Elegant way to copy only a part of an object [dupl

问题:

This question already has answers here:

One-liner to take some properties from object in ES 6 (11 answers)

Closed 2 years ago.

I would like to create an new object from a bigger one, by copying only a few properties from it. All the solutions I know are not very elegant, I wonder if there is a better choice, native if possible (no additional function like at the end of the following code)?

Here is what I usually do for now:

// I want to keep only x, y, and z properties:
let source = {
    x: 120,
    y: 200,
    z: 150,
    radius: 10,
    color: 'red',
};

// 1st method (not elegant, especially with even more properties):
let coords1 = {
    x: source.x,
    y: source.y,
    z: source.z,
};

// 2nd method (problem: it pollutes the current scope):
let {x, y, z} = source, coords2 = {x, y, z};

// 3rd method (quite hard to read for such simple task):
let coords3 = {};
for (let attr of ['x','y','z']) coords3[attr] = source[attr];

// Similar to the 3rd method, using a function:
function extract(src, ...props) {
    let obj = {};
    props.map(prop => obj[prop] = src[prop]);
    return obj;
}
let coords4 = extract(source, 'x', 'y', 'z');

回答1:

One way to do it is through object destructuring and an arrow function:

let source = {
    x: 120,
    y: 200,
    z: 150,
    radius: 10,
    color: 'red',
};

let result = (({ x, y, z }) => ({ x, y, z }))(source);

console.log(result);

The way this works is that the arrow function (({ x, y, z }) => ({ x, y, z })) is immediately called with source as the parameter. It destructures source into x, y, and z, and then immediately returns those as a new object.

回答2:

Just take a function.

const extract = ({ x, y, z }) => ({ x, y, z });

let source = { x: 120, y: 200, z: 150, radius: 10, color: 'red' };

console.log(extract(source));

Another solution is a destructuring to a target object with target properties.

let source = { x: 120, y: 200, z: 150, radius: 10, color: 'red' }, 
    target = {};

({ x: target.x, y: target.y, z: target.z } = source);

console.log(target);

回答3:

IIFE with destructuring maybe?:

const coords = (({x, y, z}) => ({x, y, z}))(source);

回答4:

You can do it like below via Spread Operator

let source = {
    x: 120,
    y: 200,
    z: 150,
    radius: 10,
    color: 'red',
};

let {radius, color, ...newObj} = source;
console.log(newObj);

回答5:

For simple cases like this the object destructuring mentioned in other answers is very neat but tends to look a bit cumbersome when dealing with larger structures as you double up on property names.

Expanding on your own answer - if you were going to write an extract utility (I'll roll my own for fun)... you can make it more flexible by currying it - allowing you to swap the order of the arguments (notably putting the data source last) while still being variadic in accepting property names.

I'd consider this signature: extract = (...props) => src => { ... } more elegant as it allows for a greater degree of reuse in composing new, named functions:

const extract = (...props) => src => 
    Object.entries(src).reduce(
        (obj, [key, val]) => (
            props.includes(key) && (obj[key] = val), 
            obj
    ), {})

const getCoords = extract('x', 'y', 'z')

const source = {
    x: 120,
    y: 200,
    z: 150,
    radius: 10,
    color: 'red'
}

console.log(getCoords(source))

回答6:

1st method is elegant and readable.

Please don't obfuscate simple operations by some workarounds. Other people who will need to maintain this code, including future yourself will be very thankful in the future.

回答7:

You can try reduce over [x,y,z] array:

let source = {
    x: 120,
    y: 200,
    z: 150,
    radius: 10,
    color: 'red',
};

const coords = ['x','y','z'].reduce((a,c) => Object.assign(a,{[c]: source[c]}), {});

console.log(coords);