如何将时间四舍五入到最接近的 X 分钟？

Question

是否有一个简单的函数可以将 DateTime 向上舍入到最接近的 15 分钟？

例如

2011-08-11 16:59 变为 2011-08-11 17:00

2011-08-11 17:00 保持为 <代码>2011-08-11 17:00

2011-08-11 17:01 变为2011-08-11 17:15

Answer 1

DateTime RoundUp(DateTime dt, TimeSpan d)
{
    return new DateTime((dt.Ticks + d.Ticks - 1) / d.Ticks * d.Ticks, dt.Kind);
}

示例：

var dt1 = RoundUp(DateTime.Parse("2011-08-11 16:59"), TimeSpan.FromMinutes(15));
// dt1 == {11/08/2011 17:00:00}

var dt2 = RoundUp(DateTime.Parse("2011-08-11 17:00"), TimeSpan.FromMinutes(15));
// dt2 == {11/08/2011 17:00:00}

var dt3 = RoundUp(DateTime.Parse("2011-08-11 17:01"), TimeSpan.FromMinutes(15));
// dt3 == {11/08/2011 17:15:00}

Answer 2

想出了一个不涉及乘法和除法长数字的解决方案。

public static DateTime RoundUp(this DateTime dt, TimeSpan d)
{
    var modTicks = dt.Ticks % d.Ticks;
    var delta = modTicks != 0 ? d.Ticks - modTicks : 0;
    return new DateTime(dt.Ticks + delta, dt.Kind);
}

public static DateTime RoundDown(this DateTime dt, TimeSpan d)
{
    var delta = dt.Ticks % d.Ticks;
    return new DateTime(dt.Ticks - delta, dt.Kind);
}

public static DateTime RoundToNearest(this DateTime dt, TimeSpan d)
{
    var delta = dt.Ticks % d.Ticks;
    bool roundUp = delta > d.Ticks / 2;
    var offset = roundUp ? d.Ticks : 0;

    return new DateTime(dt.Ticks + offset - delta, dt.Kind);
}

用法：

var date = new DateTime(2010, 02, 05, 10, 35, 25, 450); // 2010/02/05 10:35:25
var roundedUp = date.RoundUp(TimeSpan.FromMinutes(15)); // 2010/02/05 10:45:00
var roundedDown = date.RoundDown(TimeSpan.FromMinutes(15)); // 2010/02/05 10:30:00
var roundedToNearest = date.RoundToNearest(TimeSpan.FromMinutes(15)); // 2010/02/05 10:30:00

Answer 3

我已经看到了许多有用的实现，例如来自 @dtb 或 @redent84 的实现。
由于性能差异可以忽略不计，因此我远离位移并简单地创建了可读代码。我经常在我的实用程序库中使用这些扩展。

public static class DateTimeExtensions
{
    public static DateTime RoundToTicks(this DateTime target, long ticks) => new DateTime((target.Ticks + ticks / 2) / ticks * ticks, target.Kind);
    public static DateTime RoundUpToTicks(this DateTime target, long ticks) => new DateTime((target.Ticks + ticks - 1) / ticks * ticks, target.Kind);
    public static DateTime RoundDownToTicks(this DateTime target, long ticks) => new DateTime(target.Ticks / ticks * ticks, target.Kind);

    public static DateTime Round(this DateTime target, TimeSpan round) => RoundToTicks(target, round.Ticks);
    public static DateTime RoundUp(this DateTime target, TimeSpan round) => RoundUpToTicks(target, round.Ticks);
    public static DateTime RoundDown(this DateTime target, TimeSpan round) => RoundDownToTicks(target, round.Ticks);

    public static DateTime RoundToMinutes(this DateTime target, int minutes = 1) => RoundToTicks(target, minutes * TimeSpan.TicksPerMinute);
    public static DateTime RoundUpToMinutes(this DateTime target, int minutes = 1) => RoundUpToTicks(target, minutes * TimeSpan.TicksPerMinute);
    public static DateTime RoundDownToMinutes(this DateTime target, int minutes = 1) => RoundDownToTicks(target, minutes * TimeSpan.TicksPerMinute);

    public static DateTime RoundToHours(this DateTime target, int hours = 1) => RoundToTicks(target, hours * TimeSpan.TicksPerHour);
    public static DateTime RoundUpToHours(this DateTime target, int hours = 1) => RoundUpToTicks(target, hours * TimeSpan.TicksPerHour);
    public static DateTime RoundDownToHours(this DateTime target, int hours = 1) => RoundDownToTicks(target, hours * TimeSpan.TicksPerHour);

    public static DateTime RoundToDays(this DateTime target, int days = 1) => RoundToTicks(target, days * TimeSpan.TicksPerDay);
    public static DateTime RoundUpToDays(this DateTime target, int days = 1) => RoundUpToTicks(target, days * TimeSpan.TicksPerDay);
    public static DateTime RoundDownToDays(this DateTime target, int days = 1) => RoundDownToTicks(target, days * TimeSpan.TicksPerDay);
}

