在 Javascript 中，从 object 中删除不在数组中的键

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.

Answer 1

A couple of improvements.

1. 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.

2. Put the properties you want to keep in a Set instead of an Array for potentially faster lookup.

3. 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);

Answer 2

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);