问题描述
大家好,我在尝试创建函数时遇到一些问题。如代码所示。我想做interestMatch函数。该功能需要做的是查看所有“用户”并找到那些具有相同兴趣的用户-我的代码中的“ bob”和“ jack”。我认为应该带有if语句。 Kinda:“如果有任何用户有相同的兴趣”“返回此用户,并且此用户为匹配对象!” 有人能帮我吗!? 非常感谢!
class User {
constructor(username,password,firstname,lastname,email,birthday,gender,interest){
this.username = username;
this.password = password;
this.firstname = firstname;
this.lastname = lastname;
this.email = email;
this.birthday = birthday;
this.gender = gender;
this.interest = interest
}
}
let InterrestArray = ["Netflix","Sports","Party","Business","Children","Hygge"]
let bob = new User ("bob123","12345","Bob","bobiski","bob@bob.com","2000-10-10","male",interrest[0]);
let jack = new User ("jack20","340302","Jack","jackiski","jack@jack.com","2000-06-10",interrest[0]);
let thif = new User ("thif20","204903","Thifanny","Thifiski","thif@jack.com","2000-09-12","famale",interrest[1]);
function interestmatch(){
???
}
console.log(interestmatch())
// I want it to console.log - "bob maches with jack!" - because the have the same interrest
```
解决方法
一种非常简单的方法是拥有两个嵌套的for
循环,这些循环遍历用户并比较他们的兴趣:
function interestMatch(users){
for (let i = 0; i < users.length; i++) {
const user1 = users[i];
// No need to iterate over any user that comes "before" user1.
for (let j = i+1; j < users.length; j++) {
const user2 = users[j];
if (user1.interest === user2.interest) {
return `${user1.username} matches with ${user2.username}`;
}
}
}
}
当然,如果用户有多个兴趣,或者您想返回所有匹配项,事情就会变得更加复杂。在这种情况下,创建一个interest -> list of users
映射可能很有意义。
OP寻找的任何方法都比
它应该是带有if语句的东西。
一种有效的方法是浏览用户列表,查找用户的每个兴趣并将其推入与兴趣相关的数组。因此,基本方法使用特定兴趣的用户列表创建基于<DataGrid x:Name="dgListings" HeadersVisibility="Column" ItemsSource="{Binding LoadedListings}" CanUserAddRows="False" AutoGenerateColumns="False">
的地图或索引。
这只是第一部分,提供了一种创建可以进一步处理的数据结构的方法。
掌握了所需的数据结构(一个键值存储,其中interest
是键,一个特定的用户列表是每个键的值),其中一个接近第二部分,即输出/呈现。
对于每个键值对(兴趣和用户列表),确实会调用特定于用户列表的呈现功能。在这一步骤中,也确实将interest
列表映射到一个列表,例如每个用户的名字。
然后,最终的渲染功能需要处理边缘情况……例如……有一个用户,而其他任何人都没有共享的兴趣……或者……有一个共同兴趣并且拥有相同兴趣的两个用户...
user
class User {
constructor(
username,password,firstname,lastname,email,birthday,gender,interest
) {
Object.assign(this,{
username,interest
});
}
}
function groupUserByInterest(index,user) {
const { interest } = user;
const sameInterestList = index[interest] || (index[interest] = []);
sameInterestList.push(user);
return index;
}
const interestList = ["Netflix","Sports","Party","Business","Children","Hygge"];
const userList = [
new User ("bob123","12345","Bob","bobiski","bob@bob.com","2000-10-10","male",interestList[0]),new User ("jack20","340302","Jack","jackiski","jack@jack.com","2000-06-10",new User ("jane20","340303","Jane","janinski","jane@jane.com","famale",new User ("thif20","204903","Thifanny","Thifiski","thif@jack.com","2000-09-12",interestList[1])
];
// const [ bob,jack,jane,thif ] = userList;
function renderUsersOfSameInterest(interest,userList) {
const [ first,...rest ] = userList;
let template;
if (!rest.length) {
template = `There is no other user who matches ${ first }'s interest (${ interest }).`;
} else {
template = `${ first }'s interest (${ interest }) maches with the one of ${ rest.join(' and ') }!`;
}
console.log(template);
}
function renderUsersOfSameInterestsByFirstname(index) {
Object.entries(index).forEach(([interest,userList]) =>
renderUsersOfSameInterest(interest,userList.map(user =>
user.firstname))
);
// console.log('index',index);
}
renderUsersOfSameInterestsByFirstname(
userList.reduce(groupUserByInterest,{})
);