javascript - 如何删除数组中重复出现的内容？

Question

我有任何看起来像这样的数组:

[
    { "tempValue": "Sunday  CLOSE to CLOSE" }, 
    { "tempValue": "Sunday  CLOSE to 08:30" }, 
    { "tempValue": "Sunday  CLOSE to 08:30" }, 
    { "tempValue": "Tuesday  CLOSE to 08:30" }, 
    { "tempValue": "Thursday  CLOSE to 08:30" }
]

我需要从该数组中删除所有重复项。在上面的示例中，重复项是Sunday...

所以我尝试了这样的事情:

function unique(list) {
    var result = [];
    $.each(list, function(i, e) {
        if ($.inArray(e, result) == -1) result.push(e);
    });
    return result;
}


unique(hours);

变量hours是我的数组。

但是我上面的代码不起作用!

有人可以就这个问题提出建议吗？

编辑:

所需的输出是这样的:

[
    { "tempValue": "Tuesday  CLOSE to 08:30" }, 
    { "tempValue": "Thursday  CLOSE to 08:30" }
]

Answer 1

我首先计算每天出现的次数，然后创建一个仅包含出现多次的日期的数组。然后，您可以通过数组项中的日期是否包含在该数组中来.filter:

const input = [{
    "tempValue": "Sunday  CLOSE to CLOSE"
}, {
    "tempValue": "Sunday  CLOSE to 08:30"
}, {
    "tempValue": "Sunday  CLOSE to 08:30"
}, {
    "tempValue": "Tuesday  CLOSE to 08:30"
}, {
    "tempValue": "Thursday  CLOSE to 08:30"
}];

const dayCountObj = input.reduce((a, { tempValue }) => {
  const day = tempValue.match(/^\S+/)[0];
  a[day] = (a[day] || 0) + 1;
  return a;
}, {});
const daysToRemove = Object.entries(dayCountObj)
  .filter(([, count]) => count > 1)
  .map(([day]) => day);
  
const output = input.filter(({ tempValue }) => {
  const day = tempValue.match(/^\S+/)[0];
  return !daysToRemove.includes(day);
});
console.log(output);

如果您想将计算复杂度降低到 O(N) 而不是 O(N^2)，请使用 Set 而不是数组作为 删除天数:

const input = [{
    "tempValue": "Sunday  CLOSE to CLOSE"
}, {
    "tempValue": "Sunday  CLOSE to 08:30"
}, {
    "tempValue": "Sunday  CLOSE to 08:30"
}, {
    "tempValue": "Tuesday  CLOSE to 08:30"
}, {
    "tempValue": "Thursday  CLOSE to 08:30"
}];

const dayCountObj = input.reduce((a, { tempValue }) => {
  const day = tempValue.match(/^\S+/)[0];
  a[day] = (a[day] || 0) + 1;
  return a;
}, {});
const daysToRemove = new Set(
  Object.entries(dayCountObj)
    .filter(([, count]) => count > 1)
    .map(([day]) => day)
);
  
const output = input.filter(({ tempValue }) => {
  const day = tempValue.match(/^\S+/)[0];
  return !daysToRemove.has(day);
});
console.log(output);