161

How one can write a function, which takes only few attributes in most-compact way in ES6?

I've came up with solution using destructuring + simplified object literal, but I don't like that list of fields is repeated in the code.

Is there an even slimmer solution?

(v) => {
    let { id, title } = v;
    return { id, title };
}
Michał Perłakowski
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kirilloid
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11 Answers11

135

Here's something slimmer, although it doesn't avoid repeating the list of fields. It uses "parameter destructuring" to avoid the need for the v parameter.

({id, title}) => ({id, title})

(See a runnable example in this other answer).

@EthanBrown's solution is more general. Here is a more idiomatic version of it which uses Object.assign, and computed properties (the [p] part):

function pick(o, ...props) {
    return Object.assign({}, ...props.map(prop => ({[prop]: o[prop]})));
}

If we want to preserve the properties' attributes, such as configurable and getters and setters, while also omitting non-enumerable properties, then:

function pick(o, ...props) {
    var has = p => o.propertyIsEnumerable(p),
        get = p => Object.getOwnPropertyDescriptor(o, p);

    return Object.defineProperties({},
        Object.assign({}, ...props
            .filter(prop => has(prop))
            .map(prop => ({prop: get(props)})))
    );
}
Dan Dascalescu
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    +1 nice answer, torazaburo; thanks for making me aware of `Object.assign`; es6 is like a Christmas tree with so many presents under it I'm still finding gifts months after the holiday – Ethan Brown May 27 '15 at 20:57
  • Got an error: Property description must be an object: undefined. Shouldn't it be `filter(...).map(prop => ({[prop]: get(prop)})))`? – Endless Nov 22 '17 at 03:20
  • For your first `pick()` implementation you could also do something like `return props.reduce((r, prop) => (r[prop] = o[prop], r), {})` – Patrick Roberts Jan 31 '18 at 19:47
  • unfortunately that version of pick won't be type safe in flow or typescript. if you want type safety, there's no way around destructure assignment of original object, then assigning each into a new object. – duhseekoh Mar 07 '18 at 23:08
  • When a property doesn't exist in an object, you get `undefined`. Sometimes it matters. Other than that, nice idea. – x-yuri Feb 10 '19 at 18:09
  • `lodash` doesn't have this peculiarity. I think I'll go with it. – x-yuri Feb 10 '19 at 20:44
  • FYI a typescript variant of this is: export function pick(o: T, ...props: D[]) { return Object.assign({}, ...props.map(prop => ({[prop]: o[prop]}))); } – Guy Gascoigne-Piggford Sep 20 '20 at 21:34
44

I don't think there's any way to make it much more compact than your answer (or torazburo's), but essentially what you're trying to do is emulate Underscore's pick operation. It would be easy enough to re-implement that in ES6:

function pick(o, ...fields) {
    return fields.reduce((a, x) => {
        if(o.hasOwnProperty(x)) a[x] = o[x];
        return a;
    }, {});
}

Then you have a handy re-usable function:

var stuff = { name: 'Thing', color: 'blue', age: 17 };
var picked = pick(stuff, 'name', 'age');
Ethan Brown
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  • Thanks. This is not an answer for my question, but very nice addition. – kirilloid Sep 18 '14 at 11:48
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    (shrug) I feel like it _is_ an answer for your solution; there is no slimmer _general_ solution (torazaburo's solution removes from extra verbage, but the essential problem -- that all property names have to be written twice -- means it doesn't scale any better than your solution). My solution at least scales well...right the `pick` function once, and you can pick as many properties you want and it won't double them. – Ethan Brown Sep 18 '14 at 15:58
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    Why do you use `hasOwnProperty`? If the fields are hand-selected, even `in` seems to be more appropriate; although I'd go for omitting the check completely and just let them default to `undefined`. – Bergi Nov 24 '15 at 00:19
  • Bergi, it's a reasonable point...I just consider properties (not methods) on a prototype chain to be weird and "smelly" (as in they are a code smell), and I prefer to filter them out by default. If there's an application that needs prototype properties, well...there can be an option for that. – Ethan Brown Nov 24 '15 at 03:34
  • what about json arrays ! – Rizwan Patel Sep 14 '17 at 12:48
22

The trick to solving this as a one-liner is to flip the approach taken: Instead of starting from original object orig, one can start from the keys they want to extract.

