在C#中,我想按列表中每个字符串的长度对List<KeyValuePair<int, string>>
进行排序。在Psuedo-Java中,这将是一个匿名的,看起来像:
Collections.Sort(someList, new Comparator<KeyValuePair<int, string>>( {
public int compare(KeyValuePair<int, string> s1, KeyValuePair<int, string> s2)
{
return (s1.Value.Length > s2.Value.Length) ? 1 : 0; //specify my sorting criteria here
}
});
答案 0 :(得分:29)
C#中的等价物是使用lambda表达式和Sort
方法:
someList.Sort((x, y) => x.Value.Length.CompareTo(y.Value.Length));
您还可以使用OrderBy
扩展方法。它的代码略少,但它增加了更多的开销,因为它创建了列表的副本而不是将其排序到位:
someList = someList.OrderBy(x => x.Value.Length).ToList();
答案 1 :(得分:10)
您可以使用linq调用OrderBy
list.OrderBy(o => o.Value.Length);
有关@Guffa指出查找Linq and Deferred Execution的更多信息,基本上它只会在需要时执行。因此,要立即从此行返回一个列表,您需要添加一个.ToList()
,这将使表达式返回一个列表。
答案 2 :(得分:3)
using System;
using System.Collections.Generic;
class Program
{
static int Compare1(KeyValuePair<string, int> a, KeyValuePair<string, int> b)
{
return a.Key.CompareTo(b.Key);
}
static int Compare2(KeyValuePair<string, int> a, KeyValuePair<string, int> b)
{
return a.Value.CompareTo(b.Value);
}
static void Main()
{
var list = new List<KeyValuePair<string, int>>();
list.Add(new KeyValuePair<string, int>("Perl", 7));
list.Add(new KeyValuePair<string, int>("Net", 9));
list.Add(new KeyValuePair<string, int>("Dot", 8));
// Use Compare1 as comparison delegate.
list.Sort(Compare1);
foreach (var pair in list)
{
Console.WriteLine(pair);
}
Console.WriteLine();
// Use Compare2 as comparison delegate.
list.Sort(Compare2);
foreach (var pair in list)
{
Console.WriteLine(pair);
}
}
}