C# 在使用双重分组后,如何展平Linq查询?
假设我有一个表示订单行的类,例如C# 在使用双重分组后,如何展平Linq查询?,c#,linq,C#,Linq,假设我有一个表示订单行的类,例如 public class Line { public string Code ; public string No ; // Invoice Number public DateTime Date ; public string Product ; public decimal Quantity ; } 和一系列行,如 List<Line> myList = new List<Line>(); m
public class Line
{
public string Code ;
public string No ; // Invoice Number
public DateTime Date ;
public string Product ;
public decimal Quantity ;
}
和一系列行,如
List<Line> myList = new List<Line>();
myList.Add(new Line() { Code = "ABC001", No = "1001" ,Date = new DateTime(2012,4,1) , Product = "X", Quantity= 1m});
myList.Add(new Line() { Code = "ABC001", No = "1001" ,Date = new DateTime(2012,4,1) , Product = "Y", Quantity= 1m});
myList.Add(new Line() { Code = "ABC002", No = "1002" ,Date = new DateTime(2012,4,2) , Product = "X", Quantity= 1m});
myList.Add(new Line() { Code = "ABC002", No = "1002" ,Date = new DateTime(2012,4,2) , Product = "Y", Quantity= 1m});
myList.Add(new Line() { Code = "ABC002", No = "1003" ,Date = new DateTime(2012,4,3) , Product = "Z", Quantity= 1m});
myList.Add(new Line() { Code = "ABC002", No = "1004" ,Date = new DateTime(2012,4,4) , Product = "X", Quantity= 1m});
myList.Add(new Line() { Code = "ABC003", No = "1005" ,Date = new DateTime(2012,4,4) , Product = "X", Quantity= 1m});
myList.Add(new Line() { Code = "ABC003", No = "1006" ,Date = new DateTime(2012,4,4) , Product = "X", Quantity= 1m});
myList.Add(new Line() { Code = "ABC003", No = "1006" ,Date = new DateTime(2012,4,4) , Product = "Y", Quantity= 1m});
query1现在包含正确的记录,但被包装在iGroup中,我很难将结果作为列表导出
我试过了
query1.SelectMany(r=>r.Results.ToList()
但这仍然让我不得不分组,这就是我被困的地方
我可以使用嵌套for循环,如中所示
List<Line> output = new List<Line>();
foreach (var r1 in query1)
{
foreach(var r2 in r1.Results)
foreach(var r3 in r2)
output.Add(r3);
}
你要踢自己了:
query1.SelectMany(q => q);
ABC002 1003 3/04/2012 12:00:00 AM Z 1
ABC002 1004 4/04/2012 12:00:00 AM X 1
ABC003 1006 4/04/2012 12:00:00 AM X 1
ABC003 1006 4/04/2012 12:00:00 AM Y 1
来自query1
的返回是IGrouping
的可枚举(我删除了您的列表),而IGrouping
本身就是可枚举的,因此我们可以直接将其展平
请看这里:
编辑:记得我还简化了您的代码:
var query1 =
(from r in myList
group r by new { r.Code , r.No , r.Date } into results
group results by new { results.Key.Code } into results2
where results2.Count() > 1
from result in results2.OrderBy(i=>i.Key.Date).Skip(1)
select result
);
试试这个:
var flattenedLines = from outerGroup in query1
from innerGroup in outerGroup.Results
from line in innerGroup
select line;
或
此代码:
List<Line> output = new List<Line>();
foreach (var r1 in query1)
foreach(var r2 in r1.Results)
foreach(var r3 in r2)
output.Add(r3);
我会将results2.ToList().OrderBy(I=>I.Key.Date).跳过(1).ToList()
更改为results2.OrderBy(I=>I.Key.Date).跳过(1).ToList()
。没有理由第一次调用ToList()
。@phoog我已经删除了第一个ToList(),因为你是正确的,因为它不需要,但也不会影响结果,所以我仍然有我原来的问题。是的,我把它作为一个注释发布,因为它没有回答你的问题。我还在下面贴了一个答案。@NiklasB。否,因为outerGroup.Results是IEnumerable
,其中TKey
是定义键的匿名类型。我认为这是OP最初的问题。Yamen展示了如何通过简化query1
来简化扁平化查询,但是如果我们保持query1
不变,我们必须有两个SelectMany调用,或者从
子句中有三个。如果query1.SelectMany(q=>q)在简化我的查询之前起作用,那么我肯定会受到严厉的惩罚,但说实话,我不认为我自己能做到这一点。谢谢。是的,我忘了我简化了查询。在此之前,@phoog给出了正确的答案。不管怎样,诀窍是认识到i分组
实现了i编号
。
var flattenedLines = from outerGroup in query1
from innerGroup in outerGroup.Results
from line in innerGroup
select line;
var flattenedLines = query1
.SelectMany(outerGroup => outerGroup.Results, (outerGroup, innerGroup) => innerGroup)
.SelectMany(x => x);
List<Line> output = new List<Line>();
foreach (var r1 in query1)
foreach(var r2 in r1.Results)
foreach(var r3 in r2)
output.Add(r3);
var q2 = from r1 in query1
from r2 in r1.Results
from r3 in r2
select r3;
var output = q2.ToList();