如何在JS中將兩個陣列與物件進行比較期間找到相似項的索引

我有 2 個帶有物件的陣列。

const users1 = [
    { name: 'John', age: 15, gender: 'M', city: 'London', country: 'UK' },
    { name: 'Jack', age: 20, gender: 'M', city: 'New York', country: 'USA' },
    { name: 'Jinny', age: 30, gender: 'F', city: 'London', country: 'UK' },
    { name: 'Key', age: 15, gender: 'M', city: 'Leeds', country: 'UK' },
  ],

const users2 = [
        { name: 'Jack', age: 20, gender: 'M', city: 'New York', country: 'USA' },
        { name: 'Key', age: 15, gender: 'M', city: 'Leeds', country: 'UK' },
      ],

我需要從users1中包含的那些部分獲取索引users2。

結果我應該得到[{1: true}, {3: true}]. 我以這種方式嘗試，但它不正確

const arr = user1.filter((item, index) => user2.map(i) => {
    if(i.id === item.id) return {index: true}
     return null;
    })

uj5u.com熱心網友回復：

Set您可以從所有 id 中創建一個users2（一個 Set 有點像一個陣列，但允許您快速檢查一個值是否在其中使用.has()并且它只存盤唯一值）。獲得集合后，您可以使用.reduce()創建一個新陣列，.concat()如果集合具有來自的物件的 id，則可以在其中使用users1：

const users1 = [{ id: 1, name: 'John', age: 15, gender: 'M', city: 'London', country: 'UK' }, { id: 2, name: 'Jack', age: 20, gender: 'M', city: 'New York', country: 'USA' }, { id: 3, name: 'Jinny', age: 30, gender: 'F', city: 'London', country: 'UK' }, { id: 4, name: 'Key', age: 15, gender: 'M', city: 'Leeds', country: 'UK' }, ];
const users2 = [{ id: 2, name: 'Jack', age: 20, gender: 'M', city: 'New York', country: 'USA' }, { id: 4, name: 'Key', age: 15, gender: 'M', city: 'Leeds', country: 'UK' }, ];

const u2Ids = new Set(users2.map(user => user.id));
const res = users1.reduce((acc, obj, i) => u2Ids.has(obj.id) ? acc.concat({[i]: true}) : acc, []);
console.log(res);

創建物件時，您需要確保在將索引添加為鍵時使用計算屬性名稱{[i]: true}，以便使用值 ati而不是文字i名稱。上面的代碼也可以用for..of回圈（通過訪問Array.prototype.entries()）來撰寫，這樣更有效并且（在我看來）更具可讀性：

const users1 = [{ id: 1, name: 'John', age: 15, gender: 'M', city: 'London', country: 'UK' }, { id: 2, name: 'Jack', age: 20, gender: 'M', city: 'New York', country: 'USA' }, { id: 3, name: 'Jinny', age: 30, gender: 'F', city: 'London', country: 'UK' }, { id: 4, name: 'Key', age: 15, gender: 'M', city: 'Leeds', country: 'UK' }, ];
const users2 = [{ id: 2, name: 'Jack', age: 20, gender: 'M', city: 'New York', country: 'USA' }, { id: 4, name: 'Key', age: 15, gender: 'M', city: 'Leeds', country: 'UK' }, ];

const res = [];
const u2Ids = new Set(users2.map(user => user.id));
for(const [i, obj] of users1.entries()) {
  if(u2Ids.has(obj.id))
    res.push({[i]: true});
}
console.log(res);

uj5u.com熱心網友回復：

首先，您的陣列不包含我現在過濾的名稱的 id。如果您使用過濾器將只過濾您的陣列并且索引將不再相同，那么過濾器將在 bcs 中不起作用，請嘗試以下操作

const users1 = [
    { name: 'John', age: 15, gender: 'M', city: 'London', country: 'UK' },
    { name: 'Jack', age: 20, gender: 'M', city: 'New York', country: 'USA' },
    { name: 'Jinny', age: 30, gender: 'F', city: 'London', country: 'UK' },
    { name: 'Key', age: 15, gender: 'M', city: 'Leeds', country: 'UK' }
  ];

const users2 = [
        { name: 'Jack', age: 20, gender: 'M', city: 'New York', country: 'USA' },
        { name: 'Key', age: 15, gender: 'M', city: 'Leeds', country: 'UK' }
      ];

const arr = [];
users1.forEach((item, idx) =>{
    if(users2.some(u=> u.name == item.name)){
    arr.push({[idx]:true})
    }
    });
    console.log(arr)

uj5u.com熱心網友回復：

只是另一種方法 - 與Array.reduce()

const users1 = [
  { name: 'John', age: 15, gender: 'M', city: 'London', country: 'UK' },
  { name: 'Jack', age: 20, gender: 'M', city: 'New York', country: 'USA' },
  { name: 'Jinny', age: 30, gender: 'F', city: 'London', country: 'UK' },
  { name: 'Key', age: 15, gender: 'M', city: 'Leeds', country: 'UK' },
];

const users2 = [
  { name: 'Jack', age: 20, gender: 'M', city: 'New York', country: 'USA' },
  { name: 'Key', age: 15, gender: 'M', city: 'Leeds', country: 'UK' },
];

const arr = users1.
  reduce((previousValue, currentValue, index) => {
    // change here to find by id 
    if (users2.some(u2 => u2.name === currentValue.name) ){
      // map to whatever you want
      const valueToSave = {[index] : true};
      return [...previousValue, valueToSave];
    } else {
      return previousValue;
    }
  }, []);

console.log(arr);

uj5u.com熱心網友回復：

@festina，為了簡單起見，我們在 users2 上進行映射，因為我們試圖在 users2 中查找 users1 的物件，我們使用 findIndex 使用（id/name 任何可用的唯一鍵）獲取索引，然后按索引 > 0 過濾，并格式化產生所需的格式

const users1 = [
 { name: 'John', age: 15, gender: 'M', city: 'London', country: 'UK' },
 { name: 'Jack', age: 20, gender: 'M', city: 'New York', country: 'USA' },
 { name: 'Jinny', age: 30, gender: 'F', city: 'London', country: 'UK' },
 { name: 'Key', age: 15, gender: 'M', city: 'Leeds', country: 'UK' },
];

const users2 = [
    { name: 'Jack', age: 20, gender: 'M', city: 'New York', country: 'USA' },
    { name: 'Key', age: 15, gender: 'M', city: 'Leeds', country: 'UK' },
  ];
// your solution
const result = users2.map((user, index) => users1.findIndex(u => u.name === user.name))
.filter(index => index > 0)
.map((i => { return { [i]:  true} }));
console.log(result)

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