Underscore的pick方法就是你要找的。

var obj = { "id": 'kl45wkfj1k4j34', "firstname": "Arun", "lastname": "K" };
var filter = 'firstname, lastname',
var result = _.pick(obj, filter.split(', '));

有很多方法可以做到这一点。到目前为止的答案假设您要修改现有对象，但问题未指定;"过滤器"这个词暗示可能不是。因此，如果要创建一个新的过滤对象，而不是改变现有对象，则可以使用reduce函数。你说你的键列表是一个字符串，但为了保持示例干净，让我们假设你只是做str.split(',')或类似操作，所以在将其传递给这些函数之前它已经是一个数组。

ES5

function createObjectFilter(keys) {
  return function(obj) {
    return keys.reduce(function(acc, key) {
      acc[key] = obj[key];
      return acc;
    }, {});
  };
}
var myFilter = createObjectFilter([ 'a', 'b' ]);
var filteredObject = myFilter(object);

ES6

const createObjectFilter = keys => obj => keys.reduce((acc, key) => {
  acc[key] = obj[key];
  return acc;
}, {});
const myFilter = createObjectFilter([ 'a', 'b' ]);
const filteredObject = myFilter(object);

现在，createObjectFilter 函数根据给定的键列表返回实际的筛选器函数。您可以改为"多合一"，但这种方法的优点是可以在更多情况下重用您的过滤器。例如：

const filteredObjects = unfilteredObjects.map(myFilter);

键/值对数组在BA中派上用场。

为此，您首先拆分字符串并映射对。

const
    object = { id: 'kl45wkfj1k4j34', firstname: "Arun", lastname: "K" },
    filter = 'firstname, lastname',
    keys = filter.split(', ');
    result = Object.fromEntries(keys.map(k => [k, object[k]]));
console.log(result);

var toBeFilteredObject = {...}; // {"id":"kl45wkfj1k4j34", "firstname":"Arun", "lastname":"K"}
var filteredObject = {};
'comma, seperated, string'.split(',').forEach(function(key) {
  key = key.trim();
  filteredObject[key] = toBeFilteredObject[key];
});

使用 _.reduce

var data = {"id":"kl45wkfj1k4j34", "firstname":"Arun", "lastname":"K"}
var filters = 'firstname, lastname'.split(', ');
_.reduce(data,function(result,value,key){
 if(filters.indexOf(key)!=-1){
   result[key] =value
 }
 return result;
},{})

如果你有这个对象：

 var obj = {"id":"kl45wkfj1k4j34", "firstname":"Arun", "lastname":"K"};  

然后你可以这样做：

delete obj.id;
console.log(obj);

现在根据评论：

var newlist = listarray.map(function (obj){
      delete obj.id;
       return obj;
});

这将创建一个没有id的新列表数组。

或使用特定键：

var newlist = listarray.map(function (obj){
       var o = {};
        o.firstname = obj.firstname;
        o.lastname = obj.lastname;
       return o;
});

如果您有数组，则更通用的解决方案

var obj = [{ "id": 'kl45wkfj1k4j34', "firstname": "Arun", "lastname": "K" }, { "id": '34234', "firstname": "kajshd", "lastname": "ajsdh" }, { "id": '263742', "firstname": "asdjasd", "lastname": "asdjahs" }],
    filter = 'firstname, lastname'.split(', '),
    result = {}; 
var output = [];
obj.forEach(function(i, j) {
  filter.forEach(function (k) {
    // console.log(obj[j]);
    result[k] = obj[j][k];
  }); 
  output.push(result); 
  result = {};
});

我已经将解决方案从@Nina调整为TypeScript，并对其进行了改进。

/**
 * Constructs a new object by picking certain properties of the given object.
 * Example: `const publicUser = pick(user, ['id', 'name'])`.
 * Adapted from https://stackoverflow.com/a/35667463/4034572.
 */
export function pick<T extends object, K extends keyof T>(
  object: T,
  keys: K[]
): Pick<T, K> {
  const entries = keys.filter((k) => k in object).map((k) => [k, object[k]])
  return Object.fromEntries(entries)
}

使用示例：

type User = {
  readonly id: number
  name: string
  email: string
  password: string
}
type PublicUser = Pick<User, 'id' | 'name'>
const user: User = {
  id: 2,
  name: 'Albert',
  email: 'a@a.co',
  password: 'pwd',
}
const publicUser: PublicUser = pick(user, ['id', 'name'])
// publicUser is { id: 2, name: 'Albert' }

好消息是 TypeScript 不允许您使用无效键。例如，执行pick(user, ['xyz'])会引发错误 TS2322：类型"xyz"不可分配给类型"keyof User"。

而且您也会获得自动完成功能:)

起色

我添加了filter((k) => k in object)，因为如果不这样做pick(user, ['xyz'])会给出{ xyz: undefined }，添加一个类型User上不存在的属性。

但是，对于filter，它正确地给出了{}，这是有意义的，因为属性xyz在User上不存在。

如果你有对象数组，那么使用这个

 data = [{
   "id": "kl45wkfj1k4j34",
   "firstname": "Arun1",
   "lastname": "K1"
 }, {
   "id": "kl45wkfj1k4j14",
   "firstname": "Arun2",
   "lastname": "K2"
 }, {
   "id": "2",
   "firstname": "Arun3",
   "lastname": "K3"
 }];
 data = data.map(function(o) {
   delete o.id;
   return o
 })
 document.write('<pre>' + JSON.stringify(data, 0, 4) + '</pre>');