我有一个进程,在运行时输出一个唯一的int ID和一个double值,它可能不一定是唯一的。例如:
ID,价值 23,56000 25,67000 26,67000 45,54000
我必须捕获这些并通过增加值(从小到大)对ID进行排名,然后形成一个形式的字符串:id1,id2,id3等... 因此,在上面的情况下,输出将是:45; 26; 25; 23
永远不会有大量的ID - 但每次传递可以说10个。
我的方法是使用哈希表来捕获值。排序代码如下:
/// <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。评论/建议/代码样本非常感谢。 谢谢。 学家
答案 0 :(得分:4)
您只需使用一些LINQ和字符串实用程序就可以做到这一点:
public static string SortAndConvertToString(Hashtable ht)
{
var keysOrderedByValue = ht.Cast<DictionaryEntry>()
.OrderBy(x => x.Value)
.Select(x => x.Key);
return string.Join(";", keysOrderedByValue);
}
有关正常工作的演示,请参阅this fiddle。
我建议您使用通用Dictionary<int, double>
而不是Hashtable
。请参阅this related question。