我想在这里提到一个非常好的策展:

pick-es2019.js

Object.fromEntries(
  Object.entries(obj)
  .filter(([key]) => ['whitelisted', 'keys'].includes(key))
);

pick-es2017.js

Object.entries(obj)
.filter(([key]) => ['whitelisted', 'keys'].includes(key))
.reduce((obj, [key, val]) => Object.assign(obj, { [key]: val }), {});

pick-es2015.js

Object.keys(obj)
.filter((key) => ['whitelisted', 'keys'].indexOf(key) >= 0)
.reduce((newObj, key) => Object.assign(newObj, { [key]: obj[key] }), {})

omit-es2019.js

Object.fromEntries(
  Object.entries(obj)
  .filter(([key]) => !['blacklisted', 'keys'].includes(key))
);

omit-es2017.js

Object.entries(obj)
.filter(([key]) => !['blacklisted', 'keys'].includes(key))
.reduce((obj, [key, val]) => Object.assign(obj, { [key]: val }), {});

omit-es2015.js

Object.keys(obj)
.filter((key) => ['blacklisted', 'keys'].indexOf(key) < 0)
.reduce((newObj, key) => Object.assign(newObj, { [key]: obj[key] }), {})

我建议看看Lodash;它有很多实用函数。

例如，pick()就是你要找的东西:

var subset = _.pick(elmo, ['color', 'height']);

小提琴

如果你想保留比你想删除的属性更多的属性，你可以使用rest参数语法:

const obj = {
  a:1,
  b:2,
  c:3,
  d:4
};
const { a, ...newObj } = obj;
console.log(newObj); // {b: 2, c: 3, d: 4}

虽然它有点啰嗦，但通过使用Array.prototype.reduce，你可以完成其他人在2年前推荐的underscore/lodash。

var subset = ['color', 'height'].reduce(function(o, k) { o[k] = elmo[k]; return o; }, {});

这种方法从另一方面解决了这个问题:而不是获取一个对象并将属性名传递给它来提取，而是获取一个属性名数组并将它们简化为一个新对象。

While it's more verbose in the simplest case, a callback here is pretty handy, since you can easily meet some common requirements, e.g. change the 'color' property to 'colour' on the new object, flatten arrays, etc. -- any of the things you need to do when receiving an object from one service/library and building a new object needed somewhere else. While underscore/lodash are excellent, well-implemented libs, this is my preferred approach for less vendor-reliance, and a simpler, more consistent approach when my subset-building logic gets more complex.

编辑:es7版本相同:

const subset = ['color', 'height'].reduce((a, e) => (a[e] = elmo[e], a), {});

编辑:这也是咖喱的一个很好的例子!让'pick'函数返回另一个函数。

const pick = (...props) => o => props.reduce((a, e) => ({ ...a, [e]: o[e] }), {});

上面的方法非常接近于另一种方法，除了它让你在运行中构建一个“picker”。如。

pick('color', 'height')(elmo);

这种方法的特别之处在于，你可以很容易地将所选的“picks”传递到任何带有函数的对象中，例如Array#map:

[elmo, grover, bigBird].map(pick('color', 'height'));
// [
//   { color: 'red', height: 'short' },
//   { color: 'blue', height: 'medium' },
//   { color: 'yellow', height: 'tall' },
// ]

我知道它不是最干净的，但它简单易懂。

function obj_multi_select(obj, keys){
    let return_obj = {};
    for (let k = 0; k < keys.length; k++){
        return_obj[keys[k]] = obj[keys[k]];
    };
    return return_obj;
};

我发现最简单的方法，它不会创建不必要的变量，是一个函数，你可以调用，工作原理与lodash相同，如下所示:

pick(obj, keys){
    return  Object.assign({}, ...keys.map(key => ({ [key]: obj[key] })))
}

例如:

pick(obj, keys){
    return  Object.assign({}, ...keys.map(key => ({ [key]: obj[key] })))
}
const obj = {a:1, b:2, c:3, d:4}
const keys = ['a', 'c', 'f']
const picked = pick(obj,keys)
console.log(picked)

Pick = (obj, keys) => { 返回对象。分配({},…键。Map (key => ({ (例子):obj(例子) }))) ｝ Const obj = { 答:1, b: 2 c: 3, d: 4 ｝ Const keys = ['a'， 'c'， 'f'] Const selected = pick(obj, keys) console.log(选)