基于javascript中的两个属性获取数组中重复对象的列表

Question

我正在尝试根据 2 个属性在数组中获取重复的对象。假设对象如下所示。

let arry = [
    {Level: "A-1", Status: "approved"},
    {Level: "A-2", Status: "approved"},
    {Level: "A-3", Status: "approved"},
    {Level: "A-4", Status: "reject"},
    {Level: "A-5", Status: "reject"},
    {Level: "A-1", Status: "approved"},
    {Level: "A-2", Status: "approved"},
    {Level: "A-6", Status: "approved"},
    {Level: "A-7", Status: "reject"},
    {Level: "A-1", Status: "approved"},
    {Level: "A-6", Status: "approved"},
]

应该返回重复的对象，并且只有一次“已批准”状态，如下所示：

[
{Level: "A-1", Status: "approved"},
 {Level: "A-2", Status: "approved"}
]

Answer 1

您可以使用 Array.reduce 在遍历数组时跟踪所有之前的项目，同时只保留之前看到的那些（Status 是 "approved"）。

为了不输出重复的重复，它还会检查当前重复是否已经被找到为重复，如果是则忽略它。

const array = [
  { Level: "A-1", Status: "approved" },
  { Level: "A-2", Status: "approved" },
  { Level: "A-3", Status: "approved" },
  { Level: "A-4", Status: "reject" },
  { Level: "A-5", Status: "reject" },
  { Level: "A-1", Status: "approved" },
  { Level: "A-2", Status: "approved" },
  { Level: "A-6", Status: "approved" },
  { Level: "A-7", Status: "reject" },
  { Level: "A-1", Status: "approved" },
  { Level: "A-6", Status: "approved" },
];

const dupes = array.reduce((dupes, { Level, Status }, i) => {
  const matches = ({ Level: l, Status: s }) => l == Level && s == Status;
  const match = dupes.memory.find(matches);
  const found = dupes.dupes.find(matches);
  (match?.Status == 'approved' && !found)
    ? dupes.dupes.push({ Level, Status })
    : dupes.memory.push({ Level, Status });
  return (i == array.length - 1) ? dupes.dupes : dupes;
}, { dupes: [], memory: [] });

console.log(dupes);

一个不太通用但更简洁的策略可能是：

const dupes = [...new Set(array
  .filter(({ Status: s }) => s == 'approved')
  .map(({ Level: l }) => +l.match(/\d+/)[0])
  .filter((num, _, arr) => arr.filter((n) => n == num).length > 1))]
  .map((n) => ({ Level: 'A-' + n, Status: 'approved' }));

例如：

const array = [
  { Level: "A-1", Status: "approved" },
  { Level: "A-2", Status: "approved" },
  { Level: "A-3", Status: "approved" },
  { Level: "A-4", Status: "reject" },
  { Level: "A-5", Status: "reject" },
  { Level: "A-1", Status: "approved" },
  { Level: "A-2", Status: "approved" },
  { Level: "A-6", Status: "approved" },
  { Level: "A-7", Status: "reject" },
  { Level: "A-1", Status: "approved" },
  { Level: "A-6", Status: "approved" },
];

const dupes = [...new Set(array
  .filter(({ Status: s }) => s == 'approved')
  .map(({ Level: l }) => +l.match(/\d+/)[0])
  .filter((num, _, arr) => arr.filter((n) => n == num).length > 1))]
  .map((n) => ({ Level: 'A-' + n, Status: 'approved' }));

console.log(dupes);

Answer 2

let arry = [
    {Level: "A-1", Status: "approved"},
    {Level: "A-2", Status: "approved"},
    {Level: "A-3", Status: "approved"},
    {Level: "A-4", Status: "reject"},
    {Level: "A-5", Status: "reject"},
    {Level: "A-1", Status: "approved"},
    {Level: "A-2", Status: "approved"},
    {Level: "A-6", Status: "approved"},
    {Level: "A-7", Status: "reject"},
    {Level: "A-1", Status: "approved"},
    {Level: "A-6", Status: "approved"},
]

function dups(arry) {
  const hashMap = {};

  return arry.reduce((acc, curr) => {
    const level = curr.Level;
    if(hashMap[level] && hashMap[level] === 1 && curr.Status === 'approved') {
      acc.push(curr);
      hashMap[level]++;
    } else {
      hashMap[level] = 1;
    }

    return acc;
  }, []);
}

console.log(dups(arry));