获取泛型字典的指定值的多个键？
获取泛型字典的指定值的多个键？
提问于 2008-11-01 00:49:08
从.NET通用字典中很容易获得键的值：
Dictionary<int, string> greek = new Dictionary<int, string>();
greek.Add(1, "Alpha");
greek.Add(2, "Beta");
string secondGreek = greek[2]; // Beta但是，尝试为键指定一个值并不简单，因为可能有多个键：
int[] betaKeys = greek.WhatDoIPutHere("Beta"); // expecting single 2回答 11
Stack Overflow用户
回答已采纳
发布于 2008-11-01 08:03:09
好的，这是多个双向版本：
using System;
using System.Collections.Generic;
using System.Text;
class BiDictionary<TFirst, TSecond>
{
IDictionary<TFirst, IList<TSecond>> firstToSecond = new Dictionary<TFirst, IList<TSecond>>();
IDictionary<TSecond, IList<TFirst>> secondToFirst = new Dictionary<TSecond, IList<TFirst>>();
private static IList<TFirst> EmptyFirstList = new TFirst[0];
private static IList<TSecond> EmptySecondList = new TSecond[0];
public void Add(TFirst first, TSecond second)
{
IList<TFirst> firsts;
IList<TSecond> seconds;
if (!firstToSecond.TryGetValue(first, out seconds))
{
seconds = new List<TSecond>();
firstToSecond[first] = seconds;
}
if (!secondToFirst.TryGetValue(second, out firsts))
{
firsts = new List<TFirst>();
secondToFirst[second] = firsts;
}
seconds.Add(second);
firsts.Add(first);
}
// Note potential ambiguity using indexers (e.g. mapping from int to int)
// Hence the methods as well...
public IList<TSecond> this[TFirst first]
{
get { return GetByFirst(first); }
}
public IList<TFirst> this[TSecond second]
{
get { return GetBySecond(second); }
}
public IList<TSecond> GetByFirst(TFirst first)
{
IList<TSecond> list;
if (!firstToSecond.TryGetValue(first, out list))
{
return EmptySecondList;
}
return new List<TSecond>(list); // Create a copy for sanity
}
public IList<TFirst> GetBySecond(TSecond second)
{
IList<TFirst> list;
if (!secondToFirst.TryGetValue(second, out list))
{
return EmptyFirstList;
}
return new List<TFirst>(list); // Create a copy for sanity
}
}
class Test
{
static void Main()
{
BiDictionary<int, string> greek = new BiDictionary<int, string>();
greek.Add(1, "Alpha");
greek.Add(2, "Beta");
greek.Add(5, "Beta");
ShowEntries(greek, "Alpha");
ShowEntries(greek, "Beta");
ShowEntries(greek, "Gamma");
}
static void ShowEntries(BiDictionary<int, string> dict, string key)
{
IList<int> values = dict[key];
StringBuilder builder = new StringBuilder();
foreach (int value in values)
{
if (builder.Length != 0)
{
builder.Append(", ");
}
builder.Append(value);
}
Console.WriteLine("{0}: [{1}]", key, builder);
}
}Stack Overflow用户
发布于 2008-11-01 01:03:00
字典实际上并不是这样工作的，因为键的唯一性是有保证的，值的唯一性是不能保证的。
var greek = new Dictionary<int, string> { { 1, "Alpha" }, { 2, "Alpha" } };您希望从greek.WhatDoIPutHere("Alpha")中获得什么
因此，你不能期望像这样的东西被加入到框架中。你需要你自己的方法来实现你自己的独特用途-你想返回一个数组(或IEnumerable<T>)吗？如果有多个键具有给定值，是否要抛出异常？如果没有呢？
就我个人而言，我会选择一个可枚举的，如下所示：
IEnumerable<TKey> KeysFromValue<TKey, TValue>(this Dictionary<TKey, TValue> dict, TValue val)
{
if (dict == null)
{
throw new ArgumentNullException("dict");
}
return dict.Keys.Where(k => dict[k] == val);
}
var keys = greek.KeysFromValue("Beta");
int exceptionIfNotExactlyOne = greek.KeysFromValue("Beta").Single();Stack Overflow用户
发布于 2008-11-01 01:03:34
也许最简单的方法是在没有Linq的情况下循环遍历这些对：
int betaKey;
foreach (KeyValuePair<int, string> pair in lookup)
{
if (pair.Value == value)
{
betaKey = pair.Key; // Found
break;
}
}
betaKey = -1; // Not found如果你有Linq，它可以很容易地这样做：
int betaKey = greek.SingleOrDefault(x => x.Value == "Beta").Key;页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/255341
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