使用 IComparable 对对象进行排序

    public int CompareTo(object obj)
    {
        //Int reference table:
        //1 or greater means the current instance occurs after obj
        //0 means both elements occur in the same position
        //-1 or less means the current instance occurs before obj
        if (obj == null)
            return 1;
        Upgrade otherUpgrade = obj as Upgrade;
        if (otherUpgrade != null)
            return _name.CompareTo(otherUpgrade.Name);
        else
            throw new ArgumentException("Passed object is not an Upgrade.");
    }

你想要"自然顺序"——熟悉英语惯例的人会选择的排序规则——而不是你所拥有的，即"词典"排序：为每个字母分配严格的排序，然后依次按每个字母排序。

Jeff有一篇关于这里一些细节的好文章，其中包含指向尝试解决问题的不同算法的链接：

http://www.codinghorror.com/blog/2007/12/sorting-for-humans-natural-sort-order.html

Raymond 在这里讨论了 Windows 如何处理它：

http://technet.microsoft.com/en-us/magazine/hh475812.aspx

基本上问题是：自然顺序整理需要解决人工智能问题;你试图模仿人类会做什么，这可能令人惊讶地棘手。祝你好运！

字符串按字典顺序排序。 您必须将所有数字格式化为具有相同的长度（例如：测试升级 02），或者解析比较器中的数字并将其合并到比较逻辑中。

发生这种情况的原因是您正在进行字符串比较，它没有明确的数字知识。它按每个字符的相应字符代码对每个字符串进行排序。

要获得您想要的效果，需要更多的工作。请参阅此问题：对可能包含数字的字符串进行排序

字母数字排序

public class AlphanumComparatorFast : IComparer
{
    public int Compare(object x, object y)
    {
    string s1 = x as string;
    if (s1 == null)
    {
        return 0;
    }
    string s2 = y as string;
    if (s2 == null)
    {
        return 0;
    }
    int len1 = s1.Length;
    int len2 = s2.Length;
    int marker1 = 0;
    int marker2 = 0;
    // Walk through two the strings with two markers.
    while (marker1 < len1 && marker2 < len2)
    {
        char ch1 = s1[marker1];
        char ch2 = s2[marker2];
        // Some buffers we can build up characters in for each chunk.
        char[] space1 = new char[len1];
        int loc1 = 0;
        char[] space2 = new char[len2];
        int loc2 = 0;
        // Walk through all following characters that are digits or
        // characters in BOTH strings starting at the appropriate marker.
        // Collect char arrays.
        do
        {
        space1[loc1++] = ch1;
        marker1++;
        if (marker1 < len1)
        {
            ch1 = s1[marker1];
        }
        else
        {
            break;
        }
        } while (char.IsDigit(ch1) == char.IsDigit(space1[0]));
        do
        {
        space2[loc2++] = ch2;
        marker2++;
        if (marker2 < len2)
        {
            ch2 = s2[marker2];
        }
        else
        {
            break;
        }
        } while (char.IsDigit(ch2) == char.IsDigit(space2[0]));
        // If we have collected numbers, compare them numerically.
        // Otherwise, if we have strings, compare them alphabetically.
        string str1 = new string(space1);
        string str2 = new string(space2);
        int result;
        if (char.IsDigit(space1[0]) && char.IsDigit(space2[0]))
        {
        int thisNumericChunk = int.Parse(str1);
        int thatNumericChunk = int.Parse(str2);
        result = thisNumericChunk.CompareTo(thatNumericChunk);
        }
        else
        {
        result = str1.CompareTo(str2);
        }
        if (result != 0)
        {
        return result;
        }
    }
    return len1 - len2;
    }
}

用法：

using System;
using System.Collections;
class Program
{
    static void Main()
    {
    string[] highways = new string[]
    {
        "100F",
        "50F",
        "SR100",
        "SR9"
    };
    //
    // We want to sort a string array called highways in an
    // alphanumeric way. Call the static Array.Sort method.
    //
    Array.Sort(highways, new AlphanumComparatorFast());
    //
    // Display the results
    //
    foreach (string h in highways)
    {
        Console.WriteLine(h);
    }
    }
}

输出

50F 100F

SR9
SR100

感谢所有的回复。我做了自己的方法，似乎工作正常。它不适用于所有情况，但它适用于我的方案。以下是任何感兴趣的人的代码：

    /// <summary>
    /// Compares the upgrade to another upgrade
    /// </summary>
    /// <param name="obj"></param>
    /// <returns></returns>
    public int CompareTo(object obj)
    {
        //Int reference table:
        //1 or greater means the current instance occurs after obj
        //0 means both elements occur in the same position
        //-1 or less means the current instance occurs before obj
        if (obj == null)
            return 1;
        Upgrade otherUpgrade = obj as Upgrade;
        if (otherUpgrade != null)
        {
            //Split strings into arrays
            string[] splitStringOne = _name.Split(new char[] { ' ' });
            string[] splitStringTwo = otherUpgrade.Name.Split(new char[] { ' ' });
            //Will hold checks to see which comparer will be used
            bool sameWords = false, sameLength = false, bothInt = false;
            //Will hold the last part of the string if it is an int
            int intOne = 0, intTwo = 0;
            //Check if they have the same length
            sameLength = (splitStringOne.Length == splitStringTwo.Length);
            if (sameLength)
            {
                //Check to see if they both end in an int
                bothInt = (int.TryParse(splitStringOne[splitStringOne.Length - 1], out intOne) && int.TryParse(splitStringTwo[splitStringTwo.Length - 1], out intTwo));
                if (bothInt)
                {
                    //Check to see if the previous parts of the string are equal
                    for (int i = 0; i < splitStringOne.Length - 2; i++)
                    {
                        sameWords = (splitStringOne[i].ToLower().Equals(splitStringTwo[i].ToLower()));
                        if (!sameWords)
                            break;
                    }
                }
            }
            //If all criteria is met, use the customk comparer
            if (sameWords && sameLength && bothInt)
            {
                if (intOne < intTwo)
                    return -1;
                else if (intOne > intTwo)
                    return 1;
                else //Both equal
                    return 0;
            }
            //Else use the default string comparer
            else
                return _name.CompareTo(otherUpgrade.Name);            
        }
        else
            throw new ArgumentException("Passed object is not an Upgrade.");
    }

适用于使用" "字符间隔的字符串，如下所示：

测试数据：

• 你好 11
• 你好2
• 你好 13

结果

• 你好2
• 你好 11
• 你好 13

不适用于Hello11和Hello2等数据，因为它无法拆分它们。不区分大小写。