按组日期范围,计数和排序LINQ中的每个组

时间:2013-09-17 14:07:35

标签: c# linq

我有一个存储在我的对象中的日期集合。这是样本数据。实时,日期将来自服务电话,我将不知道将返回多少日期和数量:

        var ListHeader = new List<ListHeaderData>
        {
            new ListHeaderData
            {
                EntryDate = new DateTime(2013, 8, 26)
            },
            new ListHeaderData
            {
                EntryDate = new DateTime(2013, 9, 11)
            },
            new ListHeaderData
            {
                EntryDate = new DateTime(2013, 1, 1)
            },
            new ListHeaderData
            {
                EntryDate = new DateTime(2013, 9, 15)
            },
            new ListHeaderData
            {
                EntryDate = new DateTime(2013, 9, 17)
            },
            new ListHeaderData
            {
                EntryDate = new DateTime(2013, 9, 5)
            },
        };

我现在需要按照日期范围进行分组:

   Today (1) <- contains the date 9/17/2013 and count of 1
   within 2 weeks (3) <- contains dates 9/15,9/11,9/5 and count of 3
   More than 2 weeks (2) <- contains dates 8/26, 1/1 and count of 2

这是我的LINQ声明,它没有实现我的需要,但我认为我在球场(如果我不是那么善良):

                   var defaultGroups = from l in ListHeader
                                    group l by l.EntryDate into g
                                    orderby g.Min(x => x.EntryDate)
                                    select new { GroupBy = g };

按个别日期分组,所以我有6个组,每个组有1个日期。如何按每个组中的日期范围,计数和排序进行分组?提前感谢。

4 个答案:

答案 0 :(得分:5)

这样做怎么样?

引入一个新属性进行分组和分组。

class ListHeaderData
{
    public DateTime EntryDate;
    public int DateDifferenceFromToday
    {
        get
        {
            TimeSpan difference = DateTime.Today - EntryDate.Date;
            if (difference.TotalDays == 0)//today
            {
                return 1;
            }
            else if (difference.TotalDays <= 14)//less than 2 weeks
            {
                return 2;
            }
            else
            {
                return 3;//something else
            }
        }
    }
}

编辑:正如@servy在评论中指出的那样,其他开发人员可能会使用int混淆enum将更具可读性。

所以,你的类的修改版本看起来像这样

class ListHeaderData
{
    public DateTime EntryDate;
    public DateRange DateDifferenceFromToday
    {
        get
        {
            //I think for this version no comments needed names are self explanatory
            TimeSpan difference = DateTime.Today - EntryDate.Date;
            if (difference.TotalDays == 0)
            {
                return DateRange.Today;
            }
            else if (difference.TotalDays <= 14)
            {
                return DateRange.LessThanTwoWeeks;
            }
            else
            {
                return DateRange.MoreThanTwoWeeks;
            }
        }
    }
}

enum DateRange
{ 
    None = 0,
    Today = 1,
    LessThanTwoWeeks = 2,
    MoreThanTwoWeeks = 3
}

并像这样使用

 var defaultGroups = from l in ListHeader
     group l by l.DateDifferenceFromToday into g // <--Note group by DateDifferenceFromToday
     orderby g.Min(x => x.EntryDate)
     select new { GroupBy = g };

答案 1 :(得分:5)

介绍数组,其中包含您要分组的范围。这是两个范围 - 今天(零天)和14天(两周):

var today = DateTime.Today;
var ranges = new List<int?> { 0, 14 };

现在按照它所属的范围对您的项目进行分组。如果没有合适的范围(所有日期超过两周),则将使用默认的null范围值:

var defaultGroups = 
      from h in ListHeader
      let daysFromToday = (int)(today - h.EntryDate).TotalDays
      group h by ranges.FirstOrDefault(range => daysFromToday <= range) into g
      orderby g.Min(x => x.EntryDate)
      select g;

更新:添加自定义范围以进行分组:

var ranges = new List<int?>();
ranges.Add(0); // today
ranges.Add(7*2); // two weeks
ranges.Add(DateTime.Today.Day); // within current month
ranges.Add(DateTime.Today.DayOfYear); // within current year
ranges.Sort();

答案 2 :(得分:1)

您是否特意想以这种方式实现解决方案?您是否真的想在课堂上引入虚假属性以满足这些要求?

这三行将满足您的要求,对于大型集合来说,性能会更高。

var todays = listHeader.Where(item => item.EntryDate == DateTime.Today);

var twoWeeks = listHeader.Where(item => item.EntryDate < DateTime.Today.AddDays(-1) 
                                && item.EntryDate >= DateTime.Today.AddDays(-14));

var later = listHeader.Where(item => item.EntryDate < DateTime.Today.AddDays(-14));

您也可以在不影响课程的情况下获得不同分组的灵活性。

[编辑:响应订购查询]

使用上面提供的Enum,您可以应用Union子句和OrderBy子句Linq扩展方法,如下所示:

var ord = todays.Select(item => new {Group = DateRange.Today, item.EntryDate})
          .Union(
          twoWeeks.Select(item => new {Group = DateRange.LessThanTwoWeeks, item.EntryDate}))
          .Union(
          later.Select(item => new {Group = DateRange.MoreThanTwoWeeks, item.EntryDate}))
          .OrderBy(item => item.Group);

请注意,我正在通过Linq Select和匿名类添加分组,以再次动态推送Group属性而不影响原始类。这将根据原始帖子生成以下输出:

Group            EntryDate
Today            17/09/2013 00:00:00
LessThanTwoWeeks 11/09/2013 00:00:00
LessThanTwoWeeks 15/09/2013 00:00:00 
LessThanTwoWeeks 05/09/2013 00:00:00 
MoreThanTwoWeeks 26/08/2013 00:00:00 
MoreThanTwoWeeks 01/01/2013 00:00:00 

并获取带有计数的分组日期范围:

var ord = todays.Select(item => new {Group = DateRange.Today, Count=todays.Count()})
          .Union(
          twoWeeks.Select(item => new {Group = DateRange.LessThanTwoWeeks, Count=twoWeeks.Count()}))
          .Union(
          later.Select(item => new {Group = DateRange.MoreThanTwoWeeks, Count=later.Count()}))
          .OrderBy(item => item.Group);

输出是:


Group            Count
Today            1 
LessThanTwoWeeks 3 
MoreThanTwoWeeks 2 

答案 3 :(得分:0)

我想这取决于你计划使用它的程度。我生成了很多报告,所以我创建了一个模型IncrementDateRangeStartTimeEndTimeTimeIncrement作为枚举。

时间增量处理程序有很多基于开关的功能,根据小时/天/周/月/季度/年等等,在开始和结束范围之间吐出一系列时间。

然后你得到IncrementDateRange的列表和linq中的任何一个:

TotalsList = times.Select(t => new RetailSalesTotalsListItem()
{
    IncrementDateRange = t,
    Total = storeSales.Where(s => s.DatePlaced >= t.StartTime && s.DatePlaced <= t.EndTime).Sum(s => s.Subtotal),
})

TotalsList = storeSales.GroupBy(g => g.IncrementDateRange.StartTime).Select(gg => new RetailSalesTotalsListItem()
{
    IncrementDateRange = times.First(t => t.StartTime == gg.Key),
    Total = gg.Sum(rs => rs.Subtotal),
}).ToList(),