Using Array#reduce one can then store each needed key on the empty object which is passed in as the initialValue for said function.

Like so:

const orig = {
  id: 123456789,
  name: 'test',
  description: '…',
  url: 'https://…',
};

const filtered = ['id', 'name'].reduce((result, key) => { result[key] = orig[key]; return result; }, {});

console.log(filtered); // Object {id: 123456789, name: "test"}

alternatively...

const filtered = ['id', 'name'].reduce((result, key) => ({
    ...result, 
    [key]: orig[key] 
}), {});

console.log(filtered); // Object {id: 123456789, name: "test"}
Community
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Bramus
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14

A tiny bit shorter solution using the comma operator:

const pick = (O, ...K) => K.reduce((o, k) => (o[k]=O[k], o), {})

console.log(
  pick({ name: 'John', age: 29, height: 198 }, 'name', 'age')
)  
mplungjan
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shesek
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8

TC39's object rest/spread properties proposal will make this pretty slick:

let { x, y, ...z } = { x: 1, y: 2, a: 3, b: 4 };
z; // { a: 3, b: 4 }

(It does have the downside of creating the x and y variables which you may not need.)

alxndr
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4

You can use object destructuring to unpack properties from the existing object and assign them to variables with different names - fields of a new, initially empty object.

const person = {
  fname: 'tom',
  lname: 'jerry',
  aage: 100,
}

let newPerson = {};

({fname: newPerson.fname, lname: newPerson.lname} = person);

console.log(newPerson);
Dan Dascalescu
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Saksham
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4

ES6 was the latest spec at the time when the question was written. As explained in this answer, key picking is significantly shorter in ES2019 than in ES6:

Object.fromEntries(
  Object.entries(obj)
  .filter(([key]) => ['foo', 'bar'].includes(key))
)
Estus Flask
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2

There's currently a strawman proposal for improving JavaScript's object shorthand syntax, which would enable "picking" of named properties without repetition:

const source = {id: "68646", genre: "crime", title: "Scarface"};
const target = {};
Object.assign(target, {source.title, source.id});

console.log(picked);
// {id: "68646", title: "Scarface"}

Unfortunately, the proposal doesn't seem to be going anywhere any time soon. Last edited in July 2017 and still a draft at Stage 0, suggesting the author may have ditched or forgotten about it.

ES5 and earlier (non-strict mode)

The concisest possible shorthand I can think of involves an ancient language feature nobody uses anymore:

Object.assign(target, {...(o => {
    with(o) return { id, title };
})(source)});

with statements are forbidden in strict mode, making this approach useless for 99.999% of modern JavaScript. Bit of a shame, because this is the only halfway-decent use I've found for the with feature.

1

I have similar to Ethan Brown's solution, but even shorter - pick function. Another function pick2 is a bit longer (and slower), but allows to rename properties in the similar to ES6 manner.

const pick = (o, ...props) => props.reduce((r, p) => p in o ? {...r, [p]: o[p]} : r, {})

const pick2 = (o, ...props) => props.reduce((r, expr) => {
  const [p, np] = expr.split(":").map( e => e.trim() )
  return p in o ? {...r, [np || p]: o[p]} : r
}, {}) 

Here is the usage example:

const d = { a: "1", c: "2" }

console.log(pick(d, "a", "b", "c"))        // -> { a: "1", c: "2" }
console.log(pick2(d, "a: x", "b: y", "c")) // -> { x: "1", c: "2" }
0

I required this sollution but I didn't knew if the proposed keys were available. So, I took @torazaburo answer and improved for my use case:

function pick(o, ...props) {
  return Object.assign({}, ...props.map(prop => {
    if (o[prop]) return {[prop]: o[prop]};
  }));
}

// Example:
var person = { name: 'John', age: 29 };
var myObj = pick(person, 'name', 'sex'); // { name: 'John' }
Alwin Kesler
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-1

inspired by the reduce approach of https://stackoverflow.com/users/865693/shesek:

const pick = (orig, ...keys) => keys.reduce((acc, key) => ({...acc, [key]: orig[key]}), {})

usage:

pick({ model : 'F40', manufacturer: 'Ferrari', productionYear: 1987 }, 'model', 'productionYear') results in: {model: "F40", productionYear: 1987}

Kevin K.
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