How to get multiple key values from array with lodash?

Question

I have below object in javascript:

var users = [{
    'user': 'barney',
    'age': 36,
    'active': true
}, {
    'user': 'fred',
    'age': 40,
    'active': false
}, {
    'user': 'pebbles',
    'age': 1,
    'active': true
}];

I want to get a new object from above object but remove active key like below:

var users = [{
        'user': 'barney',
        'age': 36,
    }, {
        'user': 'fred',
        'age': 40,
    }, {
        'user': 'pebbles',
        'age': 1,
    }];

I know that I can create the object by a for-loop, but I am looking for a better way to do that. Whether it can be done by one line of lodash code?

Why do you favor a library solution over a loop?

1252748
– 1252748

2017-02-03 03:10:45 +00:00
Commented Feb 3, 2017 at 3:10 — 1252748
– 1252748, Commented Feb 3, 2017 at 3:10
_.map(users, user => _.omit(user, 'active'))

Tushar
– Tushar

2017-02-03 03:13:02 +00:00
Commented Feb 3, 2017 at 3:13 — Tushar
– Tushar, Commented Feb 3, 2017 at 3:13

Tom Coughlin · Accepted Answer · 2017-02-03 03:19:22Z

5

If you can use the latest JavaScript features...

users = users.map(({ active, ...rest }) => rest);

https://jsfiddle.net/22kqjvc5/1/

answered Feb 3, 2017 at 3:19

Tom Coughlin

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1 Comment

Josh Crozier Over a year ago

1+ - Nice! That is a really clever usage of the spread operator.

bereket gebredingle · Accepted Answer · 2019-10-12 09:11:34Z

4

An ES6 way to do this

var users = [{
    'user': 'barney',
    'age': 36,
    'active': true
  }, {
    'user': 'pebbles',
    'age': 1,
    'active': true
}];

console.log ( _.map(users, i => _.pick(i, 'user', 'active'))  )

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.15/lodash.core.js"></script>

Inspired by this

answered Oct 12, 2019 at 9:11

bereket gebredingle

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Comments

1252748 · Accepted Answer · 2017-02-03 03:19:35Z

3

You don't need a library.

var result = users.map(el => {
   delete el.active;
   return el;
});

answered Feb 3, 2017 at 3:19

1252748

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Comments

Josh Crozier · Accepted Answer · 2017-02-03 03:12:13Z

3

You could use _.map in combination with _.omit in order to omit the active property:

Example Here

_.map(users, user => _.omit(user, 'active'))

answered Feb 3, 2017 at 3:12

Josh Crozier

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