如何从 2 个数组中仅返回不重复的对象

Question

我发现了很多类似的问题，但它们解释了如何删除重复的对象。在这种情况下，我需要创建一个不包含在两个数组中找到的对象的新数组。

const firstArray = [
  {firstName: 'John', lastName: 'Doe'}, 
  {firstName: 'Sara', lastName: 'Connor'},
  {firstName: 'Mike', lastName: 'Hunt'}, 
  {firstName: 'Steve', lastName: 'Irvine'}
];

const secondArray = [ 
  {firstName: 'John', lastName: 'Doe'},  
  {firstName: 'Sara', lastName: 'Connor'} 
];

上一个数据样本的预期结果应该是：

const result = [
  {firstName: 'Mike', lastName: 'Hunt'},
  {firstName: 'Steve', lastName: 'Irvine'}
];

Answer 1

我将首先连接两个数组，然后 filter() 使用 some() 对每个数组进行搜索的那些没有出现在两个数组上的元素。请注意，此方法考虑了firstArray 和secondArray 都可以包含非重复对象的可能性。

const firstArray = [
  {firstName: 'John', lastName: 'Doe'}, 
  {firstName: 'Sara', lastName: 'Connor'},
  {firstName: 'Mike', lastName: 'Hunt'}, 
  {firstName: 'Steve', lastName: 'Irvine'}
];

const secondArray = [ 
  {firstName: 'John', lastName: 'Doe'},  
  {firstName: 'Sara', lastName: 'Connor'},
  {firstName: 'Joseph', lastName: 'Muguera'}
];

let res = firstArray.concat(secondArray).filter(({firstName, lastName}) =>
{
    let foundOnFirst = firstArray.some(
        x => x.firstName === firstName && x.lastName === lastName
    );
    let foundOnSecond = secondArray.some(
        y => y.firstName === firstName && y.lastName === lastName
    );
    return !(foundOnFirst && foundOnSecond);
});

console.log(res);

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Answer 2

您可以通过迭代 secondArray 来过滤 firstArray 并检查想要的属性以进行比较。

const
    firstArray = [{ firstName: 'John', lastName: 'Doe' }, { firstName: 'Sara', lastName: 'Connor' }, { firstName: 'Mike', lastName: 'Hunt' }, { firstName: 'Steve', lastName: 'Irvine' }],
    secondArray = [{ firstName: 'John', lastName: 'Doe' }, { firstName: 'Sara', lastName: 'Connor' }],
    result = firstArray.filter(o => 
        secondArray.every(p =>
            !['firstName', 'lastName'].some(k => o[k] === p[k])));

console.log(result);

一种具有Set 的联合属性的方法。

const
    firstArray = [{ firstName: 'John', lastName: 'Doe' }, { firstName: 'Sara', lastName: 'Connor' }, { firstName: 'Mike', lastName: 'Hunt' }, { firstName: 'Steve', lastName: 'Irvine' }],
    secondArray = [{ firstName: 'John', lastName: 'Doe' }, { firstName: 'Sara', lastName: 'Connor' }],
    getKey = ({ firstName, lastName }) => [firstName, lastName].join('|'),
    secondSet = new Set(firstArray.map(getKey)),
    result = firstArray.filter(o => !set.has(getKey(o)));

console.log(result);

Answer 3

您可以过滤一个数组并将对象与JSON.stringify 与第二个数组进行比较，

const firstArray = [
    {firstName: 'John', lastName: 'Doe'}, 
    {firstName: 'Sara', lastName: 'Connor'},
    {firstName: 'Mike', lastName: 'Hunt'}, 
    {firstName: 'Steve', lastName: 'Irvine'}
    ];

const secondArray = [ 
    {firstName: 'John', lastName: 'Doe'},  
    {firstName: 'Sara', lastName: 'Connor'} 
    ];
    
const res=firstArray.filter(function(obj) {
 // JSON.stringify(obj)==JSON.stringify(obj)
  for( var i=0, len=secondArray.length; i<len; i++ ){
          if(JSON.stringify(obj)==JSON.stringify(secondArray[i]) ) {
              return false;
          }
      }
	return true; 
});
console.log(res);

Answer 4

另一种方法可以基于.findIndex()和JSON.stringify()：

const firstArray = [
    {firstName: 'John', lastName: 'Doe'},
    {firstName: 'Sara', lastName: 'Connor'},
    {firstName: 'Mike', lastName: 'Hunt'},
    {firstName: 'Steve', lastName: 'Irvine'}
];

const secondArray = [
    {firstName: 'John', lastName: 'Doe'},
    {firstName: 'Sara', lastName: 'Connor'}
];
const result = firstArray.filter(ele =>
    secondArray.findIndex(ele1 =>
        JSON.stringify(ele1) == JSON.stringify(ele)) == -1
    );

console.log(result);