043

4

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

const { regex, ...other } = myObject;

console.log(myObject)
console.log(regex)
console.log(other)

xiang

Posted 2008-10-16T10:57:45.830

Reputation: 353

2

Very simple:

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

delete myObject.regex;

Behnam Mohammadi

Posted 2008-10-16T10:57:45.830

Reputation: 6 724

2

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}

johndavedecano

Posted 2008-10-16T10:57:45.830

Reputation: 329

1

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

  1. Method which returns property path as 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"));

BEJGAM SHIVA PRASAD

Posted 2008-10-16T10:57:45.830

Reputation: 746

1

you can use the delete operator as of below.

 var multiverse = {
        earth1: "Silver Age",
        earth2: "Golden Age"
    };

delete multiverse.earth2;//will return true if it finds 

// Outputs: { earth1: "Silver Age" }
console.log(multiverse);

The delete operator also has a return value. If it succeeds in deleting a property, it will return true. If it fails to delete a property because the property is unwritable it will return false, or if in strict mode it will throw an error.

MANISH MOHAN

Posted 2008-10-16T10:57:45.830

Reputation: 44

1

For

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

delete myObject["regex"]; Also you know index of key , for Eg. You need to delete 2nd key(regex) then

var key =  Object.keys(myObject)[2]
delete myObject[key];

Many time we require index based removing in object But Remember keys in object may be up-down position so please careful during using it

RITESH ARORA

Posted 2008-10-16T10:57:45.830

Reputation: 343

1

function removeProperty(obj, prop) {
    if(prop in obj){

    Reflect.deleteProperty(obj, prop);
    return true;
   } else {
    return false;
  }
}

diayn geogiev

Posted 2008-10-16T10:57:45.830

Reputation: 41

1

@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 }

hygull

Posted 2008-10-16T10:57:45.830

Reputation: 2 846

0

function removeElement(elementId) {
    // Removes an element from the document
    var element = document.getElementById(elementId);
    element.parentNode.removeChild(element);
}

function addElement(parentId, elementTag, elementId, html) {
    // Adds an element to the document
    var p = document.getElementById(parentId);
    var newElement = document.createElement(elementTag);
    newElement.setAttribute('id', elementId);
    newElement.innerHTML = html;
    p.appendChild(newElement);
}

for more info click this

hossein.sharifipour

Posted 2008-10-16T10:57:45.830

Reputation: 27

-1

delete someObject["keyName"];
someObject["keyName"] // undefined

C. Morgan

Posted 2008-10-16T10:57:45.830

Reputation: 59