合并和轉換物件中的鍵和陣列的最佳方法

我有一個看起來像這樣的物件：

{ role1: ["id1", "id2", "id3"], role2: ["id1", "id2"], role3: ["id2", "id4"] }

我想將其轉換為物件陣列，使 ids 成為唯一識別符號，并將角色轉換為陣列，如下所示：

[
  { id: "id1", roles: ["role1", "role2"] },
  { id: "id2", roles: ["role1", "role2", "role3"] },
  { id: "id3", roles: ["role1"] },
  { id: "id4", roles: ["role3"] },
]

我現在就是這樣做的，但我不確定是否有更好的方法。感覺好像我把事情復雜化了。

const obj = {
  role1: ["id1", "id2", "id3"],
  role2: ["id1", "id2"],
  role3: ["id2", "id4"]
};
const users = [];

Object.entries(obj).forEach(([key, ids]) => {
  ids.forEach(id => {
    const user = users.find(x => x.id === id);
    if (user) {
      user.roles.push(key);
      return;
    }
    users.push({
      id,
      roles: [key]
    });
  });
});

console.log(users);

uj5u.com熱心網友回復：

您可能只需將其轉換為如下物件：

const obj = {
  role1: ["id1", "id2", "id3"],
  role2: ["id1", "id2"],
  role3: ["id2", "id4"]
};
let userDict = {};

Object.entries(obj).forEach(([key, ids]) => {
  ids.forEach(id => {
    userDict[id] = (userDict[id] || []);
    userDict[id].push(key);
  });
});

let users = Object.entries(userDict).map(([id, roles]) => ({ id, roles }));

console.log(users);

.as-console-wrapper { max-height: 100% !important; top: auto; }

甚至reduce像這樣使用：

顯示代碼片段

const obj = {
  role1: ["id1", "id2", "id3"],
  role2: ["id1", "id2"],
  role3: ["id2", "id4"]
};
let userDict = Object.entries(obj).reduce((acc, [role, ids]) => {
  ids.forEach(id => {
    acc[id] = (acc[id] || []);
    acc[id].push(role);
  });
  return acc;
}, {});

let users = Object.entries(userDict).map(([id, roles]) => ({ id, roles }));

console.log(users);

.as-console-wrapper { max-height: 100% !important; top: auto; }

任何一個選項都比一直使用陣列更簡潔和簡單。

uj5u.com熱心網友回復：

首先，我通過使用Set（以避免重復）和flat從該物件中獲取所有 id

const obj = {
  role1: ["id1", "id2", "id3"],
  role2: ["id1", "id2"],
  role3: ["id2", "id4"],
};
const objsId = [
  ...new Set(
    Object.values(obj) //[["id1","id2","id3"],["id1","id2"],["id2","id4"]]
      .flat() //['id1', 'id2', 'id3', 'id1', 'id2', 'id2', 'id4']
  ),
]; 
console.log(objsId); //['id1', 'id2', 'id3', 'id4']

第二件事是通過使用從該陣列創建一個物件Object.fromEntries

方法將鍵值對串列轉換為物件。

const objsId = ['id1', 'id2', 'id3', 'id4'];
const idRoles = Object.fromEntries(
  objsId.map((id) => [id, []]) //[["id1",[]],["id2",[]],["id3",[]],["id4",[]]]
); 
console.log(idRoles) //{"id1":[],"id2":[],"id3":[],"id4":[]}

最后回圈遍歷我們的陣列 objsId和每個 id，使用Object.fromEntries() 和for..of回圈遍歷您的物件 obj條目 如果當前 id 存在于角色中，那么將影響該角色添加到idRoles

const obj = {
  role1: ["id1", "id2", "id3"],
  role2: ["id1", "id2"],
  role3: ["id2", "id4"],
};
const objsId = ['id1', 'id2', 'id3', 'id4'];
const idRoles = { id1: [], id2: [], id3: [], id4: [] };
for (const id of objsId) {
  for (const [role, ids] of Object.entries(obj)) {
    if (ids.includes(id)) idRoles[id].push(role);
  }
}
console.log(idRoles);

顯示代碼片段

const obj = {
  role1: ["id1", "id2", "id3"],
  role2: ["id1", "id2"],
  role3: ["id2", "id4"],
};
const objsId = [
  ...new Set(
    Object.values(obj) //[["id1","id2","id3"],["id1","id2"],["id2","id4"]]
      .flat() //['id1', 'id2', 'id3', 'id1', 'id2', 'id2', 'id4']
  ),
]; //['id1', 'id2', 'id3', 'id4']

const idRoles = Object.fromEntries(
  objsId.map((id) => [id, []]) //[["id1",[]],["id2",[]],["id3",[]],["id4",[]]]
); //{"id1":[],"id2":[],"id3":[],"id4":[]}

for (const id of objsId) {
  for (const [role, ids] of Object.entries(obj)) {
    if (ids.includes(id)) idRoles[id].push(role);
  }
}
const result = [];
for (let [id, roles] of Object.entries(idRoles)) {
  result.push({ id, roles });
}
console.log(JSON.stringify(result, null, 4));

uj5u.com熱心網友回復：

只是 Lodash 庫的另一個版本

const obj = {
  role1: ["id1", "id2", "id3"],
  role2: ["id1", "id2"],
  role3: ["id2", "id4"]
};

const users = _.chain(obj)
  .entries()
  .reduce((accm, [role, ids]) => {
    ids.forEach((id) => (accm[id] || (accm[id] = [])).push(role));
    return accm;
  }, {})
  .entries()
  .map(([id, roles]) => ({ id, roles }));

標籤：javascript 数组 目的 合并

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