对象的两个javascript数组的区别

时间：2019-02-26 17:31:12

标签： javascript

我有两个这样的数组：

owners: [
  {
    userID: "58c4d7ac",
    username: "John.Doe",
    firstName: "John",
    lastName: "Doe",
    email: "John.Doe@acme.com"
  },
  {
    userID: "68c4d7ac",
    username: "User2.Name2",
    firstName: "User2",
    lastName: "Name2",
    email: "dh@acme.com"
  }
]


users: [
  {
    userID: "58c4d7ac",
    username: "John.Doe",
    firstName: "John",
    lastName: "Doe",
    email: "John.Doe@acme.com"
  },
  {
    userID: "68c4d7ac",
    username: "User2.Name2",
    firstName: "User2",
    lastName: "Name2",
    email: "dh@acme.com"
  },
  {
    userID: "88c4d7ac",
    username: "User3.Name3",
    firstName: "User3",
    lastName: "Name3",
    email: "dh@acme.com"
  }
]

我想得到一个用户数组，其中只包含不在owners数组中的元素。

我尝试了不同的方法。最后，我得出了解决方案：

const usersItems = users.map(user => {
    // Check whether the user is already an owner
    if (owners.findIndex(owner => owner.userID === user.userID) === -1) {
        return  owner
    } else {
        return null;
    }
});

console.log(usersItems);

// Filter out all items which are null
const newUsersItems = usersItems.filter(user => {
    if (user) return user;
});

console.log(usersItems);

在我看来，这似乎不是一个干净的解决方案。有没有更清洁，更轻松的方法来做到这一点？结果，我希望拥有：

newUsers: [
  {
    userID: "88c4d7ac",
    username: "User3.Name3",
    firstName: "User3",
    lastName: "Name3",
    email: "dh@acme.com"
  }
]

7 个答案:

答案 0 :(得分：6)

您可以摆脱map，而只需使用过滤器（这正是过滤器的目的）

const filtered = users.filter(user => {
    // Check whether the user is already an owner
   return owners.findIndex(owner => owner.userID === user.userID) === -1

});

可能会起作用

答案 1 :(得分：2)

首先，您可以使用userID's数组的owners创建 Set() ，然后可以使用 Array.filter() < / {strong}（在users数组上）以过滤userID不属于先前创建的集合的用户。

const owners = [
  {userID: "58c4d7ac", username: "John.Doe", firstName: "John", lastName: "Doe", email: "John.Doe@acme.com"},
  {userID: "68c4d7ac", username: "User2.Name2", firstName: "User2", lastName: "Name2", email: "dh@acme.com"}
];

const users = [
  {userID: "58c4d7ac", username: "John.Doe", firstName: "John", lastName: "Doe", email: "John.Doe@acme.com"},
  {userID: "68c4d7ac", username: "User2.Name2", firstName: "User2", lastName: "Name2", email: "dh@acme.com"},
  {userID: "88c4d7ac", username: "User3.Name3", firstName: "User3", lastName: "Name3", email: "dh@acme.com"}
];

let ownerIdsSet = new Set(owners.map(x => x.userID));
let res = users.filter(x => !ownerIdsSet.has(x.userID));
console.log(res);
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但是为什么要先构建一个Set？

总而言之，它将提高过滤过程的性能，尤其是在owners数组较大的情况下。您应注意，findIndex()，find()和some()之类的方法需要遍历数组以检查相关条件，同时检查userID是否属于{{1} }是 Set 计算。但是，当然，在创建上述Set时，初始化时会产生额外的负担。

答案 2 :(得分：2)

