将数组的值与对象数组的键进行比较

As the question indicates, I need to compare the values of an array of this type [1,2,3], with the values of a key of an array of objects [{id: 1, name: 'jhon'}, {id: 2 , name: 'max'}] in this case I want to compare it with the value of the id key, I write the following example of what I need:

如问题所示，我需要将此类型数组的值[1,2,3]与对象数组的键值[{id：1，name：'jhon'}，{进行比较id：2，name：'max'}]在这种情况下，我想将它与id键的值进行比较，我写下我需要的示例：

if I have this array of objects:

如果我有这个对象数组：

[
   {
      id: 1
      name: 'jhon'
   },
   {
     id: 2,
     name: 'max'
   },
   {
     id: 3,
     name: 'fer'
   }
]

and I have this array:

我有这个数组：

[1,2,3,4,5]

I need to compare them and obtain the values that are not present in the array of objects, in the previous example the values 4 and 5 are not present in the array of objects, so I need to obtain a new array or the same, but with the values that do not they were, [4,5]

我需要比较它们并获得对象数组中不存在的值，在前面的例子中，值4和5不存在于对象数组中，所以我需要获得一个新的数组或者相同，但是他们没有的价值观，[4,5]

NOTE: this should also work if I have an array, only with numbers that are not present, for example [8,9], they should return those same values.

注意：如果我有一个数组，只有不存在的数字，例如[8,9]，它也应该有效，它们应该返回相同的值。

EDIT: it is possible to obtain in a new array those values that are only present in the array of objects. For example, with the arrays of the previous code:

编辑：可以在新数组中获得仅存在于对象数组中的值。例如，使用前面代码的数组：

[{id: 1, name: 'jhon'}, {id: 2, name: 'max'}, {id: 3, name: 'fer'}]

now this array would have the following values [1,4,5] with the previous answers I get the following array [4,5] what is correct. how can I get the values [2,3] in a new array.

现在这个数组将具有以下值[1,4,5]与之前的答案我得到以下数组[4,5]什么是正确的。如何在新数组中获取值[2,3]。

5 个解决方案

#1

You can use filter and find

您可以使用过滤器并查找

let arr1 = [{id: 1,name: 'jhon'},{id: 2,name: 'max'},{id: 3,name: 'fer'}]
let arr2 = [1, 2, 3, 4, 5];

let result = arr2.filter(o => !arr1.find(x => x.id === o));

console.log(result);

Update: This is basically the reverse of the first example. You filter the arr2 and check if the id exists on arr1. Then, use map to return the id

更新：这基本上与第一个示例相反。您过滤arr2并检查arr1上是否存在id。然后，使用map返回id

let arr1 = [{id: 1,name: 'jhon'},{id: 2,name: 'max'},{id: 3,name: 'fer'}]
let arr2 = [1,4,5];

let result = arr1.filter(o => !arr2.find(x => x === o.id)).map(o=>o.id);

console.log(result);

Doc: filter(), find()

Doc：filter（），find（）

#2

Create a Set from the ids of arr1 using Array.map(). Then filter arr2 by checking if the Set has the number:

使用Array.map（）从arr1的id创建一个Set。然后通过检查Set是否具有数字来过滤arr2：

const arr1 = [{"id":1,"name":"jhon"},{"id":2,"name":"max"},{"id":3,"name":"fer"}];
const arr2 = [1,2,3,4,5];
const arr3 = [1, 5];

const difference = (arr1, arr2, key) => {
  const arr1Values = new Set(key ? arr1.map(({ [key]: v }) => v) : arr1);

  return arr2.filter((n) => !arr1Values.has(n));
};

// array of primitives and array of objects
console.log(difference(arr1, arr2, 'id'));

// two arrays of primitives 
console.log(difference(arr3, arr2));

#3

You can use filter to check if id property of element in arr2 matches to the current element in arr2

您可以使用filter来检查arr2中元素的id属性是否与arr2中的当前元素匹配

const arr1 = [{"id":1,"name":"jhon"},{"id":2,"name":"max"},{"id":3,"name":"fer"}];

const arr2 = [8,9];
var diffArray = arr2.filter(x => !arr1.filter(y => y.id === x).length);
console.log(diffArray);

Answer [ 8,9 ]

回答[8,9]

#4

You can use .filter() and .includes():

您可以使用.filter（）和.includes（）：

let arr1 = [{id: 1,name: 'jhon'},{id: 2, name: 'max'},{id: 3,name: 'fer'}],
    arr2 = [1, 2, 3, 4, 5];

let result = (
    ids => arr2.filter(n => !ids.includes(n))
)(arr1.map(({id}) => id));

console.log(result);

Description:

描述：

Create a new array by extracting only ids from first array using .map().
通过使用.map（）仅从第一个数组中提取ID来创建新数组。
Filter elements using .filter() and .includes() to filter only those elements of second array that doesn't exists in newly created array.
使用.filter（）和.includes（）过滤元素，仅过滤新创建的数组中不存在的第二个数组的元素。

Docs:

文档：

Array.prototype.map()
Array.prototype.map（）
Array.prototype.filter()
Array.prototype.filter（）
Array.prototype.includes()
Array.prototype.includes（）
Object Destructuring
对象解构
Arrow Functions
箭头功能

#5

You could take a set with the given ids and then delete the ones of the object. Later return the array with the leftover ids.

您可以使用给定的ID获取一个集合，然后删除该对象的集合。稍后使用剩余的id返回数组。

var data = [{ id: 1, name: 'jhon' }, { id: 2, name: 'max' }, { id: 3, name: 'fer' }],
    ids = [1, 2, 3, 4, 5],
    result = Array.from(data.reduce((s, { id }) => (s.delete(id), s), new Set(ids)));

console.log(result);

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#1