这是我的类结构的简化版本:
public class TransferObject
{
public List<Event> events { get; set; }
public TransferObject()
{
events = new List<Event>();
}
}
public class Event
{
public List<Data> eventData { get; set; }
public Event()
{
eventData = new List<Data>();
}
}
public class Data
{
public int someData {get; set;}
public DateTime date { get; set; }
}
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我想要做的是获取列表Data中的所有对象eventData,并将它们放入一个新的分组列表中date.
我考虑将所有Data对象放入临时列表,然后将该列表与Linq分组,但我想知道是否有更优雅的方式来执行它而不使用临时列表.
如果我明白你想要什么,这样的事情应该有效:
var results =
from e in transferObject.events
from d in e.eventData
group d by d.date;
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或者用流利的语法:
var results = transferObject.events
.SelectMany(e => e.eventData)
.GroupBy(d => d.date);
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嗯,这取决于您要查找的数据。如果您不关心实际的日期值,只关心整数按它们分组(我怀疑也按它们排序),那么从对 Event 类的引用中您将拥有:
List<List<int>> byDate = evt.eventData
.GroupBy(d => d.date)
.OrderBy(g=>g.Key)
.Select(g => g.Select(v => v.someData)
.ToList())
.ToList();
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但是,如果您关心日期,则可以采用两种方法。您最终可以放弃对 Data 类的任何引用:
Dictionary<DateTime, List<int>> byDate2 = evt.eventData
.GroupBy(d => d.date)
.OrderBy(g=>g.Key)
.ToDictionary(
g=>g.Key,
g => g.Select(v => v.someData)
.ToList());
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或者您可以保留数据对象:
List<List<Data>> byDate = evt.eventData
.GroupBy(d => d.date)
.OrderBy(g => g.Key)
.Select(g => g.ToList())
.ToList();
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但假设您想通过对 TransferObject 类的引用来执行此操作。然后,您将执行几乎相同的操作,但需要使用 SelectMany 函数:
TransferObject txobj = new TransferObject(); // however this happens
List<List<int>> byDate = txobj.events
.SelectMany(evt=>evt.eventData
.GroupBy(d => d.date)
.OrderBy(g => g.Key)
.Select(g => g.Select(v => v.someData)
.ToList()))
.ToList();
Dictionary<DateTime, List<int>> byDate2 = txobj.events
.SelectMany(e=>e.eventData)
.GroupBy(d => d.date)
.OrderBy(g=>g.Key)
.ToDictionary(
g=>g.Key,
g => g.Select(v => v.someData)
.ToList());
List<List<Data>> byDate3 = txobj.events
.SelectMany(evt=>evt.eventData
.GroupBy(d => d.date)
.OrderBy(g => g.Key)
.Select(g => g.ToList()))
.ToList();
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这里的关键是 SelectMany 函数本质上连接了普通 Select 将返回的一级集合,即
any IEnumerable<IEnumerable<T>> becomes an IEnumerable<T>.
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