我有两个对象数组:
array1 = [
{id:1, name: 'one'},
{id:4, name: 'four'}
]
array2 = [
{id:1, name: 'one'},
{id:2, name: 'two'},
{id:3, name: 'three'},
{id:5, name: 'five'},
{id:6, name: 'six'},
{id:7, name: 'seven'}
]
我想从array1
中删除任何对象,id
array2
中不存在。
所以我的预期结果是:
array1 = [
{id:1, name:'one'}
]
使用 lodash 的_.intersectionBy()
:
var array1 = [
{id:1, name: 'one'},
{id:4, name: 'four'}
];
array2 = [
{id:1, name: 'one'},
{id:2, name: 'two'},
{id:3, name: 'three'},
{id:5, name: 'five'},
{id:6, name: 'six'},
{id:7, name: 'seven'}
];
var result = _.intersectionBy(array1, array2, 'id');
console.log(result);
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.4/lodash.min.js"></script>
一个快速且可读的选项是:
var referenceKeys = array2.map(function(entity) { return entity.id; });
var result = array1.filter(function(entity) {
return referenceKeys.indexOf(entity.id) !== -1;
});
但不能保证它是所有维度中最快的。(重复次数、数组长度 1、数组长度 2(。
您可以使用标准方法,即使用哈希表,该哈希表仅对两个数组使用一次迭代。
var array1 = [{ id: 1, name: 'one' }, { id: 4, name: 'four' }],
array2 = [{ id: 1, name: 'one' }, { id: 2, name: 'two' }, { id: 3, name: 'three' }, { id: 5, name: 'five' }, { id: 6, name: 'six' }, { id: 7, name: 'seven' }],
hash = Object.create(null),
result;
array2.forEach(function (o) {
hash[o.id] = true;
});
result = array1.filter(function (o) {
return hash[o.id];
});
console.log(result);
您可以使用
Set
:
const seenIds = array2.reduce((set, o) => set.add(o.id), new Set());
const result = array1.filter(o => seenIds.has(o.id));