How to omit specific properties from an object in JavaScript

Question

Is there a clean way to return a new object that omits certain properties that the original object contains without having to use something like lodash?

Answer 1

const { bar, baz, ...qux } = foo

Now your object qux has all of the properties of foo except for bar and baz.

Answer 2

If you know the list of the properties that you want preserved as well as omitted, the following "whitelisting" approach should work:

const exampleFilter = ({ keepMe, keepMeToo }) => ({ keepMe, keepMeToo })

console.log(
  exampleFilter({
    keepMe: 'keepMe',
    keepMeToo: 'keepMeToo',
    omitMe: 'omitMe',
    omitMeToo: 'omitMeToo'
  })
)

Answer 3

In modern environments you can use this code snippet:

function omit(key, obj) {
  const { [key]: omitted, ...rest } = obj;
  return rest;
}

Answer 4

const {omittedPropertyA, omittedPropertyB, ...remainingObject} = originalObject;

---

Explanation:

With ES7

you could use object destructuring and the spread operator to omit properties from a javascript object:

const animalObject = { 'bear': 1, 'snake': '2', 'cow': 3 };
 
const {bear, ...other} = animalObject

// other is: { 'snake': '2', 'cow:'  3}

source: https://remotedevdaily.com/two-ways-to-omit-properties-from-a-javascript-object/

Answer 5

There's the blacklist package on npm which has a very flexible api.

Also a situational trick using the object rest-spread proposal (stage-3).

const {a, b, ...rest} = {a: 1, b: 2, c: 3, d: 4};
a // 1
b // 2
rest // {c: 3, d: 4}

This is often used in react components where you want to use a few properties and pass the rest as props to a <div {...props} /> or similar.

Answer 6

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


const {obj1,obj2, ...rest} = x;
console.log(rest)

<script src="https://cdnjs.cloudflare.com/ajax/libs/d3/4.9.1/d3.min.js"></script>

Answer 7

Omit and array of keys, using ES7 w/ recursion.

function omit(keys, obj) {
  if (!keys.length) return obj
  const { [keys.pop()]: omitted, ...rest } = obj;
  return omit(keys, rest);
}

This builds on top of @Eddie Cooro answer.

Answer 8

You can use Object.assign(), delete

var not = ["a", "b"]; // properties to delete from obj object
var o = Object.assign({}, obj);
for (let n of not) delete o[n];

Alternatively

var props = ["c", "d"];
let o = Object.assign({}, ...props.map(prop => ({[prop]:obj[prop]})));

Answer 9

Sure, why not something like:

var original = {
  name: 'Rory',
  state: 'Bored',
  age: '27'
};

var copied = Object.assign({}, original);
delete copied.age;

console.log(copied);

https://jsfiddle.net/4nL08zk4/

Answer 10

function omitKeys(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; 
    }

Answer 11

A solution that hasn't been mentioned yet:

Omit single

o = {a: 5, b: 6, c: 7}
Object.fromEntries(Object.entries(o).filter(e => e[0] != 'b'))
// Object { a: 5, c: 7 }

Omit multiple

o = {a: 1, b: 2, c: 3, d: 4}
exclude = new Set(['a', 'b'])
Object.fromEntries(Object.entries(o).filter(e => !exclude.has(e[0])))
// Object { c: 3, d: 4 }

The Set above is used because it leads to linearithmic complexity even if the number of elements in exclude is in the same asymptotic equivalence class as the number of elements in o.

Answer 12

One solution, I'm sure many others exist.

const testObj = {
    prop1: 'value1',
    prop2: 'value2',
    prop3: 'value3'
};

const removeProps = (...propsToFilter) => obj => {
   return Object.keys(obj)
   .filter(key => !propsToFilter.includes(key))
   .reduce((newObj, key) => {
       newObj[key] = obj[key];
       return newObj;
   }, {});
};


console.log(removeProps('prop3')(testObj));
console.log(removeProps('prop1', 'prop2')(testObj));

Edit: I always forget about delete...

const testObj = {
    prop1: 'value1',
    prop2: 'value2',
    prop3: 'value3'
};

const removeProps = (...propsToFilter) => obj => {
   const newObj = Object.assign({}, obj);
   propsToFilter.forEach(key => delete newObj[key]);
   return newObj;
};


console.log(removeProps('prop3')(testObj));
console.log(removeProps('prop1', 'prop2')(testObj));

Answer 13

var obj = {
  a: 1,
  b: 2,
  c: 3,
  d: {
    c: 3,
    e: 5
  }
};

Object.extract = function(obj, keys, exclude) {
  var clone = Object.assign({}, obj);
  if (keys && (keys instanceof Array)) {
    for (var k in clone) {
      var hasKey = keys.indexOf(k) >= 0;
      if ((!hasKey && !exclude) || (hasKey && exclude)) {
        delete clone[k];
      }
    }
  }
  return clone;
};

console.log('Extract specified keys: \n-----------------------');
var clone1 = Object.extract(obj, ['a', 'd']);
console.log(clone1);
console.log('Exclude specified keys: \n-----------------------');
var clone2 = Object.extract(obj, ['a', 'd'], true);
console.log(clone2);

Answer 14

If you already use lodash, you may also do omit(obj, ["properties", "to", "omit"]) to get a new Object without the properties provided in the array.

Answer 15

I saw this question and I wanted to remove 1 specific key, not a full method, so here's my suggestion:

const originalObj = {wantedKey: 111, anotherWantedKey: 222, unwantedKey: 1010};
const cleanedObj = Object.assign(originalObj, {unwantedKey: undefined});