Answer 4

如果您需要四舍五入到最近的时间间隔（而不是向上）
那么我建议使用以下内容

    static DateTime RoundToNearestInterval(DateTime dt, TimeSpan d)
    {
        int f=0;
        double m = (double)(dt.Ticks % d.Ticks) / d.Ticks;
        if (m >= 0.5)
            f=1;            
        return new DateTime(((dt.Ticks/ d.Ticks)+f) * d.Ticks);
    }

Answer 5

void Main()
{
    var date1 = new DateTime(2011, 8, 11, 16, 59, 00);
    date1.Round15().Dump();

    var date2 = new DateTime(2011, 8, 11, 17, 00, 02);
    date2.Round15().Dump();

    var date3 = new DateTime(2011, 8, 11, 17, 01, 23);
    date3.Round15().Dump();

    var date4 = new DateTime(2011, 8, 11, 17, 00, 00);
    date4.Round15().Dump();
}

public static class Extentions
{
    public static DateTime Round15(this DateTime value)
    {   
        var ticksIn15Mins = TimeSpan.FromMinutes(15).Ticks;

        return (value.Ticks % ticksIn15Mins == 0) ? value : new DateTime((value.Ticks / ticksIn15Mins + 1) * ticksIn15Mins);
    }
}

结果：

8/11/2011 5:00:00 PM
8/11/2011 5:15:00 PM
8/11/2011 5:15:00 PM
8/11/2011 5:00:00 PM

Answer 6

由于我讨厌重新发明轮子，因此我可能会遵循此算法将 DateTime 值舍入到指定的时间增量 (Timespan)：

• 将要舍入的 DateTime 值转换为十进制浮点值表示 TimeSpan 单位的整数和小数。
• 使用 Math.Round() 将其舍入为整数。
• 通过将四舍五入的整数乘以 TimeSpan 单位中的刻度数来缩放回刻度。
• 根据四舍五入的刻度数实例化一个新的 DateTime 值并将其返回给调用者。

这是代码：

public static class DateTimeExtensions
{

    public static DateTime Round( this DateTime value , TimeSpan unit )
    {
        return Round( value , unit , default(MidpointRounding) ) ;
    }

    public static DateTime Round( this DateTime value , TimeSpan unit , MidpointRounding style )
    {
        if ( unit <= TimeSpan.Zero ) throw new ArgumentOutOfRangeException("unit" , "value must be positive") ;

        Decimal  units        = (decimal) value.Ticks / (decimal) unit.Ticks ;
        Decimal  roundedUnits = Math.Round( units , style ) ;
        long     roundedTicks = (long) roundedUnits * unit.Ticks ;
        DateTime instance     = new DateTime( roundedTicks ) ;

        return instance ;
    }

}

Answer 7

我的版本

DateTime newDateTimeObject = oldDateTimeObject.AddMinutes(15 - oldDateTimeObject.Minute % 15);

作为一种方法，它会像这样锁定

public static DateTime GetNextQuarterHour(DateTime oldDateTimeObject)
{
    return oldDateTimeObject.AddMinutes(15 - oldDateTimeObject.Minute % 15);
}

并被这样调用

DateTime thisIsNow = DateTime.Now;
DateTime nextQuarterHour = GetNextQuarterHour(thisIsNow);

Answer 8

更详细的解决方案，使用模并避免不必要的计算。

public static class DateTimeExtensions
{
    public static DateTime RoundUp(this DateTime dt, TimeSpan ts)
    {
        return Round(dt, ts, true);
    }

    public static DateTime RoundDown(this DateTime dt, TimeSpan ts)
    {
        return Round(dt, ts, false);
    }

    private static DateTime Round(DateTime dt, TimeSpan ts, bool up)
    {
        var remainder = dt.Ticks % ts.Ticks;
        if (remainder == 0)
        {
            return dt;
        }

        long delta;
        if (up)
        {
            delta = ts.Ticks - remainder;
        }
        else
        {
            delta = -remainder;
        }

        return dt.AddTicks(delta);
    }
}

Answer 9

我的 DateTimeOffset 版本，基于 Ramon 的回答：

public static class DateExtensions
{
    public static DateTimeOffset RoundUp(this DateTimeOffset dt, TimeSpan d)
    {
        return new DateTimeOffset((dt.Ticks + d.Ticks - 1) / d.Ticks * d.Ticks, dt.Offset);
    }
}

