Question

是否有一个简单的功能可以将 UP a DateTime四舍五入到最近的15分钟？

E.g。

2011-08-11 16:59变为2011-08-11 17:00

2011-08-11 17:00保持为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：的

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

提出的解决方案不涉及乘以和划分 long数字。

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

如果你需要舍入到最近的时间间隔（不是）那么我建议使用以下

    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 4

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 5

由于我讨厌重新发明轮子，我可能会遵循这个算法将DateTime值四舍五入到指定的时间增量（Timespan）：

将要舍入的DateTime值转换为表示TimeSpan单位的整数和小数的十进制浮点值。
使用Math.Round()将其舍入为整数。
通过将舍入整数乘以TimeSpan单位中的刻度数来缩放回刻度。
从舍入的舍入次数中实例化一个新的DateTime值并将其返回给调用者。

以下是代码：

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 6

我的版本

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

作为一种方法，它会像这样

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

被这样称呼

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

Answer 7

优雅？

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

Answer 8

警告：上面的公式不正确，即以下内容：

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

应改写为：

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

Answer 9

更详细的解决方案，它使用模数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 10

这是一个简单的解决方案，可以最接近1分钟。它保留DateTime的TimeZone和Kind信息。它可以进一步修改以满足您自己的需要（如果您需要舍入到最近的5分钟等）。

try humidity = int(humidity):
    if humidity>75:
        subList[2]="high"
    else:
        subList[2]="low"
except:
    return 'bad values in file - humidity is not a number'

Answer 11

您可以使用此方法，它使用指定的日期来确保它维护以前在datetime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

如果您想使用TimeSpan进行舍入，可以使用它。

/// <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);
}

TimeSpan Fiddle

Answer 12

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

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 13

我的DateTimeOffset版本，基于拉蒙的答案：

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 14

您可以尝试以下方法：

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）必须增加并调整为溢出的适当值，例如59分钟> 00，然后将hr加1，如果为23，则小时变为00。例如将7:34:57舍入为07:35，将9:59:45舍入为10:00，将23:59:45舍入为00:00，这是第二天的时间。

Answer 15

我见过许多有用的实现，比如来自 @dtb 或 @redent84 的实现。由于性能差异可以忽略不计，我远离位移并简单地创建可读代码。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);
}

如何将时间向上舍入到最接近的X分钟？

15 个答案: