I m试图了解如何利用LINQ每隔一段时间对数据进行分类;然后理想的是,将每一组数据集中起来。
寻找许多有明确日期范围的实例,我试图按5分钟、1小时、1天等时间分组。
例如,我有一个总结一个有价值的日期的班子:
public class Sample
{
public DateTime timestamp;
public double value;
}
这些意见载于一份清单中:
List<Sample> series;
因此,按小时计算,按平均总价值计算,Im试图做以下事情:
var grouped = from s in series
group s by new TimeSpan(1, 0, 0) into g
select new { timestamp = g.Key, value = g.Average(s => s.value };
这从根本上说是有缺陷的,因为它把时间范围本身混为一谈。 我无法理解在询问中如何使用时间Span(或代表间隔的任何数据类型)。
你们可以把时间印记到下一个边界(即过去接近5分钟的边界)上,并以此为主:
var groups = series.GroupBy(x =>
{
var stamp = x.timestamp;
stamp = stamp.AddMinutes(-(stamp.Minute % 5));
stamp = stamp.AddMilliseconds(-stamp.Millisecond - 1000 * stamp.Second);
return stamp;
})
.Select(g => new { TimeStamp = g.Key, Value = g.Average(s => s.value) })
.ToList();
更重要的是,通过在组别中使用经过修改的时间印章,将会议记录排在前5分钟的边界上,并删除第二和小秒。 当然,其他时间,即小时和日,也可以采用同样的方法。
<><>Edit>:
根据这一构成的抽样投入:
var series = new List<Sample>();
series.Add(new Sample() { timestamp = DateTime.Now.AddMinutes(3) });
series.Add(new Sample() { timestamp = DateTime.Now.AddMinutes(4) });
series.Add(new Sample() { timestamp = DateTime.Now.AddMinutes(5) });
series.Add(new Sample() { timestamp = DateTime.Now.AddMinutes(6) });
series.Add(new Sample() { timestamp = DateTime.Now.AddMinutes(7) });
series.Add(new Sample() { timestamp = DateTime.Now.AddMinutes(15) });
为我制作了3个小组,其中1个分组时间为3:05,1个有3:10,1个有3:20皮克(根据目前时间,结果可能有所不同)。
我很晚才参加这一游戏,但我是在寻找其他东西时走到这里的,我认为我有更好的办法。
series.GroupBy (s => s.timestamp.Ticks / TimeSpan.FromHours(1).Ticks)
.Select (s => new {
series = s
,timestamp = s.First ().timestamp
,average = s.Average (x => x.value )
}).Dump();
这里是一个样本子板方案,以便你能够验证和测试。
void Main()
{
List<Sample> series = new List<Sample>();
Random random = new Random(DateTime.Now.Millisecond);
for (DateTime i = DateTime.Now.AddDays(-5); i < DateTime.Now; i += TimeSpan.FromMinutes(1))
{
series.Add(new UserQuery.Sample(){ timestamp = i, value = random.NextDouble() * 100 });
}
//series.Dump();
series.GroupBy (s => s.timestamp.Ticks / TimeSpan.FromHours(1).Ticks)
.Select (s => new {
series = s
,timestamp = s.First ().timestamp
,average = s.Average (x => x.value )
}).Dump();
}
// Define other methods and classes here
public class Sample
{
public DateTime timestamp;
public double value;
}
For grouping by hour you need to group by the hour part of your timestamp which could be done as so:
var groups = from s in series
let groupKey = new DateTime(s.timestamp.Year, s.timestamp.Month, s.timestamp.Day, s.timestamp.Hour, 0, 0)
group s by groupKey into g select new
{
TimeStamp = g.Key,
Value = g.Average(a=>a.value)
};
I d建议使用newtime()至, 避免 任何 限值低于秒的问题。
var versionsGroupedByRoundedTimeAndAuthor = db.Versions.GroupBy(g =>
new
{
UserID = g.Author.ID,
Time = RoundUp(g.Timestamp, TimeSpan.FromMinutes(2))
});
页: 1
private DateTime RoundUp(DateTime dt, TimeSpan d)
{
return new DateTime(((dt.Ticks + d.Ticks - 1) / d.Ticks) * d.Ticks);
}
N.B.,我在这里由作者归类。 ID as well as the Rounded TimeStamp.
圆桌会议 https://stackoverflow.com/a/7029464/661584“>。
Read about how equality down to the millisecond doesn t always mean equality here Why does this unit test fail when testing DateTime equality?
我改进了《布罗肯格拉斯》的答案,使之更加笼统,增加了保障。 如果你选择了9年的间隔,那么你将不做预期的事情。 任何60人的情况都不相同。 例如,Im使用9,从午夜开始(0:00)。
For me, this is a massive issue.
I m not sure how to fix that, but you can add safeguards.
Changes:
工作时间也不同。
double minIntervalAsDouble = Convert.ToDouble(minInterval);
if (minIntervalAsDouble <= 0)
{
string message = "minInterval must be a positive number, exiting";
Log.getInstance().Info(message);
throw new Exception(message);
}
else if (minIntervalAsDouble < 60.0 && 60.0 % minIntervalAsDouble != 0)
{
string message = "60 must be divisible by minInterval...exiting";
Log.getInstance().Info(message);
throw new Exception(message);
}
else if (minIntervalAsDouble >= 60.0 && (24.0 % (minIntervalAsDouble / 60.0)) != 0 && (24.0 % (minIntervalAsDouble / 60.0) != 24.0))
{
//hour part must be divisible...
string message = "If minInterval is greater than 60, 24 must be divisible by minInterval/60 (hour value)...exiting";
Log.getInstance().Info(message);
throw new Exception(message);
}
var groups = datas.GroupBy(x =>
{
if (minInterval < 60)
{
var stamp = x.Created;
stamp = stamp.AddMinutes(-(stamp.Minute % minInterval));
stamp = stamp.AddMilliseconds(-stamp.Millisecond);
stamp = stamp.AddSeconds(-stamp.Second);
return stamp;
}
else
{
var stamp = x.Created;
int hourValue = minInterval / 60;
stamp = stamp.AddHours(-(stamp.Hour % hourValue));
stamp = stamp.AddMilliseconds(-stamp.Millisecond);
stamp = stamp.AddSeconds(-stamp.Second);
stamp = stamp.AddMinutes(-stamp.Minute);
return stamp;
}
}).Select(o => new
{
o.Key,
min = o.Min(f=>f.Created),
max = o.Max(f=>f.Created),
o
}).ToList();
选择发言中你喜欢的东西! 我说得min,因为比较容易测试。
尽管我确实很晚,但这里是我的两点:
我想四舍五入到五分钟的间隔时间:
10:31 --> 10:30
10:33 --> 10:35
10:36 --> 10:35
可以通过转换为时段来实现这一点。 计算和转换为日期,使用Math.Round():
public DateTime GetShiftedTimeStamp(DateTime timeStamp, int minutes)
{
return
new DateTime(
Convert.ToInt64(
Math.Round(timeStamp.Ticks / (decimal)TimeSpan.FromMinutes(minutes).Ticks, 0, MidpointRounding.AwayFromZero)
* TimeSpan.FromMinutes(minutes).Ticks));
}
如上所示,改换时Stamp可用于气球组群。
我知道,这只是直接回答了问题,但我正在寻找一个非常相似的解决办法,从一个小的时期到一个更长的时期(5、10、15、30)对库存/密码货币的中值数据进行汇总。 你们只能从现在的十分钟回来,因为总合时期的顶点是稳定的。 你们还必须看到,在清单的开始和结束时,有足够的数据来全面标明较长时间。 有鉴于此,我提出的解决办法如下。 (它假设,如原产地所示,较小时期的蜡dle按装饰时序排列)
public class Candle
{
public long Id { get; set; }
public Period Period { get; set; }
public DateTime Timestamp { get; set; }
public double High { get; set; }
public double Low { get; set; }
public double Open { get; set; }
public double Close { get; set; }
public double BuyVolume { get; set; }
public double SellVolume { get; set; }
}
public enum Period
{
Minute = 1,
FiveMinutes = 5,
QuarterOfAnHour = 15,
HalfAnHour = 30
}
private List<Candle> AggregateCandlesIntoRequestedTimePeriod(Period rawPeriod, Period requestedPeriod, List<Candle> candles)
{
if (rawPeriod != requestedPeriod)
{
int rawPeriodDivisor = (int) requestedPeriod;
candles = candles
.GroupBy(g => new { TimeBoundary = new DateTime(g.Timestamp.Year, g.Timestamp.Month, g.Timestamp.Day, g.Timestamp.Hour, (g.Timestamp.Minute / rawPeriodDivisor) * rawPeriodDivisor , 0) })
.Where(g => g.Count() == rawPeriodDivisor )
.Select(s => new Candle
{
Period = requestedPeriod,
Timestamp = s.Key.TimeBoundary,
High = s.Max(z => z.High),
Low = s.Min(z => z.Low),
Open = s.First().Open,
Close = s.Last().Close,
BuyVolume = s.Sum(z => z.BuyVolume),
SellVolume = s.Sum(z => z.SellVolume),
})
.OrderBy(o => o.Timestamp)
.ToList();
}
return candles;
}
A generalised solution:
static IEnumerable<IGrouping<DateRange, T>> GroupBy<T>(this IOrderedEnumerable<T> enumerable, TimeSpan timeSpan, Func<T, DateTime> predicate)
{
Grouping<T> grouping = null;
foreach (var (a, dt) in from b in enumerable select (b, predicate.Invoke(b)))
{
if (grouping == null || dt > grouping.Key.End)
yield return grouping = new Grouping<T>(new DateRange(dt, dt + timeSpan), a);
else
grouping.Add(a);
}
}
class Grouping<T> : IGrouping<DateRange, T>
{
readonly List<T> elements = new List<T>();
public DateRange Key { get; }
public Grouping(DateRange key) => Key = key;
public Grouping(DateRange key, T element) : this(key) => Add(element);
public void Add(T element) => elements.Add(element);
public IEnumerator<T> GetEnumerator()=> this.elements.GetEnumerator();
IEnumerator IEnumerable.GetEnumerator() => GetEnumerator();
}
class DateRange
{
public DateRange(DateTime start, DateTime end)
{
this.Start = start;
this.End = end;
}
public DateTime Start { get; set; }
public DateTime End { get; set; }
}
基于问题的测试(使用自动图书馆)
void Test()
{
var many = new Fixture().CreateMany<Sample>(100);
var groups = many.OrderBy(a => a.timestamp).GroupBy(TimeSpan.FromDays(365), a => a.timestamp).Select(a => a.Average(b => b.value)).ToArray();
}
public class Sample
{
public DateTime timestamp;
public double value;
}
如果《蓝色地图册》没有发挥作用,那么你也可以在价值上打上标准。 本文是一份工作版本:
var groups = series.GroupBy(x =>
{
var stamp = x.timestamp;
stamp = stamp.AddMinutes(-(stamp.Minute % 5));
stamp = stamp.AddTicks(-(stamp.Ticks % TimeSpan.TicksPerMinute));
return stamp;
})
.Select(g => new { TimeStamp = g.Key, Value = g.Average(s => s.value) })
.ToList();
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