可以使用扩展方法对 observable 进行排序并返回已排序的相同对象。对于较大的集合，请注意集合更改通知的数量。
我已更新我的代码以提高性能（感谢 nawfal）并处理在撰写本文时此处没有其他答案的重复项。 observable 被划分为左排序的一半和右未排序的一半，每次最小项（如在排序列表中找到的）从未排序的部分移动到排序分区的末尾。最坏情况 O(n)。本质上是一个选择排序（见下文输出）。
public static void Sort<T>(this ObservableCollection<T> collection)
where T : IComparable<T>, IEquatable<T>
{
List<T> sorted = collection.OrderBy(x => x).ToList();
int ptr = 0;
while (ptr < sorted.Count - 1)
{
if (!collection[ptr].Equals(sorted[ptr]))
{
int idx = search(collection, ptr+1, sorted[ptr]);
collection.Move(idx, ptr);
}
ptr++;
}
}
public static int search<T>(ObservableCollection<T> collection, int startIndex, T other)
{
for (int i = startIndex; i < collection.Count; i++)
{
if (other.Equals(collection[i]))
return i;
}
return -1; // decide how to handle error case
}
用法：
带有观察者的示例（使用 Person 类保持简单）
public class Person:IComparable<Person>,IEquatable<Person>
{
public string Name { get; set; }
public int Age { get; set; }
public int CompareTo(Person other)
{
if (this.Age == other.Age) return 0;
return this.Age.CompareTo(other.Age);
}
public override string ToString()
{
return Name + " aged " + Age;
}
public bool Equals(Person other)
{
if (this.Name.Equals(other.Name) && this.Age.Equals(other.Age)) return true;
return false;
}
}
static void Main(string[] args)
{
Console.WriteLine("adding items...");
var observable = new ObservableCollection<Person>()
{
new Person {Name = "Katy", Age = 51},
new Person {Name = "Jack", Age = 12},
new Person {Name = "Bob", Age = 13},
new Person {Name = "Alice", Age = 39},
new Person {Name = "John", Age = 14},
new Person {Name = "Mary", Age = 41},
new Person {Name = "Jane", Age = 20},
new Person {Name = "Jim", Age = 39},
new Person {Name = "Sue", Age = 5},
new Person {Name = "Kim", Age = 19}
};
//what do observers see?
observable.CollectionChanged += (sender, e) =>
{
Console.WriteLine(
e.OldItems[0] + " move from " + e.OldStartingIndex + " to " + e.NewStartingIndex);
int i = 0;
foreach (var person in sender as ObservableCollection<Person>)
{
if (i == e.NewStartingIndex)
{
Console.Write("(" + (person as Person).Age + "),");
}
else
{
Console.Write((person as Person).Age + ",");
}
i++;
}
Console.WriteLine();
};
显示集合如何旋转的排序进度详细信息：
Sue aged 5 move from 8 to 0
(5),51,12,13,39,14,41,20,39,19,
Jack aged 12 move from 2 to 1
5,(12),51,13,39,14,41,20,39,19,
Bob aged 13 move from 3 to 2
5,12,(13),51,39,14,41,20,39,19,
John aged 14 move from 5 to 3
5,12,13,(14),51,39,41,20,39,19,
Kim aged 19 move from 9 to 4
5,12,13,14,(19),51,39,41,20,39,
Jane aged 20 move from 8 to 5
5,12,13,14,19,(20),51,39,41,39,
Alice aged 39 move from 7 to 6
5,12,13,14,19,20,(39),51,41,39,
Jim aged 39 move from 9 to 7
5,12,13,14,19,20,39,(39),51,41,
Mary aged 41 move from 9 to 8
5,12,13,14,19,20,39,39,(41),51,
Person 类同时实现了 IComparable 和 IEquatable，后者用于最小化对集合的更改，从而减少引发的更改通知的数量
要返回 ObservableCollection，请在 *sortedOC* 上调用 .ToObservableCollection，例如使用[此实现][1]。
**** 原始答案 - 这将创建一个新集合 ****
您可以使用 linq，如下所示的 doSort 方法。一个快速的代码 sn-p: 产生
3:xy
6:fty
7:aaa
或者，您可以在集合本身上使用扩展方法
var sortedOC = _collection.OrderBy(i => i.Key);
private void doSort()
{
ObservableCollection<Pair<ushort, string>> _collection =
new ObservableCollection<Pair<ushort, string>>();
_collection.Add(new Pair<ushort,string>(7,"aaa"));
_collection.Add(new Pair<ushort, string>(3, "xey"));
_collection.Add(new Pair<ushort, string>(6, "fty"));
var sortedOC = from item in _collection
orderby item.Key
select item;
foreach (var i in sortedOC)
{
Debug.WriteLine(i);
}
}
public class Pair<TKey, TValue>
{
private TKey _key;
public TKey Key
{
get { return _key; }
set { _key = value; }
}
private TValue _value;
public TValue Value
{
get { return _value; }
set { _value = value; }
}
public Pair(TKey key, TValue value)
{
_key = key;
_value = value;
}
public override string ToString()
{
return this.Key + ":" + this.Value;
}
}