Set all Object keys to false

Question

Lets say I have an object

  filter: {
    "ID": false,
    "Name": true,
    "Role": false,
    "Sector": true,
    "Code": false
  }

I want to set all keys to false (to reset them). What's the best way to do this, I'd like to avoid looping with foreach and stuff. Any neat one liner?

How about this, `filter: {}` if you had handled the `filter` for the same? — Aruna, Nov 28 '16 at 02:22

nnnnnn · Answer 1 · 2016-11-28T02:22:35.550

71

Well here's a one-liner with vanilla JS:

Object.keys(filter).forEach(v => filter[v] = false)

It does use an implicit loop with the .forEach() method, but you'd have to loop one way or another (unless you reset by replacing the whole object with a hardcoded default object literal).

edited Nov 28 '16 at 02:22

answered Nov 28 '16 at 02:19

nnnnnn

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2

Correct me if I'm wrong but this can't be done **WITHOUT** looping. Not sure if the OP understands this. – Carl Edwards Nov 28 '16 at 02:20
@CarlEdwards that's fine carl. I didn't realise arrow functions were this useful. – chefcurry7 Nov 28 '16 at 02:22
Even without an arrow function it is still pretty short, as in the other answer. – nnnnnn Nov 28 '16 at 02:23
Yeah in this case the arrow syntax is almost identical to the traditional `function() {}`. Just a little less typing. – Carl Edwards Nov 28 '16 at 02:24

score 35 · Accepted Answer · answered Nov 28 '16 at 02:30

35

Using lodash, mapValues is a graceful, loop-free way:

filter = {
    "ID": false,
    "Name": true,
    "Role": false,
    "Sector": true,
    "Code": false
};

filter = _.mapValues(filter, () => false);

If you want to do this with Underscore.js, there is an equivalent, but with a slightly different name:

filter = _.mapObject(filter, () => false);

In either case, the value of filter will be set to:

{ ID: false, 
  Name: false, 
  Role: false, 
  Sector: false, 
  Code: false }

answered Nov 28 '16 at 02:30

Jonathan Eunice

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thanks for that. one question, if i want to obtain the property that has value "true" (only one at any time), how best to do it? – chefcurry7 Nov 28 '16 at 02:44
1

The example data belie the assertion that only one property is `true` at a time. But regardless of however many are, `_.keys(filter).filter(k => filter[k])` will give you a quick array of all keys to values that are truthy. You can use exact equivalence to `true` if you want strictly booleans. And you can always index the resulting array to get the first one: `_.keys(filter).filter(k => filter[k])[0]`. That returns `undefined` if there are none. Sorry for the three instances of "filter" in that expression, but your variable name overloads the function name. – Jonathan Eunice Nov 28 '16 at 03:05
@chefcurry7 You might also like `_.find(_.keys(filter), k => filter[k])`. It works similarly, but slighly shorter. – Jonathan Eunice Nov 28 '16 at 03:46
This is the best solution if you use lodash! – gandharv garg Oct 15 '18 at 15:26

score 12 · Answer 3 · edited May 31 '20 at 09:04

12

If you're not using ES6, here is its ES5 counterpart.

Object.keys(filter).forEach(function(key, value) {
    return filter[key] = false;
})

edited May 31 '20 at 09:04

Ryan

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answered Nov 28 '16 at 02:22

Joshua Russell

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Giovanni Benussi · Answer 4 · 2020-07-13T17:27:25.863

9

If you don't want to modify the array, here's an ES6 alternative that returns a new one:

Object.fromEntries(Object.keys(filter).map((key) => [key, false]))

Explanation:

Object.keys returns the object's keys:

Object.keys({ a: 1, b: 2 }) // returns ["a", "b"]

Then we map the result ["a", "b"] to [key, false]:

["a", "b"].map((key) => [key, false]) // returns [['a', false], ['b', false]]

And finally we call Object.fromEntries that maps an array of arrays with the form [key, value] to an Object:

Object.fromEntries([['a', false], ['b', false]]) // returns { a: false, b: false }

edited Jul 13 '20 at 17:27

answered Jul 13 '20 at 16:58

Giovanni Benussi

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this is the best answer so far! – Hila Grossbard Oct 03 '22 at 09:06

score 4 · Answer 5 · answered Nov 28 '16 at 02:23

4

A small line of code compatible with all browsers:

for(var i in your_object) your_object[i] = false;

answered Nov 28 '16 at 02:23

Pinguto

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score 3 · Answer 6 · answered Apr 16 '19 at 19:35

3

With ES6 features one-liner without mutation:

{
  ...Object.keys(filter).reduce((reduced, key) => ({ ...reduced, [key]: false }), {})
}

answered Apr 16 '19 at 19:35

RegarBoy

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` ...Object.keys(filter).reduce((reduced, key) => ({ ...reduced, [key]: false }), {})` should suffce. No need to wrap it in an object – Alan David Garcia Aug 22 '19 at 04:44
This worked for my use-case, requiring immutability, though the OP did not state that as a requirement. – Jozzhart Feb 11 '20 at 05:54

score 2 · Answer 7 · answered Sep 13 '17 at 13:50

2

hasOwnProperty must be used

```

for(var i in your_object) {
  if (Object.hasOwnProperty.call(your_object, i)) {
    your_object[i] = false;
  }
}

```

answered Sep 13 '17 at 13:50

Alex Shwarc

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score 1 · Answer 8 · answered Sep 24 '18 at 17:52

1

In case you are dealing with 'scope' variables, this might be a better solution.

Object.keys(filter).reduce(function(accObj, parseObj) {
                accObj[parseObj] = false;
                return accObj;
              }, {});

answered Sep 24 '18 at 17:52

Avinash Srivastava

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3

score 0 · Answer 9 · answered Jul 25 '20 at 02:39

Here's what i would call a neat one-liner solution:

const filtered = Object.assign(...Object.keys(filter).map(k => ({ [k]: false })));

Demo:

const filter = {
  'ID': false,
  'Name': true,
  'Role': false,
  'Sector': true,
  'Code': false
};

const filtered = Object.assign(...Object.keys(filter).map(k => ({ [k]: false })));

console.log(filtered);

We are basically converting the object into an array using Object.assign() so we can use the map() fucntion on it to set each propety value to false, then we convert the array back to an object using the Object.assign() with the spread operator :)

Set all Object keys to false

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