按值对id排序并形成字符串

我有一个进程，在运行时输出一个唯一的int ID和一个双精度值(不一定是唯一的)。例如:

ID值23日,56000年25日,6700026日,67000年54000

我必须捕获这些并通过增加值(从小到大)对id进行排序，然后形成形式的字符串:id1, id2, id3等…因此，在上面的例子中，输出将是:45;26;25;23

永远不会有大量的id -但我们假设每次传递10个。

我的方法是使用哈希表来捕获值。排序的代码如下:

    /// <summary>
    /// Converts a hashtable (key is the id; value is the amount) to a string of the 
    /// format: x;y;z; where x,y & z are the  id numbers in order of increasing amounts
    /// cf. http://stackoverflow.com/questions/3101626/sort-hashtable-by-possibly-non-unique-values for the sorting routine
    /// </summary>
    /// <param name="ht">Hashtable (key is id; value is the actual amount)</param>
    /// <returns>String of the format: x;y;z; where x,y & z are the id numbers in order of increasing amounts</returns>
    public static string SortAndConvertToString(Hashtable ht)
    {
        if (ht.Count == 1)
            return ht.Keys.OfType<String>().FirstOrDefault() +";";
        //1. Sort the HT by value (smaller to bigger). Preserve key associated with the value                                 
        var result = new List<DictionaryEntry>(ht.Count);
        foreach (DictionaryEntry entry in ht)
        {
            result.Add(entry);
        }
        result.Sort(
            (x, y) =>
            {
                IComparable comparable = x.Value as IComparable;
                if (comparable != null)
                {
                    return comparable.CompareTo(y.Value);
                }
                return 0;
            });
        string str = "";
        foreach (DictionaryEntry entry in result)
        {
            str += ht.Keys.OfType<String>().FirstOrDefault(s => ht[s] == entry.Value) + ";";
        }
        //2. Extract keys to form string of the form: x;y;z;
        return str;
    }

我只是想知道这是最有效的做事方式还是有更快的方法。非常感谢评论/建议/代码示例。谢谢。j .

public static string SortAndConvertToString(Hashtable ht)
{
    var keysOrderedByValue = ht.Cast<DictionaryEntry>()
        .OrderBy(x => x.Value)
        .Select(x => x.Key);
    return string.Join(";", keysOrderedByValue);
}