@Tim:应该是注释，但是注释不允许代码编辑。

Dictionary<string, string> t1 = new Dictionary<string, string>();
t1.Add("a", "aaa");
Dictionary<string, string> t2 = new Dictionary<string, string>();
t2.Add("b", "bee");
Dictionary<string, string> t3 = new Dictionary<string, string>();
t3.Add("c", "cee");
t3.Add("d", "dee");
t3.Add("b", "bee");
Dictionary<string, string> merged = t1.MergeLeft(t2, t2, t3);

注意:我应用了@ANeves对@Andrew Orsich的解决方案的修改，所以mergleft现在看起来像这样:

public static Dictionary<K, V> MergeLeft<K, V>(this Dictionary<K, V> me, params IDictionary<K, V>[] others)
    {
        var newMap = new Dictionary<K, V>(me, me.Comparer);
        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;
    }

其平凡解为:

using System.Collections.Generic;
...
public static Dictionary<TKey, TValue>
    Merge<TKey,TValue>(IEnumerable<Dictionary<TKey, TValue>> dictionaries)
{
    var result = new Dictionary<TKey, TValue>();
    foreach (var dict in dictionaries)
        foreach (var x in dict)
            result[x.Key] = x.Value;
    return result;
}

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" }
};

or :

public static IDictionary<TKey, TValue> Merge<TKey, TValue>( IDictionary<TKey, TValue> x, IDictionary<TKey, TValue> y)
    {
        return x
            .Except(x.Join(y, z => z.Key, z => z.Key, (a, b) => a))
            .Concat(y)
            .ToDictionary(z => z.Key, z => z.Value);
    }

结果是一个联合，对于重复的条目，“y”胜出。

试试下面的方法

static Dictionary<TKey, TValue>
    Merge<TKey, TValue>(this IEnumerable<Dictionary<TKey, TValue>> enumerable)
{
    return enumerable.SelectMany(x => x).ToDictionary(x => x.Key, y => y.Value);
}

@user166390的回答版本增加了一个IEqualityComparer参数，以允许不区分大小写的键比较。

    public static T MergeLeft<T, K, V>(this T me, params Dictionary<K, V>[] others)
        where T : Dictionary<K, V>, new()
    {
        return me.MergeLeft(me.Comparer, others);
    }

    public static T MergeLeft<T, K, V>(this T me, IEqualityComparer<K> comparer, params Dictionary<K, V>[] others)
        where T : Dictionary<K, V>, new()
    {
        T newMap = Activator.CreateInstance(typeof(T), new object[] { comparer }) as T;

        foreach (Dictionary<K, V> src in 
            (new List<Dictionary<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;
    }