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，我试图从字典中删除重复的元素。 我正在使用 double for（我不记得其他低成本算法可以做到这一点），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：需要的结果：

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？ 有什么更有效的算法吗？ 谢谢阅读！

Answer 1

您可以在Set的帮助下过滤数组。

如果哈希值（从hour_from和hour_to构建）在集合中，则该项目将被过滤掉。 如果不是，则将散列带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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Answer 2

你可能会懒惰地使用Array.prototype.reduce()和Array.prototype.find() 。

 const src = [{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"}], dedupe = src.reduce((res, item) => ( !res.find(({hour_from, hour_to}) => hour_from == item.hour_from && hour_to == item.hour_to) ? res.push(item) : true, res ), []) console.log(dedupe)

Answer 3

在这行for (let j=0; j<=hours_array.length; j++){你可以用完数组的边界。 在您的情况下，array.length 为 10，但是当您尝试访问索引为 10 的元素时，您会得到未定义，因为您的最后一个索引是 9。您必须将<=更改为<或将长度减少 1。请参阅解决方案以下：

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];
    }
  }
}

编辑：好的，你明白这个问题，下一个问题是，如果数组仍在循环中，则删除一个元素。 有很多方法可以解决这个问题，但最简单的方法是创建一个新数组并将唯一元素推送到该新数组。 如果您需要帮助，请告诉我。

Answer 4

使用reduce方法结合Object.values 。

 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" } ]; const updated = Object.values( hours_array.reduce( (acc, curr) => ({ ...acc, [`${curr.hour_from}-${curr.hour_to}`]: { ...curr } }), {} ) ); console.log(updated);