[英]Efficient way to get unique objects from array of objects, by object property?
Here is my array of objects这是我的对象数组
const array = [
{id: 1, data: "foo"},
{id: 1, data: "bar"},
{id: 2, data: "baz"}
]
I want to remove all duplicate objects by its id and return only the array of objects that have an unique id.我想通过其 id 删除所有重复的对象,并仅返回具有唯一 id 的对象数组。
Expected result:预期结果:
[
{id: 2, data: "baz"}
]
This is what I have now: O(n^2)这就是我现在所拥有的:O(n^2)
function getUnique(array) {
const newArray = []
for (let obj of array) {
if (array.filter(x => x.id === obj.id).length === 1) {
newArray.push(obj)
}
}
return newArray
}
Whats the more efficient way to achieve this?实现这一目标的更有效方法是什么?
Is it possible to get the time-complexity to O(n) or O(n log n)?是否有可能获得 O(n) 或 O(n log n) 的时间复杂度?
const array = [{ id: 1, data: "foo" }, { id: 1, data: "bar" }, { id: 2, data: "baz" } ] let map = {}; array.forEach(x => { map[x.id] = (map[x.id] || 0) + 1 }); console.log(array.filter(x => map[x.id] === 1))
I would suggest to count the number of occurrences in your array and store them in a Map
.我建议计算数组中出现的次数并将它们存储在
Map
中。 Next you filter all element, which count is 1
接下来过滤所有元素,计数为
1
function getUnique(arr) {
const count = new Map();
arr.forEach((element) => {
count.set(element.id, (count.get(element.id) || 0) + 1);
});
return array.filter((element) => {
return count.get(element.id) === 1;
});
}
This has a runtime of 2(n)
since you have to iterate over them twice这具有
2(n)
的运行时间,因为您必须对它们进行两次迭代
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