How can I clone a JavaScript object except for one key?

Question

I have a flat JS object:

{a: 1, b: 2, c: 3, ..., z:26}

I want to clone the object except for one element:

{a: 1, c: 3, ..., z:26}

What's the easiest way to do this (preferring to use es6/7 if possible)?

Answer 1

If you use Babel you can use the following syntax to copy property b from x into variable b and then copy rest of properties into variable y:

let x = {a: 1, b: 2, c: 3, z:26};
let {b, ...y} = x;

and it will be transpiled into:

"use strict";

function _objectWithoutProperties(obj, keys) {
  var target = {};
  for (var i in obj) {
    if (keys.indexOf(i) >= 0) continue;
    if (!Object.prototype.hasOwnProperty.call(obj, i)) continue;
    target[i] = obj[i];
  }
  return target;
}

var x = { a: 1, b: 2, c: 3, z: 26 };
var b = x.b;

var y = _objectWithoutProperties(x, ["b"]);

Answer 2

var clone = Object.assign({}, {a: 1, b: 2, c: 3});
delete clone.b;

or if you accept property to be undefined:

var clone = Object.assign({}, {a: 1, b: 2, c: 3}, {b: undefined});

Answer 3

I use this ESNext one liner

const obj = { a: 1, b: 2, c: 3, d: 4 }
const clone = (({ b, c, ...o }) => o)(obj) // remove b and c
console.log(clone)

Answer 4

For those who can't use ES6, you can use lodash or underscore.

_.omit(x, 'b')

Or ramda.

R.omit('b', x)

Answer 5

To add to Ilya Palkin's answer: you can even dynamically remove keys:

const x = {a: 1, b: 2, c: 3, z:26};

const objectWithoutKey = (object, key) => {
  const {[key]: deletedKey, ...otherKeys} = object;
  return otherKeys;
}

console.log(objectWithoutKey(x, 'b')); // {a: 1, c: 3, z:26}
console.log(x); // {a: 1, b: 2, c: 3, z:26};

Demo in Babel REPL

Source:

• https://twitter.com/ydkjs/status/699845396084846592

Answer 6

You can write a simple helper function for it. Lodash has a similar function with the same name: omit

function omit(obj, omitKey) {
  return Object.keys(obj).reduce((result, key) => {
    if(key !== omitKey) {
       result[key] = obj[key];
    }
    return result;
  }, {});
}

omit({a: 1, b: 2, c: 3}, 'c')  // {a: 1, b: 2}

Also, note that it is faster than Object.assign and delete then: http://jsperf.com/omit-key

Answer 7

Maybe something like this:

var copy = Object.assign({}, {a: 1, b: 2, c: 3})
delete copy.c;

Is this good enough? Or can't c actually get copied?

Answer 8

Using Object Destructuring

const omit = (prop, { [prop]: _, ...rest }) => rest;
const obj = { a: 1, b: 2, c: 3 };
const objWithoutA = omit('a', obj);
console.log(objWithoutA); // {b: 2, c: 3}

Answer 9

Here's an option for omitting dynamic keys that I believe has not been mentioned yet:

const obj = { 1: 1, 2: 2, 3: 3, 4: 4 };
const removeMe = 1;

const { [removeMe]: removedKey, ...newObj } = obj;

removeMe is aliased as removedKey and ignored. newObj becomes { 2: 2, 3: 3, 4: 4 }. Note that the removed key does not exist, the value was not just set to undefined.

Answer 10

How about this:

let clone = Object.assign({}, value);
delete clone.unwantedKey;

Answer 11

Hey seems like you run in to reference issues when you're trying to copy an object then deleting a property. Somewhere you have to assign primitive variables so javascript makes a new value.

Simple trick (may be horrendous) I used was this

var obj = {"key1":"value1","key2":"value2","key3":"value3"};

// assign it as a new variable for javascript to cache
var copy = JSON.stringify(obj);
// reconstitute as an object
copy = JSON.parse(copy);
// now you can safely run delete on the copy with completely new values
delete copy.key2

console.log(obj)
// output: {key1: "value1", key2: "value2", key3: "value3"}
console.log(copy)
// output: {key1: "value1", key3: "value3"}

Answer 12

You also can use spread operator to do this

const source = { a: 1, b: 2, c: 3, z: 26 }
const copy = { ...source, ...{ b: undefined } } // { a: 1, c: 3, z: 26 }

Answer 13

I recently did it this very simple way:

const obj = {a: 1, b: 2, ..., z:26};

just using spread operator to separate the unwanted property:

const {b, ...rest} = obj;

...and object.assign to take only the 'rest' part:

const newObj = Object.assign({}, {...rest});

Answer 14

The solutions above using structuring do suffer from the fact that you have an used variable, which might cause complaints from ESLint if you're using that.

