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Accepted
MaBed
Asked: 2024-09-25 17:25:21 +0800 CST

根据时间戳计算索引

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我想根据毫秒（纪元）时间戳计算数组中的索引。该数组长度为 167，每个索引为一周中的小时（即索引0指向星期一的 0:00、24星期二的 0:00、28星期二的 4:00，...）。我想到这个解决方案，它有效：

function getIndex(timestamp) {
  const hoursInDay = 24;
  const date = new Date(timestamp);
  const dayIndex = (date.getUTCDay() + 6) % 7; // index from Monday
  const hourIndex = date.getUTCHours();
  return dayIndex * hoursInDay + hourIndex;
}

问题是这很慢。我们有几千个这样的时间戳，但处理它们的时间非常有限。

是否有一种纯数值的方法可以达到相同的效果，这样我们就不必Date每次都构建和使用？

javascript

1 个回答

  1. Best Answer

    由于您使用的是 UTC 日期时间，因此不存在与时区相关的复杂性，我们可以避免创建 Date 实例。

    此函数返回与您的相同的值：

    const HOUR = 3600000;
const DAY = 24;
const THREE_DAYS = 3 * DAY;
const WEEK = 7 * DAY;

const getIndex = timestamp => ((timestamp / HOUR | 0) + THREE_DAYS) % WEEK;

    在我的测试中，该函数的执行速度比原始函数快约 50 倍。

    以下是我测试和测量的方法。请注意，我将第一个const声明移出了你的函数，因为每次调用时都对其进行评估很浪费时间：

    // Original code (with one obvious optimisation)
const hoursInDay = 24;
function getIndex(timestamp) {
  const date = new Date(timestamp);
  const dayIndex = (date.getUTCDay() + 6) % 7; // index from Monday
  const hourIndex = date.getUTCHours();
  return dayIndex * hoursInDay + hourIndex;
}

// Proposed improvement
const HOUR = 3600000;
const DAY = 24;
const THREE_DAYS = 3 * DAY;
const WEEK = 7 * DAY;

const getIndex2 = timestamp => ((timestamp / HOUR | 0) + THREE_DAYS) % WEEK;

// Verify that both functions return the same values
const addTime = 2727359; // Some arbitrary slice of time (~31.5 days)
const n = 1000000 * addTime;
for (let timestamp = 0; timestamp < n; timestamp += addTime) {
    let a = getIndex2(timestamp);
    let b = getIndex2(timestamp);
    if (a !== b) throw "functions return a different result for timestamp " + timestamp;
}

// Perform time measurements
let time1 = 0, time2 = 0, start;
for (let repeat = 1; repeat < 10; repeat++) {
    start = performance.now();
    for (let timestamp = 0; timestamp < n; timestamp += addTime) {
        getIndex(timestamp);
    }
    time1 += performance.now() - start;
    
    start = performance.now();
    for (let timestamp = 0; timestamp < n; timestamp += addTime) {
        getIndex2(timestamp);
    }
    time2 += performance.now() - start;
}

console.log("speed up factor", time1 / time2);

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