Remove value from object without mutation

Question

Update:

You could remove a property from an object with a tricky Destructuring assignment:

const doSomething = (obj, prop) => {
  let {[prop]: omit, ...res} = obj
  return res
}

Though, if property name you want to remove is static, then you could remove it with a simple one-liner:

let {lastname, ...o2} = o

---

The easiest way is simply to Or you could clone your object before mutating it:

const doSomething = (obj, prop) => {
  let res = Object.assign({}, obj)
  delete res[prop]
  return res
}

Alternatively you could use omit function from lodash utility library:

let o2 = _.omit(o, 'lastname')

It's available as a part of lodash package, or as a standalone lodash.omit package.

With ES7 object destructuring:

const myObject = {
  a: 1,
  b: 2,
  c: 3
};
const { a, ...noA } = myObject;
console.log(noA); // => { b: 2, c: 3 }

one line solution

const removeKey = (key, {[key]: _, ...rest}) => rest;

Explanations:

This is a generic arrow function to remove a specific key. The first argument is the name of the key to remove, the second is the object from where you want to remove the key. Note that by restructuring it, we generate the curated result, then return it.

Example:

let example = { 
  first:"frefrze",
  second:"gergerge",
  third: "gfgfg"
}

console.log(removeKey('third', example))
/*
Object {
  first: "frefrze",
  second: "gergerge"
}
*/

To add some spice bringing in Performance. Check this thread bellow

https://github.com/googleapis/google-api-nodejs-client/issues/375

The use of the delete operator has performance negative effects for the V8 hidden classes pattern. In general it's recommended do not use it.

Alternatively, to remove object own enumerable properties, we could create a new object copy without those properties (example using lodash):

_.omit(o, 'prop', 'prop2')

Or even define the property value to null or undefined (which is implicitly ignored when serializing to JSON):

o.prop = undefined

You can use too the destructing way

const {remov1, remov2, ...new} = old;
old = new;

And a more practical exmple:

this._volumes[this._minCandle] = undefined;
{ 
     const {[this._minCandle]: remove, ...rest} = this._volumes;
     this._volumes = rest; 
}

As you can see you can use [somePropsVarForDynamicName]: scopeVarName syntax for dynamic names. And you can put all in brackets (new block) so the rest will be garbage collected after it.

Here a test:

exec:

Or we can go with some function like

function deleteProps(obj, props) {
    if (!Array.isArray(props)) props = [props];
    return Object.keys(obj).reduce((newObj, prop) => {
        if (!props.includes(prop)) {
            newObj[prop] = obj[prop];
        }
        return newObj;
    }, {});
}

for typescript

function deleteProps(obj: Object, props: string[]) {
    if (!Array.isArray(props)) props = [props];
    return Object.keys(obj).reduce((newObj, prop) => {
        if (!props.includes(prop)) {
            newObj[prop] = obj[prop];
        }
        return newObj;
    }, {});
}

Usage:

let a = {propH: 'hi', propB: 'bye', propO: 'ok'};

a = deleteProps(a, 'propB'); 

// or 

a = deleteProps(a, ['propB', 'propO']);

This way a new object is created. And the fast property of the object is kept. Which can be important or matter. If the mapping and the object will be accessed many many times.

Also associating undefined can be a good way to go with. When you can afford it. And for the keys you can too check the value. For instance to get all the active keys you do something like:

const allActiveKeys = Object.keys(myObj).filter(k => myObj[k] !== undefined);
//or
const allActiveKeys = Object.keys(myObj).filter(k => myObj[k]); // if any false evaluated value is to be stripped.

Undefined is not suited though for big list. Or development over time with many props to come in. As the memory usage will keep growing and will never get cleaned. So it depend on the usage. And just creating a new object seem to be the good way.

Then the Premature optimization is the root of all evil will kick in. So you need to be aware of the trade off. And what is needed and what's not.

Note about _.omit() from lodash

It's removed from version 5. You can't find it in the repo. And here an issue that talk about it.

https://github.com/lodash/lodash/issues/2930

v8

You can check this which is a good reading https://v8.dev/blog/fast-properties

As suggested in the comments above if you want to extend this to remove more than one item from your object I like to use filter. and reduce

eg

    const o = {
      "firstname": "Jane",
      "lastname": "Doe",
      "middlename": "Kate",
      "age": 23,
      "_id": "599ad9f8ebe5183011f70835",
      "index": 0,
      "guid": "1dbb6a4e-f82d-4e32-bb4c-15ed783c70ca",
      "isActive": true,
      "balance": "$1,510.89",
      "picture": "http://placehold.it/32x32",
      "eyeColor": "green",
      "registered": "2014-08-17T09:21:18 -10:00",
      "tags": [
        "consequat",
        "ut",
        "qui",
        "nulla",
        "do",
        "sunt",
        "anim"
      ]
    };

    const removeItems = ['balance', 'picture', 'tags']
    console.log(formatObj(o, removeItems))

    function formatObj(obj, removeItems) {
      return {
        ...Object.keys(obj)
          .filter(item => !isInArray(item, removeItems))
          .reduce((newObj, item) => {
            return {
              ...newObj, [item]: obj[item]
            }
          }, {})
      }
    }

    function isInArray(value, array) {
      return array.indexOf(value) > -1;
    }

My issue with the accepted answer, from an ESLint rule standard, if you try to destructure:

    const { notNeeded, alsoNotNeeded, ...rest } = { ...ogObject };

the 2 new variables, notNeeded and alsoNotNeeded may throw a warning or error depending on your setup since they are now unused. So why create new vars if unused?

I think you need to use the delete function truly.