假设你有一个字典，你可以直接使用下面一行对它们进行排序:

var x = (from c in dict orderby c.Value.Order ascending select c).ToDictionary(c => c.Key, c=>c.Value);

假设你有一个字典，你可以直接使用下面一行对它们进行排序:

var x = (from c in dict orderby c.Value.Order ascending select c).ToDictionary(c => c.Key, c=>c.Value);

你可以按值对Dictionary进行排序并将其保存回自身(这样当你对它进行foreach时，值就会按顺序出来):

dict = dict.OrderBy(x => x.Value).ToDictionary(x => x.Key, x => x.Value);

当然，这可能不正确，但它是有效的。海鲁姆定律意味着这很可能会继续起作用。

使用VB对SortedDictionary列表进行排序以绑定到ListView控件。NET:

Dim MyDictionary As SortedDictionary(Of String, MyDictionaryEntry)

MyDictionaryListView.ItemsSource = MyDictionary.Values.OrderByDescending(Function(entry) entry.MyValue)

Public Class MyDictionaryEntry ' Need Property for GridViewColumn DisplayMemberBinding
    Public Property MyString As String
    Public Property MyValue As Integer
End Class

XAML:

<ListView Name="MyDictionaryListView">
    <ListView.View>
        <GridView>
            <GridViewColumn DisplayMemberBinding="{Binding Path=MyString}" Header="MyStringColumnName"></GridViewColumn>
            <GridViewColumn DisplayMemberBinding="{Binding Path=MyValue}" Header="MyValueColumnName"></GridViewColumn>
         </GridView>
    </ListView.View>
</ListView>

字典根据定义是一种无序的关联结构，仅以可哈希的方式包含值和键。换句话说，没有一种可预见的方式来排序一本字典。

作为参考，请阅读这篇来自python语言的文章。

链接 Python数据结构

下面的代码片段按值对Dictionary进行排序。

代码首先创建一个字典，然后使用OrderBy方法对条目进行排序。

public void SortDictionary()  
{  
  
    // Create a dictionary with string key and Int16 value pair  
    Dictionary<string, Int16> AuthorList = new Dictionary<string, Int16>();  
    AuthorList.Add("Mahesh Chand", 35);  
    AuthorList.Add("Mike Gold", 25);  
    AuthorList.Add("Praveen Kumar", 29);  
    AuthorList.Add("Raj Beniwal", 21);  
    AuthorList.Add("Dinesh Beniwal", 84);   
  
    // Sorted by Value  
  
    Console.WriteLine("Sorted by Value");  
    Console.WriteLine("=============");  
    foreach (KeyValuePair<string, Int16> author in AuthorList.OrderBy(key => key.Value))  
    {  
        Console.WriteLine("Key: {0}, Value: {1}", author.Key, author.Value);  
    }  
}