檢查一個對象數組的字段是否從另一個對象數組中丟失

Question

假設我們有一系列這樣的對象：

var array1 = 
[
 {first_name: 'Bob', last_name: 'Jones', email: 'bobjones@gmail.com'},
 {first_name: 'Susan', last_name: 'Anderson', email: 'susananderson@gmail.com'}
]

var array2 =
[
 {first_name: 'Bob', last_name: 'Jones', email: 'bobjones@gmail.com'},
 {first_name: 'Susan', last_name: 'Anderson', email: 'susananderson@gmail.com'},
 {first_name: 'John', last_name: 'Smith', email: 'johnsmith@gmail.com'}
]

我想，如果在任何數組2排是基於電子郵件地址缺少檢測，如果它缺少設置一個新的狀態active為false，這樣最終的數組將出來這樣的：

var array3 = [
 {first_name: 'Bob', last_name: 'Jones', email: 'bobjones@gmail.com'},
 {first_name: 'Susan', last_name: 'Anderson', email: 'susananderson@gmail.com'},
 {first_name: 'John', last_name: 'Smith', email: 'johnsmith@gmail.com', active: false}
]

我嘗試遍歷兩個元素中的每個元素，但是事實證明，當數組中有很多元素時，這很慢。

Answer 1

使用forEach循環和filter

 var array1 = [ {first_name: 'Bob', last_name: 'Jones', email: 'bobjones@gmail.com'}, {first_name: 'Susan', last_name: 'Anderson', email: 'susananderson@gmail.com'} ] var array2 = [ {first_name: 'Bob', last_name: 'Jones', email: 'bobjones@gmail.com'}, {first_name: 'Susan', last_name: 'Anderson', email: 'susananderson@gmail.com'}, {first_name: 'John', last_name: 'Smith', email: 'johnsmith@gmail.com'} ] array2.forEach((e)=>{ array1.filter(k=>e.email==k.email).length==0?e.active="false":false; }) console.log(array2) 

Answer 2

您可以使用map和Map

• 將array1更改為Map ，以便可以在O（1）中完成來自array1的訪問電子郵件
• 循環訪問array2，根據newMap是否存在電子郵件有條件地添加active密鑰

 let array1 = [{first_name: 'Bob', last_name: 'Jones', email: 'bobjones@gmail.com'},{first_name: 'Susan', last_name: 'Anderson', email: 'susananderson@gmail.com'}] let array2 =[{first_name: 'Bob', last_name: 'Jones', email: 'bobjones@gmail.com'}, {first_name: 'Susan', last_name: 'Anderson', email: 'susananderson@gmail.com'}, {first_name: 'John', last_name: 'Smith', email: 'johnsmith@gmail.com'}] let newMap = new Map(array1.map(({email})=>[email,true])) let final = array2.map(obj=>({ ...obj, ...(!newMap.has(obj.email) && {active: false}) }) ) console.log(final) 

Answer 3

您可以維護一組來自array1的所有電子郵件，然后在array2上使用Array.map()獲得所需的結果：

 let array1 = [ {first_name: 'Bob', last_name: 'Jones', email: 'bobjones@gmail.com'}, {first_name: 'Susan', last_name: 'Anderson', email: 'susananderson@gmail.com'} ] let array2 =[ {first_name: 'Bob', last_name: 'Jones', email: 'bobjones@gmail.com'}, {first_name: 'Susan', last_name: 'Anderson', email: 'susananderson@gmail.com'}, {first_name: 'John', last_name: 'Smith', email: 'johnsmith@gmail.com'} ]; let set = new Set(array1.map(({email})=> email)); let array3 = array2.map(o =>{ if(!set.has(o.email)) o.active = false; return o; }); console.log(array3);