LINQ GroupBy（）由不同的算法

Question

我需要对数据进行分组，以便通过条件获得不同的结果

我的实体是：

public class MyTestCollection {

    public Int32 Id { get; set; }

    public List<MyTestItem> Tests { get; set; }

}

public class MyTestItem {

    public long Id { get; set; }

    public List<DateItem> Dates { get; set; }

    public MyTestItem(long id, List<DateItem> list) {
        Id = id;
        Dates = list;
    }

}

public class DateItem {

    public DateTime? DateValue { get; set; }

    public DateTime? DateValueOne { get; set; }

    public DateTime? DateFrom { get; set; }

    public DateTime? DateTo { get; set; }

    public RateItem Rate { get; set; }

    public DateItem(DateTime date, RateItem rate) {
        DateValue = date;
        Rate = rate;
    }

    public DateItem(KeyValuePair<DateTime, DateTime> dates , RateItem rate, bool isFormat = true) {
        if (isFormat) {
            DateValue = dates.Key;
            DateValueOne = dates.Value;
        } else {
            DateFrom = dates.Key;
            DateTo = dates.Value;
        }
        Rate = rate;
    }
}

public class RateItem {

    public Int32 Value1 { get; set; }

    public Int32 Value2 { get; set; }

    public Int32 Value3 { get; set; }

    public RateItem(int value1, int value2, int value3) {
        Value1 = value1;
        Value2 = value2;
        Value3 = value3;
    }
}

我需要以不同的方式从数据表中填充它。现在我将实体DataItem填入我从表中得到的一个日期 - 每天我有价值的意思，我想知道如果RateItem相同的话，我会知道如何填写它很多天。在我的示例中，需要实现构造函数public DateItem(KeyValuePair<DateTime, DateTime> dates , RateItem rate, bool isFormat = true)的登录。只有一个人希望它应该是一个完整的linq，顺序如何从表中获取

现在的逻辑是：

    public List<MyTestCollection> GetItems() {
        var item = GetData().AsEnumerable().GroupBy(row => new { ID = row.Field<int>("MainId") })
         .Select(
         testColGroup =>
         new MyTestCollection()
         {
             Id = testColGroup.Key.ID,
             Tests = testColGroup.GroupBy(row => new { Id = row.Field<int>("Id") })
                 .Select(itemGroup => new MyTestItem(itemGroup.Key.Id,
                     itemGroup.GroupBy(row => row.Field<DateTime>("date"))
                     .Select(dategroup => new DateItem(dategroup.Key.Date, 
                          new RateItem(dategroup.Select(i => i.Field<int>("value1")).FirstOrDefault()
                              , dategroup.Select(j => j.Field<int>("value2")).FirstOrDefault()
                              , dategroup.Select(j => j.Field<int>("value2")).FirstOrDefault())))
                     .ToList()))
                 .ToList()
         }).ToList();

        return item;  

我的测试数据表：

    static DataTable GetTable() {
        var table = new DataTable();
        table.Columns.Add("MainId", typeof(Int32));
        table.Columns.Add("Id", typeof(Int32));
        table.Columns.Add("Date",typeof(DateTime));
        table.Columns.Add("value1", typeof(Int32));
        table.Columns.Add("value2", typeof(Int32));
        table.Columns.Add("value3", typeof(Int32));

        // Row 1
        var row = table.NewRow();

        row["MainId"] = 1;
        row["Id"] = 1;
        row["Date"] = "06/10/2015";
        row["value1"] = 558;
        row["value2"] = 206;
        row["value3"] = 352;

        table.Rows.Add(row);

        // Row 2
        row = table.NewRow();
        row["MainId"] = 1;
        row["Id"] = 1;
        row["Date"] = "06/11/2015";
        row["value1"] = 558;
        row["value2"] = 206;
        row["value3"] = 352;
        table.Rows.Add(row);

        // Row 3
        row = table.NewRow();
        row["MainId"] = 1;
        row["Id"] = 1;
        row["Date"] = "06/12/2015";
        row["value1"] = 558;
        row["value2"] = 206;
        row["value3"] = 352;

        table.Rows.Add(row);

        return table;
    }

我希望找到解决方案也有帮助

UPD：

这是可能的表

MainId  Id  Date        value1  value2
1       1   6/10/2015   558     206
1       1   6/11/2015   558     206
1       1   6/12/2015   111     111
1       2   6/10/2015   111     206
1       2   6/11/2015   111     206
1       2   6/12/2015   558     206
1       2   6/13/2015   558     206

所有这一切：

    <request>
        <main>1</main>
        <sku id="1">
            **<date datefrom="6/10/2015" dateto="6/11/2015">**          
                <value1>558</value1>
                <value2>206</value2>
            </date>
            **<date value="6/12/2015">**            
                <value1>111</value1>
                <value2>111</value2>
            </date>
        </sku>
        <sku id="2">
            <date datefrom="6/10/2015" dateto="6/11/2015">
                <value1>111</value1>
                <value2>206</value2>
            </date>
            <date datefrom="6/12/2015" dateto="6/13/2015">
                <value1>558</value1>
                <value2>206</value2>
            </date>
        </sku>
    </request>