基于多个唯一值进行过滤

Question

我正在尝试确定以下最佳方法。

我有两个 arrays，如下所示：

const a = [
{
  Employee: "00001",
  Seq: "01",
  ReportDate: "2021-03-10T00:00:00.000",
  ReportName: "test report",
},
{
  Employee: "00002",
  Seq: "01",
  ReportDate: "2020-03-10T00:00:00.000",
  ReportName: "test report 2",
},
];


const b = [
{
  Employee: "00001",
  EmplName: "William Apple",
  ReportDate: "2021-03-10T00:00:00.000",
  ReportName: "test report",
  Posted: "Y",
  Seq: "01",
},
{
  Employee: "00003",
  EmplName: "John Smith",
  ReportDate: "2021-03-10T00:00:00.000",
  ReportName: "test report 3",
  Posted: "Y",
  Seq: "01",
},
];

我想过滤掉数组'b'中的任何内容，它不包含数组'a'中的一个对象中的所有4个确切值。

所以，根据上面的例子，我会留下：

// from array 'b'
const c = [
{
  Employee: "00001",
  EmplName: "William Apple",
  ReportDate: "2021-03-10T00:00:00.000",
  ReportName: "test report",
  Posted: "Y",
  Seq: "01",
}
];

我已经研究了各种方法来做到这一点。 我不能只检查对象是否相等，因为数组“a”将仅包含 4 个值/键，而数组“b”将包含更多/动态数量。

通过循环和多个 IF 语句使用 Filter 方法是最好的方法吗？

Answer 1

您可以使用.reduce遍历a的项目并将它们存储在Map中，其中Employee是key ，其他属性（Seq、ReportDate、ReportName）是value 。

之后，使用.filter从b中获取他们的Employee在 map 中存在的项目，并且其他三个属性匹配。

 const a = [ { Employee: "00001", Seq: "01", ReportDate: "2021-03-10T00:00:00.000", ReportName: "test report" }, { Employee: "00002", Seq: "01", ReportDate: "2020-03-10T00:00:00.000", ReportName: "test report 2" } ]; const b = [ { Employee: "00001", EmplName: "William Apple", ReportDate: "2021-03-10T00:00:00.000", ReportName: "test report", Posted: "Y", Seq: "01" }, { Employee: "00003", EmplName: "John Smith", ReportDate: "2021-03-10T00:00:00.000", ReportName: "test report 3", Posted: "Y", Seq: "01" } ]; const employeesMap = a.reduce((map,employee) => { const { Employee, ...attributes } = employee; map.set(Employee, attributes); return map; }, new Map); const res = b.filter(employee => { const { Employee, Seq, ReportDate, ReportName } = employee; const employeeFromA = employeesMap.get(Employee); return employeeFromA && employeeFromA.Seq === Seq && employeeFromA.ReportDate === ReportDate && employeeFromA.ReportName === ReportName; }); console.log(res);

Answer 2

我会从循环代码中分离出精确 object 匹配的测试，如下所示：

 const subObject = (a, b) => Object.entries (a).every (([k, v]) => b [k] == v) const matches = (xs, ys) => ys.filter (y => xs.some (x => subObject (x, y))) const a = [{Employee: "00001", Seq: "01", ReportDate: "2021-03-10T00:00:00.000", ReportName: "test report"}, {Employee: "00002", Seq: "01", ReportDate: "2020-03-10T00:00:00.000", ReportName: "test report 2"}] const b = [{Employee: "00001", EmplName: "William Apple", ReportDate: "2021-03-10T00:00:00.000", ReportName: "test report", Posted: "Y", Seq: "01"}, {Employee: "00003", EmplName: "John Smith", ReportDate: "2021-03-10T00:00:00.000", ReportName: "test report 3", Posted: "Y", Seq: "01"}] console.log (matches (a, b))

subObject检查a的每个属性是否在b中具有相同的值。

matches使用它根据第一个列表是否有某个元素是它的子对象来过滤第二个列表。

两者都是非常简单的功能。 虽然我们可以在matches内联subObject ，但我认为它本身很有用，并且可以生成更好的代码。

Answer 3

const results = b.filter(bv => !!a.find(av => Object.keys(av).filter(ak => av[ak] === bv[ak]).length === Object.keys(av).length));

console.log(results);

这会将“b”数组过滤到仅在“a”中也有匹配的那些元素。 “匹配”在此由“a”项中的所有键/值与“b”项中的所有相应键/值匹配来确定。 “b”项中无关的键被忽略。