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;
        }

使用equalitycompararer进行合并，它将项目映射到不同的值/类型进行比较。这里我们将从KeyValuePair(枚举字典时的项类型)映射到Key。

public class MappedEqualityComparer<T,U> : EqualityComparer<T>
{
    Func<T,U> _map;

    public MappedEqualityComparer(Func<T,U> map)
    {
        _map = map;
    }

    public override bool Equals(T x, T y)
    {
        return EqualityComparer<U>.Default.Equals(_map(x), _map(y));
    }

    public override int GetHashCode(T obj)
    {
        return _map(obj).GetHashCode();
    }
}

用法:

// if dictA and dictB are of type Dictionary<int,string>
var dict = dictA.Concat(dictB)
                .Distinct(new MappedEqualityComparer<KeyValuePair<int,string>,int>(item => item.Key))
                .ToDictionary(item => item.Key, item=> item.Value);

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;
        }

选项1:这取决于在确定两个字典中都没有重复键的情况下想要发生什么。比你能做的:

var result = dictionary1.Union(dictionary2).ToDictionary(k => k.Key, v => v.Value)

注意:如果在字典中获得任何重复的键，将抛出错误。

选项2:如果你可以有重复的键，那么你必须使用where子句来处理重复的键。

var result = dictionary1.Union(dictionary2.Where(k => !dictionary1.ContainsKey(k.Key))).ToDictionary(k => k.Key, v => v.Value)

注意:它不会得到重复的密钥。如果有任何重复的键，那么它将获得dictionary1的键。

选项3:如果你想使用ToLookup。然后您将得到一个查找，每个键可以有多个值。你可以把这个查找转换成一个字典:

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

@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;
    }

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;
    }

}