I have an object:

var obj = { a: 'test1', b: 'test2', c: 'test3', d: 'test4', e: 'test5', f: 'test6', g: 'test7', h: 'test8' }

I want to get result:

res = { a: 'test1', c: 'test3', d: 'test4' }

What is the fastest way to do it?

I have an object:

var obj = { a: 'test1', b: 'test2', c: 'test3', d: 'test4', e: 'test5', f: 'test6', g: 'test7', h: 'test8' }

I want to get result:

res = { a: 'test1', c: 'test3', d: 'test4' }

What is the fastest way to do it?

• javascript

• use delete obj.e – Just code Commented Oct 19, 2018 at 8:03
• @Justcode Hmmm, you might want to test that. Particularly test accessing other properties on obj after you do it. :-) – T.J. Crowder Commented Oct 19, 2018 at 8:03
• @T.J.Crowder shouldn't be a problem with if you are dealing with 1 property. – Just code Commented Oct 19, 2018 at 8:40
• 1 @T.J.Crowder thank you, time to do some re-work :D – Just code Commented Oct 19, 2018 at 9:46
• 1 @Justcode :-) Of course, it's a bit micro-opt. (And that synthetic benchmark doesn't reveal the problem on Firefox; the delete right after creation gets optimized away I think.) It's just, the OP did specifically say "fastest"... ;-) – T.J. Crowder Commented Oct 19, 2018 at 10:02

2 Answers 2

Directly access the fields:

const res = {a: obj.a, c: obj.c, d: obj.d};

Live Example:

const obj = {
    a: "test1",
    b: "test2",
    c: "test3",
    d: "test4",
    e: "test5",
    f: "test6",
    g: "test7",
    h: "test8",
};
const res = { a: obj.a, c: obj.c, d: obj.d };
console.log(JSON.stringify(res, null, 4));

In a ment, Himanshu Agrawal asked:

What if the key is unknown and stored in a variable? const keys = ["a", "c", "d"];

I'd probably use a for-of loop to handle that:

const res = {};
for (const key of keys) {
    res[key] = obj[key];
}

Live Example:

const obj = {
    a: "test1",
    b: "test2",
    c: "test3",
    d: "test4",
    e: "test5",
    f: "test6",
    g: "test7",
    h: "test8",
};
const keys = ["a", "b", "c"];
const res = {};
for (const key of keys) {
    res[key] = obj[key];
}
console.log(JSON.stringify(res, null, 4));

But you could also use map and Object.fromEntries:

const res = Object.fromEntries(keys.map((key) => [key, obj[key]]));

Live Example:

const obj = {
    a: "test1",
    b: "test2",
    c: "test3",
    d: "test4",
    e: "test5",
    f: "test6",
    g: "test7",
    h: "test8",
};
const keys = ["a", "b", "c"];
const res = Object.fromEntries(keys.map((key) => [key, obj[key]]));
console.log(JSON.stringify(res, null, 4));

That said, the question asks for the fastest way to do it, and the map+Object.fromEntries approach involves several temporary object allocations and function calls. In most cases, it won't matter but the for-of is probably faster (depending on the degree of optimization the JavaScript engine does). Or a boring old-fashioned for loop might be faster still:

const res = {};
for (let n = 0; n < keys.length; ++n) {
    const key = keys[n];
    res[key] = obj[key];
}

Live Example:

const obj = {
    a: "test1",
    b: "test2",
    c: "test3",
    d: "test4",
    e: "test5",
    f: "test6",
    g: "test7",
    h: "test8",
};
const res = {};
for (let n = 0; n < keys.length; ++n) {
    const key = keys[n];
    res[key] = obj[key];
}
console.log(JSON.stringify(res, null, 4));

Again, it's unlikely to matter, but it's good to have multiple approaches for situations where it may.

i think you want to delete key-value pair from the object so for that here's the solution

delete obj[b];

delete obj[e];

or you can use lodash pick

var _ = require('lodash')
_.pick( obj, [a, c, d] )

or create a new Object

var final = {a: obj.a, c: obj.c, d: obj.d}