Question

我有一个对象数组（objList），每个对象都具有“ id”属性。

我有一个字符串数组（idsToRemove），表示要从objList中删除的对象的ID。

我找到了一些解决方案，但我担心它很慢，尤其是对于具有许多属性的大量对象而言。有更有效的方法吗？

var idsToRemove = ["3", "1"];
var objList = [{
    id: "1",
    name: "aaa"
  },
  {
    id: "2",
    name: "bbb"
  },
  {
    id: "3",
    name: "ccc"
  }
];

for (var i = 0, len = idsToRemove.length; i < len; i++) {
  objList = objList.filter(o => o.id != idsToRemove[i]);
}

console.log(objList);

Answer 1

将

namespace App\Http\Controllers;



use App\Employee;
use Illuminate\Http\Request;

class EmployeeController extends Controller
{

public function create()
{
    return view('addemployee');
}

public function store(Request $request)
{
    $employee = new Employee();
    $employee->firstName = $request->get('firstName');
    $employee->middleName = $request->get('middleName');
    $employee->lastName = $request->get('lastName');
    $employee->gender = $request->get('gender');
    $employee->city = $request->get('city');
    $employee->localAddress = $request->get('localAddress');
    $employee->permanentAddress = $request->get('permanentAddress');
    $employee->emailAddress = $request->get('emailAddress');
    $employee->mobileNumber = $request->get('mobileNumber');
    $employee->email = $request->get('email');
    $employee->department = $request->get('department');
    $employee->designation = $request->get('designation');
    $employee->save();

    return redirect('addemployee')->with('Success', 'Data Inserted Successfully!');

}

转换为idsToRemove，以便可以使用Set（一项Set.prototype.has操作）和O(1) .filter一次，因此总体复杂度为objList（并且您仅对可能巨大的O(n)进行了一次迭代）：

objList

请注意，var idsToRemove = ["3", "1"]; var objList = [{ id: "1", name: "aaa" }, { id: "2", name: "bbb" }, { id: "3", name: "ccc" } ]; const set = new Set(idsToRemove); const filtered = objList.filter(({ id }) => !set.has(id)); console.log(filtered);和Array.prototype.includes操作是Array.prototype.indexOf，而不是O(N)，因此，如果使用它们代替O(1)，它们可能会花费更长的时间

Answer 2

您可以使用Array.includes来检查给定的字符串是否存在于给定的数组中，并将其与Array.filter组合。

const idsToRemove = ['3', '1'];

const objList = [{
    id: '1',
    name: 'aaa',
  },
  {
    id: '2',
    name: 'bbb',
  },
  {
    id: '3',
    name: 'ccc',
  },
];

const filteredObjList = objList.filter(x => !idsToRemove.includes(x.id));

console.log(filteredObjList);

Answer 3

如果使用内置的查找功能，则不需要两个嵌套的迭代器

   objList = objList.filter(o => idsToRemove.indexOf(o.id) < 0);

文档：

Array.prototype.indexOf()

Array.prototype.includes()

Answer 4

只需使用Array.filter()

const idsToRemove = ['3', '1'];

const objList = [{
    id: '1',
    name: 'aaa',
  },
  {
    id: '2',
    name: 'bbb',
  },
  {
    id: '3',
    name: 'ccc',
  }
];

const res = objList.filter(value => !idsToRemove.includes(value.id));

console.log("result",res);

根据对象的某些属性的数组从数组中删除对象

4 个答案: