获得通用词典的价值关键？

Question

从.Net 2.0泛型字典中获取密钥的值很容易:

Dictionary<int, string> greek = new Dictionary<int, string>();
greek.Add(1, "Alpha");
greek.Add(2, "Beta");
string secondGreek = greek[2];  // Beta

但有没有一种简单的方法来获取价值的关键？

int[] betaKeys = greek.WhatDoIPutHere("Beta");  // expecting single 2

Answer 1

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);
    }
}

Answer 2

正如其他人所说的那样,字典中没有从值到键的映射.

我刚刚注意到你想要从值到多个键映射 - 我将这个解决方案留给单值版本,但我会为多条目双向映射添加另一个答案.

这里采用的常规方法是使用两个字典 - 一个映射单向,另一个映射.将它们封装在一个单独的类中,并在您有重复的键或值时计算出您想要执行的操作(例如,抛出异常,覆盖现有条目或忽略新条目).就个人而言,我可能会抛出异常 - 这使得成功行为更容易定义.像这样的东西:

using System;
using System.Collections.Generic;

class BiDictionary<TFirst, TSecond>
{
    IDictionary<TFirst, TSecond> firstToSecond = new Dictionary<TFirst, TSecond>();
    IDictionary<TSecond, TFirst> secondToFirst = new Dictionary<TSecond, TFirst>();

    public void Add(TFirst first, TSecond second)
    {
        if (firstToSecond.ContainsKey(first) ||
            secondToFirst.ContainsKey(second))
        {
            throw new ArgumentException("Duplicate first or second");
        }
        firstToSecond.Add(first, second);
        secondToFirst.Add(second, first);
    }

    public bool TryGetByFirst(TFirst first, out TSecond second)
    {
        return firstToSecond.TryGetValue(first, out second);
    }

    public bool TryGetBySecond(TSecond second, out TFirst first)
    {
        return secondToFirst.TryGetValue(second, out first);
    }
}

class Test
{
    static void Main()
    {
        BiDictionary<int, string> greek = new BiDictionary<int, string>();
        greek.Add(1, "Alpha");
        greek.Add(2, "Beta");
        int x;
        greek.TryGetBySecond("Beta", out x);
        Console.WriteLine(x);
    }
}

Answer 3

字典并不是真的意味着像这样工作,因为虽然键的唯一性得到保证,但值的唯一性却不是.所以,例如,如果你有

var greek = new Dictionary<int, string> { { 1, "Alpha" }, { 2, "Alpha" } };

你期望得到greek.WhatDoIPutHere("Alpha")什么？

因此,你不能指望这样的东西被卷入框架.您需要自己的方法来实现自己的独特用途 - 您想要返回一个数组(或IEnumerable<T>)吗？如果有多个具有给定值的键,是否要抛出异常？如果没有怎么办？

就个人而言,我会选择一个可枚举的,如下:

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();

Answer 4

也许最简单的方法,没有Linq,可以循环对:

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,它可以通过这种方式轻松完成:

int betaKey = greek.SingleOrDefault(x => x.Value == "Beta").Key;