LINQ:根据最大值分组和从对象列表中选择
本文关键字:对象 列表 选择 最大值 LINQ | 更新日期: 2023-09-27 18:26:07
我有一个MyItem类,它有3个属性,如下所示:
class MyItem
{
private string _name;
private int _value
private DateTime _TimeStamp;
public MyItem(string name, int value, string timeStamp)
{
this._name = name;
this._value = value;
this._timeStamp = DateTime.Parse(timeStamp);
}
public string Name
{ get {return this_name; } }
public int Value
{ get {return this._value; } }
public DateTime TimeStamp
{ get {return this._timeStamp; } }
// ...
}
我还有一个MyItem的列表如下:
var myItems = new List<MyItem>() {
new MyItem("A", 123, "23/02/2012"),
new MyItem("A", 323, "22/02/2012"),
new MyItem("B", 432, "23/02/2012"),
new MyItem("B", 356, "22/02/2012"),
// ...
}
如何按我的列表分组,以便只剩下具有最大时间戳的项目?即以下结果:
"A" 123 23/02/2012<br>
"B" 432 23/02/2012<br>
提前谢谢。
myItems.GroupBy(item => item.Name)
.Select(grp => grp.Aggregate((max, cur) =>
(max == null || cur.Date > max.Date) ? cur : max))
这将在尽可能快的时间内(至少我可以想象)选择结果,而无需创建新对象并在集合上迭代最少的次数。
从组中选择最大值:
from item in MyItems
group item by item.Name into grouped
let maxTimeStamp = grouped.Max(i => i.TimeStamp)
select grouped.First(i => i.TimeStamp == maxTimeStamp)
var temp = myItems.Where(x => x.TimeStamp == myItems.Where(y => y.Name == x.Name).Max(z => z.TimeStamp)).Distinct().ToList();
var tmp = select i from myItems
group i by i.Name into g
select new MyItem
{
g.Name,
Value = g.OrderByDescending(x => x.Timestamp).First().Value,
Timestamp = g.Max(x => x.Timestamp)
};
好的,为了LINQ的方便(希望这不会引起问题),我更改了您的类MyItem
,在该类中添加了一个空白构造函数:
public MyItem() { }
在一个示例控制台程序中,此代码将起作用:
static void Main(string[] args)
{
var myItems = new List<MyItem>()
{
new MyItem("A", 123, "23/02/2012"),
new MyItem("A", 323, "22/02/2012"),
new MyItem("B", 432, "23/02/2012"),
new MyItem("B", 356, "22/02/2012")
// ...
};
var grouped = from m in myItems
group m by m.Name into g
let maxTimestamp = g.Max(t => t.TimeStamp)
select new MyItem
{
Name = g.Key,
Value = g.First(f => f.TimeStamp == maxTimestamp).Value,
TimeStamp = maxTimestamp
};
foreach (var gItem in grouped)
{
Console.WriteLine(gItem.Name + ", " + gItem.Value + ", " + gItem.TimeStamp);
}
Console.ReadLine();
}
输出符合您的预期结果。