如果有多个键(“右”键取代“左”键)，这不会爆炸，可以合并一些字典(如果需要)，并保留类型(限制它需要一个有意义的默认公共构造函数):

public static class DictionaryExtensions
{
    // Works in C#3/VS2008:
    // Returns a new dictionary of this ... others merged leftward.
    // Keeps the type of 'this', which must be default-instantiable.
    // Example: 
    //   result = map.MergeLeft(other1, other2, ...)
    public static T MergeLeft<T,K,V>(this T me, params IDictionary<K,V>[] others)
        where T : IDictionary<K,V>, new()
    {
        T newMap = new T();
        foreach (IDictionary<K,V> src in
            (new List<IDictionary<K,V>> { me }).Concat(others)) {
            // ^-- echk. Not quite there type-system.
            foreach (KeyValuePair<K,V> p in src) {
                newMap[p.Key] = p.Value;
            }
        }
        return newMap;
    }

}

注意，如果你使用一个名为“Add”的扩展方法，你可以使用集合初始化器来组合尽可能多的字典，就像这样:

public static void Add<K, V>(this Dictionary<K, V> d, Dictionary<K, V> other) {
  foreach (var kvp in other)
  {
    if (!d.ContainsKey(kvp.Key))
    {
      d.Add(kvp.Key, kvp.Value);
    }
  }
}


var s0 = new Dictionary<string, string> {
  { "A", "X"}
};
var s1 = new Dictionary<string, string> {
  { "A", "X" },
  { "B", "Y" }
};
// Combine as many dictionaries and key pairs as needed
var a = new Dictionary<string, string> {
  s0, s1, s0, s1, s1, { "C", "Z" }
};

public static IDictionary<K, V> AddRange<K, V>(this IDictionary<K, V> one, IDictionary<K, V> two)
        {
            foreach (var kvp in two)
            {
                if (one.ContainsKey(kvp.Key))
                    one[kvp.Key] = two[kvp.Key];
                else
                    one.Add(kvp.Key, kvp.Value);
            }
            return one;
        }

下面是我使用的一个helper函数:

using System.Collections.Generic;
namespace HelperMethods
{
    public static class MergeDictionaries
    {
        public static void Merge<TKey, TValue>(this IDictionary<TKey, TValue> first, IDictionary<TKey, TValue> second)
        {
            if (second == null || first == null) return;
            foreach (var item in second) 
                if (!first.ContainsKey(item.Key)) 
                    first.Add(item.Key, item.Value);
        }
    }
}

这个聚会现在几乎已经死了，但是user166390的“改进”版本已经进入了我的扩展库。 除了一些细节之外，我还添加了一个委托来计算合并的值。

/// <summary>
/// Merges a dictionary against an array of other dictionaries.
/// </summary>
/// <typeparam name="TResult">The type of the resulting dictionary.</typeparam>
/// <typeparam name="TKey">The type of the key in the resulting dictionary.</typeparam>
/// <typeparam name="TValue">The type of the value in the resulting dictionary.</typeparam>
/// <param name="source">The source dictionary.</param>
/// <param name="mergeBehavior">A delegate returning the merged value. (Parameters in order: The current key, The current value, The previous value)</param>
/// <param name="mergers">Dictionaries to merge against.</param>
/// <returns>The merged dictionary.</returns>
public static TResult MergeLeft<TResult, TKey, TValue>(
    this TResult source,
    Func<TKey, TValue, TValue, TValue> mergeBehavior,
    params IDictionary<TKey, TValue>[] mergers)
    where TResult : IDictionary<TKey, TValue>, new()
{
    var result = new TResult();
    var sources = new List<IDictionary<TKey, TValue>> { source }
        .Concat(mergers);

    foreach (var kv in sources.SelectMany(src => src))
    {
        TValue previousValue;
        result.TryGetValue(kv.Key, out previousValue);
        result[kv.Key] = mergeBehavior(kv.Key, kv.Value, previousValue);
    }

    return result;
}

这在一定程度上取决于如果遇到重复项，你希望发生什么。例如，你可以这样做:

var result = dictionaries.SelectMany(dict => dict)
                         .ToDictionary(pair => pair.Key, pair => pair.Value);

如果您获得任何重复的键，将抛出异常。

编辑:如果你使用ToLookup，那么你会得到一个查找，每个键可以有多个值。然后你可以把它转换成一个字典:

var result = dictionaries.SelectMany(dict => dict)
                         .ToLookup(pair => pair.Key, pair => pair.Value)
                         .ToDictionary(group => group.Key, group => group.First());

这有点难看——而且效率很低——但从代码的角度来说，这是最快的方法。(不得不承认，我还没有测试过它。)

当然，您也可以编写自己的ToDictionary2扩展方法(有一个更好的名字，但我现在没有时间去想)——这并不难做到，只是覆盖(或忽略)重复的键。重要的一点(在我看来)是使用SelectMany，并意识到字典支持键/值对的迭代。