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比较对象的2个数组并删除重复的javascript

[英]Comparing 2 arrays of objects and removing the duplicates javascript

I'm trying to compare 2 array of objects together and remove the duplicates. 我正在尝试将2个对象数组一起比较,并删除重复项。 For example... 例如...

var array1 = [0, 1, 2, 3, 4, 5, 6];
var array2 = [7, 8, 1, 2, 9, 10];

How can I check the array2 for any element of array1 and if true then remove that from array2 to eliminate the duplication. 我如何检查array2中是否有array1的任何元素,如果为true,则将其从array2中删除以消除重复。

The expected result: array 2 = [7, 8, 9, 10] 预期结果: array 2 = [7, 8, 9, 10]

Any help would be appreciated, thanks 任何帮助,将不胜感激,谢谢

just filter second array. 只是过滤第二个数组。

const array1 = [0, 1, 2, 3, 4, 5, 6];
const array2 = [7, 8, 1, 2, 9, 10];

const newArray = array2.filter(i => !array1.includes(i));

console.log(newArray);

If the array include primitive types you can use indexOf and array#reduce 如果数组包含基本类型,则可以使用indexOfarray#reduce

 const array1 = [0, 1, 2, 3, 4, 5, 6] const array2 = [7, 8, 1, 2, 9, 10] var result = array2.filter((num) => { return array1.indexOf(num) === -1; }); console.log(result); 
 .as-console-wrapper { max-height: 100% !important; top: 0; } 

In case of an object, you can get the unique values in second array in comparison to first, please use array#reduce and array#some . 如果是对象,则与第一个数组相比,您可以在第二个数组中获得唯一值,请使用array#reducearray#some

 const person1 = [{"name":"a", "id":0},{"name":"A", "id":1},{"name":"B", "id":2},{"name":"C", "id":3},{"name":"D", "id":4},{"name":"E", "id":5},{"name":"F", "id":6}] const person2 = [{"name":"G", "id":7},{"name":"H", "id":8},{"name":"A", "id":1},{"name":"B", "id":2},{"name":"I", "id":9}, {"name":"J", "id":10}] var unique = person2.reduce((unique, o) => { let isFound = person1.some((b) => { return b.id === o.id; }); if(!isFound) unique.push(o); return unique; },[]); console.log(unique); 
 .as-console-wrapper { max-height: 100% !important; top: 0; } 

You can use do this: 您可以使用以下方法:

var array1 = [0, 1, 2, 3, 4, 5, 6];
var array2 = [7, 8, 1, 2, 9, 10];

/* run a filter function for every value in array2 and returned the final filtered array */
var array3 = array2.filter(function(currentValue, index, arr){
      return (array1.indexOf(currentValue) === -1); /* this will check whether currentValue exists in aray1 or not. */
});

console.log(array3) /* filtered array */

This should deliver what you want. 这应该可以提供您想要的东西。 Or another way: 或另一种方式:

var array1 = [0, 1, 2, 3, 4, 5, 6];
var array2 = [7, 8, 1, 2, 9, 10];
var duplicateRemoved = [];
/* run a function for every value in array2 and collect the unique value in a new array */
array2.forEach(function(currentValue, index, arr){
      if (array1.indexOf(currentValue) === -1) {
          duplicateRemoved.push(currentValue);
      }
});

console.log(duplicateRemoved)

This should work in your situation unless there are some other external factors associated with it. 除非有其他外部因素与之相关,否则这应该可以解决您的问题。

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