In Javascript remove keys from object not in an Array

Question

Suppose I have a list of objects with many keys and I want to keep only certain keys from them. This is how I am doing it.

The Problem with other good solutions on SO are that if a key is not present in the keys to keep it still adds a key, value where the value is undefined.

let data = [{
   'a': 1,
   'b': 2,
   'c': 3
 }, 
 {
   'a': 1,
   'c': 3,
   'd': 4
 }]

const keys_to_keep = ['a', 'b']

data = data.map((obj) => {
  Object.keys(obj).forEach(function(key) {
    if(!keys_to_keep.includes(key))
      delete obj[key]
  });
  return obj;
})

Output :

[ { a: 1, b: 2 }, { a: 1} ]

Is there a better way to get this done. Any help is appreciated.

What problems are you having with the way you're doing it now? — Pointy
– Pointy, Commented Sep 9, 2020 at 23:52
If you happen to already be using lodash, see stackoverflow.com/questions/30726830/… — jarmod
– jarmod, Commented Sep 9, 2020 at 23:59
@epascarello You can run the code on the SO link you shared with my data. It gives me [ { a: 1, b: 2 }, { a: 1, b: undefined } ], whereas I wanted [ { a: 1, b: 2 }, { a: 1} ] . That is exactly what I have explained. — Akshay Hazari
– Akshay Hazari, Commented Oct 11, 2020 at 4:15

jfriend00 · Accepted Answer · 2020-09-10 00:03:11Z

10

A couple of improvements.

You're using .map() which creates a new array, but then you're just assigning it to the old variable. So, you apparently don't need to create a new array at all, you can just iterate the one you have and modify it.
Put the properties you want to keep in a Set instead of an Array for potentially faster lookup.
for/of loops are generally favored over .forEach() loops because of better flow control options (break, continue, return, etc...) and more opportunities for compiler optimization.

let kv = [{
   'a': 1,
   'b': 2,
   'c': 3
 }, 
 {
   'a': 1,
   'b': 2,
   'c': 3,
   'd': 4
 }]

const l = new Set(['a', 'b']);

for (let obj of kv) {
    for (let prop of Object.keys(obj)) {
       if (!l.has(prop)) {
           delete obj[prop];
       }
    }
}

console.log(kv);

edited Sep 10, 2020 at 0:03

answered Sep 9, 2020 at 23:57

jfriend00

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trincot · Accepted Answer · 2022-02-09 21:00:32Z

3

You can use Object.fromEntries, after map and filter to keep only the relevant keys:

let data = [{'a': 1,'b': 2,'c': 3},{'a': 1,'c': 3,'d': 4}]
const keys_to_keep = ['a', 'b']

var result = data.map(obj =>
    Object.fromEntries(keys_to_keep.map(key => 
        obj.hasOwnProperty(key) && [key, obj[key]]
    ).filter(Boolean))
);

console.log(result);

answered Feb 9, 2022 at 21:00

trincot

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