Answer 10

优雅的？

dt.AddSeconds(900 - (x.Minute * 60 + x.Second) % 900)

Answer 11

注意：上面的公式是不正确的，即：

DateTime RoundUp(DateTime dt, TimeSpan d)
{
    return new DateTime(((dt.Ticks + d.Ticks - 1) / d.Ticks) * d.Ticks);
}

应重写为：

DateTime RoundUp(DateTime dt, TimeSpan d)
{
    return new DateTime(((dt.Ticks + d.Ticks/2) / d.Ticks) * d.Ticks);
}

Answer 12

这是一个简单的解决方案，可以四舍五入到最接近的 1 分钟。它保留了 DateTime 的 TimeZone 和 Kind 信息。可以对其进行修改以进一步满足您自己的需求（如果您需要四舍五入到最近的 5 分钟等）。

DateTime dbNowExact = DateTime.Now;
DateTime dbNowRound1 = (dbNowExact.Millisecond == 0 ? dbNowExact : dbNowExact.AddMilliseconds(1000 - dbNowExact.Millisecond));
DateTime dbNowRound2 = (dbNowRound1.Second == 0 ? dbNowRound1 : dbNowRound1.AddSeconds(60 - dbNowRound1.Second));
DateTime dbNow = dbNowRound2;

Answer 13

您可以使用此方法，它使用指定的日期来确保它维护先前在日期时间对象中指定的任何全球化和日期时间类型。

const long LNG_OneMinuteInTicks = 600000000;
/// <summary>
/// Round the datetime to the nearest minute
/// </summary>
/// <param name = "dateTime"></param>
/// <param name = "numberMinutes">The number minute use to round the time to</param>
/// <returns></returns>        
public static DateTime Round(DateTime dateTime, int numberMinutes = 1)
{
    long roundedMinutesInTicks = LNG_OneMinuteInTicks * numberMinutes;
    long remainderTicks = dateTime.Ticks % roundedMinutesInTicks;
    if (remainderTicks < roundedMinutesInTicks / 2)
    {
        // round down
        return dateTime.AddTicks(-remainderTicks);
    }

    // round up
    return dateTime.AddTicks(roundedMinutesInTicks - remainderTicks);
}

.Net Fiddle Test

如果你想使用TimeSpan进行舍入，可以使用这个。

/// <summary>
/// Round the datetime
/// </summary>
/// <example>Round(dt, TimeSpan.FromMinutes(5)); => round the time to the nearest 5 minutes.</example>
/// <param name = "dateTime"></param>
/// <param name = "roundBy">The time use to round the time to</param>
/// <returns></returns>        
public static DateTime Round(DateTime dateTime, TimeSpan roundBy)
{            
    long remainderTicks = dateTime.Ticks % roundBy.Ticks;
    if (remainderTicks < roundBy.Ticks / 2)
    {
        // round down
        return dateTime.AddTicks(-remainderTicks);
    }

    // round up
    return dateTime.AddTicks(roundBy.Ticks - remainderTicks);
}

时间跨度小提琴

Answer 14

Ramon Smits 的解决方案，带有 DateTime.MaxValue 检查：

DateTime RoundUp(DateTime dt, TimeSpan d) =>
    dt switch
    {
        var max when max.Equals(DateTime.MaxValue) => max,
        var v => new DateTime((v.Ticks + d.Ticks - 1) / d.Ticks * d.Ticks, v.Kind)
    };

Answer 15

您可以尝试这样做：

string[] parts = ((DateTime)date_time.ToString("HH:mm:ss").Split(':');
int hr = Convert.ToInt32(parts[0]);
int mn = Convert.ToInt32(parts[1]);
int sec2min = (int)Math.Round(Convert.ToDouble(parts[2]) / 60.0, 0);

string adjTime = string.Format("1900-01-01 {0:00}:{1:00}:00.000",
(mn + sec2min > 59 ? (hr + 1 > 23 ? 0 : hr + 1) : hr),
 mn + sec2min > 59 ? 60 - mn + sec2min : mn + sec2min);

每个部分（小时，分钟）必须递增并调整为适当的溢出值，例如 59 分钟 > 00则hr加1，如果是23，hr就变成00。
前任。 07:34:57 舍入为 07:35，09:59:45 舍入为 10:00，23:59:45 舍入为 00:00，即第二天的时间。