So here are my solutions:

const src = { a: 1, b: 2 }
const result = Object.keys(src)
  .reduce((acc, k) => k === 'b' ? acc : { ...acc, [k]: src[k] }, {})

On most platforms (except IE unless using Babel), you could also do:

const src = { a: 1, b: 2 }
const result = Object.fromEntries(
  Object.entries(src).filter(k => k !== 'b'))

Answer 15

Lodash omit

let source = //{a: 1, b: 2, c: 3, ..., z:26}
let copySansProperty = _.omit(source, 'b');
// {a: 1, c: 3, ..., z:26}

Answer 16

What about this? I never found this patter around but I was just trying to exclude one or more properties without the need of creating an extra object. This seems to do the job but there are some side effects I'm not able to see. For sure is not very readable.

const postData = {
   token: 'secret-token',
   publicKey: 'public is safe',
   somethingElse: true,
};

const a = {
   ...(({token, ...rest} = postData) => (rest))(),
}

/**
a: {
   publicKey: 'public is safe',
   somethingElse: true,
}
*/

Answer 17

const x = {obj1: 1, pass: 2, obj2: 3, obj3:26};

const objectWithoutKey = (object, key) => {
  const {[key]: deletedKey, ...otherKeys} = object;
  return otherKeys;
}

console.log(objectWithoutKey(x, 'pass'));

Answer 18

If you're dealing with a huge variable, you don't want to copy it and then delete it, as this would be inefficient.

A simple for-loop with a hasOwnProperty check should work, and it is much more adaptable to future needs :

for(var key in someObject) {
        if(someObject.hasOwnProperty(key) && key != 'undesiredkey') {
                copyOfObject[key] = someObject[key];
        }
}

Answer 19

I accomplished it this way, as an example from my Redux reducer:

 const clone = { ...state };
 delete clone[action.id];
 return clone;

In other words:

const clone = { ...originalObject } // note: original object is not altered
delete clone[unwantedKey]           // or use clone.unwantedKey or any other applicable syntax
return clone                        // the original object without the unwanted key

Answer 20

I don't know exactly what you want to use this for, so I'm not sure if this would work for you, but I just did the following and it worked for my use case:

const newObj ={...obj, [key]: undefined}

Answer 21

If you use Ramda you can use its omit and clone functions to make a deep clone of your object and omit the unnecessary fields.

var object = {a: 1, b: 2, c: 3, y:25, z:26};
R.clone(R.omit(["z", "y"], object));

Answer 22

I had an object that I wanted to remove after utilizing its method. This is what worked for me.

/* Sample data */
let items = [
  {
    name: 'John',
    doc: {
      set: async (item) => {/*...sets docs on remote server*/}
    }
  },{
    name: 'Mary',
    doc: {
      set: async (item) => {/*...sets docs on remote server*/}
    }
  },{
    name: 'Jack',
    doc: {
      set: async (item) => {/*...sets docs on remote server*/}
    }
  }
]

/* mapping to promises */
const promises = items.map(({doc, ...item}) => doc.set(item)); 
// utilized doc and utilized `rest` of the items :)

Good Luck :)

Answer 23

Here are my two cents, on Typescript, slightly derived from @Paul's answer and using reduce instead.

function objectWithoutKey(object: object, keys: string[]) {
    return keys.reduce((previousValue, currentValue) => {
        // @ts-ignore
        const {[currentValue]: undefined, ...clean} = previousValue;
        return clean
    }, object)
}

// usage
console.log(objectWithoutKey({a: 1, b: 2, c: 3}, ['a', 'b']))

Answer 24

I have one object named: options with some keys

  let options = {       
        userDn: 'somePropertyValue',
        username: 'someValue',
        userSearchBase: 'someValue',
        usernameAttribute: 'uid',
        userPassword: 'someValue'
    }

I want to log all object excelp userPassword Property because Im testing something, I am using this code:

console.log(Object.keys(options).map(x => x + ': ' + (x === "userPassword" ? '---' : options[x])));

If you want to make it dynamic, just make a function and instead of explicitly putting userPassword you can place the value of the property you want to exclude

Answer 25

The delete keyword solution will throw a compilation error if you're using typescript, because it breaks the contract of the object you instantiated. And the ES6 spread operator solution (const {x, ...keys} = object) may throw an error depending on the linting configuration you're using on your project, since the x variable is not beign used. So I came up with this solution:

const cloneObject = Object.entries(originalObject)
    .filter(entry => entry[0] !== 'keyToRemove')
    .reduce((acc, keyValueTuple) => ({ ...acc, [keyValueTuple[0]]: keyValueTuple[1] }), {});

It solves the problem using the combination of Object.entries method (to get an array of key/value pairs of the original object) and the array methods filter and reduce. It looks verbose, but I think it's nice to have a one line chainable solution.

Answer 26

Using loadash deepclone function You can do:

const _obj = _.cloneDeep(obj);
delete _obj.key;

First clones the whole object into new object and then delete the desired key from the cloned object so that the original one doesn't get affected.