您可以像这样使用filter和some函数的组合：

const owners = [
  {
    userID: "58c4d7ac",
    username: "John.Doe",
    firstName: "John",
    lastName: "Doe",
    email: "John.Doe@acme.com"
  },
  {
    userID: "68c4d7ac",
    username: "User2.Name2",
    firstName: "User2",
    lastName: "Name2",
    email: "dh@acme.com"
  }
];

const users = [
  {
    userID: "58c4d7ac",
    username: "John.Doe",
    firstName: "John",
    lastName: "Doe",
    email: "John.Doe@acme.com"
  },
  {
    userID: "68c4d7ac",
    username: "User2.Name2",
    firstName: "User2",
    lastName: "Name2",
    email: "dh@acme.com"
  },
  {
    userID: "88c4d7ac",
    username: "User3.Name3",
    firstName: "User3",
    lastName: "Name3",
    email: "dh@acme.com"
  }
];

const result = users.filter(user => !owners.some(owner => owner.userID === user.userID));

console.log(result);

答案 3 :(得分：1)

您可以使用过滤器和一些

const owners = [{userID:"58c4d7ac",username:"John.Doe",firstName:"John",lastName:"Doe",email:"John.Doe@acme.com"},{userID:"68c4d7ac",username:"User2.Name2",firstName:"User2",lastName:"Name2",email:"dh@acme.com"}]
const users = [{userID:"58c4d7ac",username:"John.Doe",firstName:"John",lastName:"Doe",email:"John.Doe@acme.com"},{userID:"68c4d7ac",username:"User2.Name2",firstName:"User2",lastName:"Name2",email:"dh@acme.com"},{userID:"88c4d7ac",username:"User3.Name3",firstName:"User3",lastName:"Name3",email:"dh@acme.com"}]

const newUsers = users
                .filter(({userID}) => !owners.some(({userID:ownerID})=> ownerID === userID))

console.log(newUsers)

答案 4 :(得分：1)

您可以仅使用单个.filter()函数，如下所示：

let owners = [{userID: "58c4d7ac",username: "John.Doe",firstName: "John",lastName: "Doe",email: "John.Doe@acme.com"},{userID: "68c4d7ac",username: "User2.Name2",firstName: "User2",lastName: "Name2",email: "dh@acme.com"}];

let users = [{userID: "58c4d7ac",username: "John.Doe",firstName: "John",lastName: "Doe",email: "John.Doe@acme.com"},{userID: "68c4d7ac",username: "User2.Name2",firstName: "User2",lastName: "Name2",email: "dh@acme.com"},{userID: "88c4d7ac",username: "User3.Name3",firstName: "User3",lastName: "Name3",email: "dh@acme.com"}];

let newUsersItems = users.filter(user => owners.findIndex(owner => owner.userID === user.userID) === -1);

console.log(newUsersItems)

您可以仅使用单个.filter()函数，如下所示：

答案 5 :(得分：0)

您可以使用reduce函数

    const diff = users.reduce((acc, user) => { 
      if(!owners.find(owner => owner.id === user.id){
     acc.push(user);
    }
    return acc;
     }, []);

为避免在每个循环中使用find（）函数，您可以将所有者ID与地图一起存储在数组中，而只需使用include（）

 const ownersIds = owners.map(owner => owner.id);
 const diff = users.reduce((acc, user) => { 
   if(!ownersIds.includes(user.id){
     acc.push(user);
    }
    return acc;
     }, []);

答案 6 :(得分：0)

您可以使用函数some或find，这种方法使用函数find 

let owners = [   {     userID: "58c4d7ac",     username: "John.Doe",     firstName: "John",     lastName: "Doe",     email: "John.Doe@acme.com"   },   {     userID: "68c4d7ac",     username: "User2.Name2",     firstName: "User2",     lastName: "Name2",     email: "dh@acme.com"   } ],
    users = [   {     userID: "58c4d7ac",     username: "John.Doe",     firstName: "John",     lastName: "Doe",     email: "John.Doe@acme.com"   },   {     userID: "68c4d7ac",     username: "User2.Name2",     firstName: "User2",     lastName: "Name2",     email: "dh@acme.com"   },   {     userID: "88c4d7ac",     username: "User3.Name3",     firstName: "User3",     lastName: "Name3",     email: "dh@acme.com"   } ],
    result = users.filter(({userID}) => !owners.find(o => o.userID === userID));

console.log(result);
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