如何将Joda图书馆的DateTime舍入到最近的X分钟？

Question

如何圆DateTime的Joda库到最近X分钟？
例如:

X = 10 minutes
Jun 27, 11:32 -> Jun 27, 11:30
Jun 27, 11:33 -> Jun 27, 11:30
Jun 27, 11:34 -> Jun 27, 11:30
Jun 27, 11:35 -> Jun 27, 11:40
Jun 27, 11:36 -> Jun 27, 11:40
Jun 27, 11:37 -> Jun 27, 11:40

Answer 1

接受的答案无法正确处理设置为秒或毫秒的日期时间.为了完整性,这是一个正确处理的版本:

private DateTime roundDate(final DateTime dateTime, final int minutes) {
    if (minutes < 1 || 60 % minutes != 0) {
        throw new IllegalArgumentException("minutes must be a factor of 60");
    }

    final DateTime hour = dateTime.hourOfDay().roundFloorCopy();
    final long millisSinceHour = new Duration(hour, dateTime).getMillis();
    final int roundedMinutes = ((int)Math.round(
        millisSinceHour / 60000.0 / minutes)) * minutes;
    return hour.plusMinutes(roundedMinutes);
}

Answer 2

使用纯DateTime(Joda)Java库:

DateTime dt = new DateTime(1385577373517L, DateTimeZone.UTC);
// Prints 2013-11-27T18:36:13.517Z
System.out.println(dt);

// Prints 2013-11-27T18:36:00.000Z (Floor rounded to a minute)
System.out.println(dt.minuteOfDay().roundFloorCopy());

// Prints 2013-11-27T18:30:00.000Z (Rounded to custom minute Window)
int windowMinutes = 10;
System.out.println(
    dt.withMinuteOfHour((dt.getMinuteOfHour() / windowMinutes) * windowMinutes)
        .minuteOfDay().roundFloorCopy()
    );

Answer 3

我曾经攻击过这种方法来做类似的事情.它没有以任何方式进行优化,但它确实做了我想要的.从未在任何生产环境中制造过,我无法告诉你有关性能的任何信息.

@Test
     public void test() {
         System.out.println(roundDate(new DateTime().withMinuteOfHour(13)));
         System.out.println(roundDate(new DateTime().withMinuteOfHour(48)));
         System.out.println(roundDate(new DateTime().withMinuteOfHour(0)));
         System.out.println(roundDate(new DateTime().withMinuteOfHour(59)));
         System.out.println(roundDate(new DateTime().withMinuteOfHour(22)));
         System.out.println(roundDate(new DateTime().withMinuteOfHour(37)));
     }

    private DateTime roundDate(final DateTime dateTime) {
        final double minuteOfHour = dateTime.getMinuteOfHour();
        final double tenth = minuteOfHour / 10;
        final long round = Math.round(tenth);
        final int i = (int) (round * 10);

        if (i == 60) {
            return dateTime.plusHours(1).withMinuteOfHour(0);
        } else {
            return dateTime.withMinuteOfHour(i);
        }

    }

Answer 4

这是另一种在Unix时间使用算术来完成的方法:

(为了清楚起见,在Scala中实现.)

import org.joda.time.{DateTime, Duration}

def roundDateTime(t: DateTime, d: Duration) = {
  t minus (t.getMillis - (t.getMillis.toDouble / d.getMillis).round * d.getMillis)
}

用法示例:

roundDateTime(new DateTime("2013-06-27T11:32:00"), Duration.standardMinutes(10))
// => 2013-06-27T11:30:00.000+02:00

roundDateTime(new DateTime("2013-06-27T11:37:00"), Duration.standardMinutes(10))
// => 2013-06-27T11:40:00.000+02:00

Answer 5

当你想要围绕Joda DateTime时,最好的解决方案是IMHO使用内置roundHalfCeilingCopy和roundHalfFloorCopy方法:

DateTime dateTime = DateTime.now();
DateTime newDateTime = dateTime.minuteOfHour().roundHalfCeilingCopy();

请注意,roundHalfCeilingCopy如果中途将有利于上限.你可以使用roundHalfFloorCopy,以便在中途的情况下支持地板.