Question

lodash是否有办法，结果我没有满足特定条件的对象。例如，

o1 = [
  {name: "a", id: 2, key: 33},
  ..,
]

o2 = [
 {name: "ab", id: 2, key: 133}
]

lodash是否有一种方法，其中结果数组仅包含o2中不存在ID的对象。例如，比较o1和o2之后的结果对象必须不具有o2中的对象，因为id=2中已经存在o1。

Answer 1

您可以使用_.differenceBy()并使用参考点的id：

const o1 = [{id: 1, name: 'a'},{id: 2, name: 'b'},{id: 3, name: 'c'}]
const o2 = [{id: 1, name: 'b'}]

const result = _.differenceBy(o1, o2, 'id')

console.log(result)

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.min.js"></script>

Answer 2

您可以通过使用.filter()并创建一个Set来实现而不用破折号。首先，您可以在o2中创建一组所有ID。然后，您可以从o1中过滤出集合中具有ID的所有对象。通过将集合与.has()配合使用，我们可以提高算法的效率（比对数组使用.includes()更高。）

请参见以下示例：

const o1 = [
  {name: "a", id: 2, key: 33},
  {name: "b", id: 3, key: 34},
  {name: "c", id: 4, key: 34}
]

const o2 = [
  {name: "d", id: 2, key: 134}
]

const o2_ids = new Set(o2.map(({id}) => id));
const result = o1.filter(({id}) => !o2_ids.has(id));
console.log(result); // includes objects with id's that appear in o1 but not in o2

Answer 3

也许是_.differenceWith？

const o1 = [
  {id: 1, name: "a"},
  {id: 2, name: "b"},
  {id: 3, name: "c"},
]

const o2 = [
  {id: 1, name: "b"},
]

const diffed = _.differenceWith(o1, o2, (o1, o2) => o1.id === o2.id)

console.log(diffed)

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.min.js"></script>

Answer 4

<!DOCTYPE html>
<html>
<head>

</head>

<body>


<script>
var o1 = [
  {name: "a", id: 77, key: 55},
  {name: "a", id: 2, key: 33},
  {name: "a", id: 1, key: 55}
]

var o2 = [
 {name: "ab", id: 88, key: 133},
 {name: "ab", id: 2, key: 133},
 {name: "ab", id: 99, key: 133}
]
//sort first

o1.sort((a, b) => {
	return a.id-b.id;//sort by id
});

o2.sort((a, b) => {
	return a.id-b.id;//sort by id
});


//then compare one by one
function SearchX(OO1,OO2){
	var o1_compare_place=0;
	var o2_compare_place=0;


	while(OO2.length>o2_compare_place && OO1.length>o1_compare_place ){
		if(OO2[o2_compare_place].id<OO1[o1_compare_place].id){
			o2_compare_place+=1;
		}else if(OO2[o2_compare_place].id>OO1[o1_compare_place].id){
			o1_compare_place+=1;
		}else{
			return "Exist Same!";
		}
	}
	return "Different!";

}

document.body.innerHTML = SearchX(o1,o2)







</script>

</body>
</html>

您在这里

比较2个对象数组中的特定键

4 个答案: