Javascript从2个数组中返回数组，删除重复项

Searched and tried and no luck so far.

搜索和尝试，到目前为止没有运气。

var newUsers = [{name: 'rich', id: 25}, {name: 'lauren', id: 35}, {name: 'dave', id: 28} ]

var likedUsers = [{name: 'derek', id: 39}, {name: 'rich', id: 25}, {name: 'brian', id: 38} ]

What I want returned is:

我想要的是：

var leftUsers = [{name: 'lauren', id: 35}, {name: 'dave', id: 28} ]

basically without the rich object as this is a duplicate. I only care about the id key.

基本上没有富对象，因为这是重复的。我只关心id键。

I have tried:

我努力了：

newUsers.forEach((nUser) => {
    likedUsers.forEach((lUser) => {
        if (nUser.id !== lUser.id){
            leftUsers.push(nUser)
        }
    })
})

but obviously this won't work as this will just add them all as soon as they don't match.

但显然这不会起作用，因为只要它们不匹配就会将它们全部添加。

if possible would like an es6 solution using forEach/map/filter

如果可能的话，想要使用forEach / map / filter的es6解决方案

thanks

谢谢

4 个解决方案

#1

With array.prototype.filter to filter out items that exists in likedUsers and array.prototype.findIndex to check the existence, it should be:

使用array.prototype.filter过滤掉preferUsers和array.prototype.findIndex中存在的项以检查是否存在，它应该是：

var newUsers = [{name: 'rich', id: 25}, {name: 'lauren', id: 35}, {name: 'dave', id: 28} ];
var likedUsers = [{name: 'derek', id: 39}, {name: 'rich', id: 25}, {name: 'brian', id: 38} ];

var leftUsers = newUsers.filter(u => likedUsers.findIndex(lu => lu.id === u.id) === -1);

console.log(leftUsers);

#2

You can do this with filter() and some() methods.

您可以使用filter（）和some（）方法执行此操作。

var newUsers = [{name: 'rich', id: 25}, {name: 'lauren', id: 35}, {name: 'dave', id: 28} ]
var likedUsers = [{name: 'derek', id: 39}, {name: 'rich', id: 25}, {name: 'brian', id: 38} ]

const result = newUsers.filter(e => !likedUsers.some(a => a.id == e.id));
console.log(result)

#3

var newUsers = [{name: 'rich', id: 25}, {name: 'lauren', id: 35}, {name: 'dave', id: 28} ];
var likedUsers = [{name: 'derek', id: 39}, {name: 'rich', id: 25}, {name: 'brian', id: 38} ];

var leftusers = newUsers.filter( item => !likedUsers.find(item2 => item.id == item2.id));

console.log(leftusers);

#4

You can create a Set of ids that are found in likedUsers, and filter the newUsers by checking if an id is in the Set:

您可以创建在preferUsers中找到的一组ID，并通过检查ID是否在Set中来过滤newUsers：

const newUsers = [{name: 'rich', id: 25}, {name: 'lauren', id: 35}, {name: 'dave', id: 28} ]
const likedUsers = [{name: 'derek', id: 39}, {name: 'rich', id: 25}, {name: 'brian', id: 38} ]

const result = newUsers.filter(function({ id }) {
  return !this.has(id) // take all users which ids is not found in the set
}, new Set(likedUsers.map(({ id }) => id))) // create a set of ids in likedUsers and assign to this

console.log(result)

#1