javascript - 通过 2 个匹配属性从对象数组中删除重复项

Question

当键值“hour_from”和“hour_to”相同时，我尝试从字典中删除重复元素。我正在使用 double for (我不记得其他成本较低的算法来做到这一点)，但我在索引值方面遇到了问题。

var hours_array = [
{day: "Mon", hour_from: "00:00", hour_to: "00:00"},
{day: "Mon", hour_from: "00:00", hour_to: "00:16"},
{day: "Mon", hour_from: "00:00", hour_to: "00:16"},
{day: "Thu", hour_from: "00:00", hour_to: "00:25"},
{day: "Mon", hour_from: "00:00", hour_to: "00:33"},
{day: "Fri", hour_from: "00:00", hour_to: "00:83"},
{day: "Sat", hour_from: "02:00", hour_to: "05:33"},
{day: "Thu", hour_from: "02:00", hour_to: "05:33"},
{day: "Wed", hour_from: "12:00", hour_to: "14:00"},
{day: "Sun", hour_from: "22:25", hour_to: "13:45"}]

for (let i=0; i< hours_array.length; i++){
 for (let j=0; j<=hours_array.length; j++){
  if ((hours_array[i]['hour_from'] == hours_array[j]['hour_from']) && (hours_array[i]['hour_to'] == hours_array[j]['hour_to'])){
  delete hours_array[j];
  }
 }
}

我认为这是索引值的错误:

编辑:所需结果:

var hours_array = [
{day: "Mon", hour_from: "00:00", hour_to: "00:00"},
{day: "Mon", hour_from: "00:00", hour_to: "00:16"},
{day: "Thu", hour_from: "00:00", hour_to: "00:25"},
{day: "Mon", hour_from: "00:00", hour_to: "00:33"},
{day: "Fri", hour_from: "00:00", hour_to: "00:83"},
{day: "Sat", hour_from: "02:00", hour_to: "05:33"},
{day: "Wed", hour_from: "12:00", hour_to: "14:00"},
{day: "Sun", hour_from: "22:25", hour_to: "13:45"}]

有什么建议吗？还有更高效的算法吗？感谢您的阅读!

Answer 1

您可以在 Set 的帮助下过滤数组.

如果哈希值(根据 hour_from 和 hour_to 构建)位于集合中，则该项目将被过滤掉。如果不是，则将哈希值添加到集合中并使用该项目。

var getKey = ({ hour_from, hour_to }) => [hour_from, hour_to].join('|'),
    hours_array = [{ day: "Mon", hour_from: "00:00", hour_to: "00:00" }, { day: "Mon", hour_from: "00:00", hour_to: "00:16" }, { day: "Mon", hour_from: "00:00", hour_to: "00:16" }, { day: "Thu", hour_from: "00:00", hour_to: "00:25" }, { day: "Mon", hour_from: "00:00", hour_to: "00:33" }, { day: "Fri", hour_from: "00:00", hour_to: "00:83" }, { day: "Sat", hour_from: "02:00", hour_to: "05:33" }, { day: "Thu", hour_from: "02:00", hour_to: "05:33" }, { day: "Wed", hour_from: "12:00", hour_to: "14:00" }, { day: "Sun", hour_from: "22:25", hour_to: "13:45" }],
    unique = hours_array.filter((s => o => !s.has(getKey(o)) && s.add(getKey(o)))(new Set));

console.log(unique);

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