“滚动您自己的”版本的字典，允许“重复键”条目，这很容易。下面是一个粗略的简单实现。您可能需要考虑添加对dictionary <T>的基本大部分(如果不是全部)的支持。

public class MultiMap<TKey,TValue>
{
    private readonly Dictionary<TKey,IList<TValue>> storage;

    public MultiMap()
    {
        storage = new Dictionary<TKey,IList<TValue>>();
    }

    public void Add(TKey key, TValue value)
    {
        if (!storage.ContainsKey(key)) storage.Add(key, new List<TValue>());
        storage[key].Add(value);
    }

    public IEnumerable<TKey> Keys
    {
        get { return storage.Keys; }
    }

    public bool ContainsKey(TKey key)
    {
        return storage.ContainsKey(key);
    }

    public IList<TValue> this[TKey key]
    {
        get
        {
            if (!storage.ContainsKey(key))
                throw new KeyNotFoundException(
                    string.Format(
                        "The given key {0} was not found in the collection.", key));
            return storage[key];
        }
    }
}

一个关于如何使用它的简单例子:

const string key = "supported_encodings";
var map = new MultiMap<string,Encoding>();
map.Add(key, Encoding.ASCII);
map.Add(key, Encoding.UTF8);
map.Add(key, Encoding.Unicode);

foreach (var existingKey in map.Keys)
{
    var values = map[existingKey];
    Console.WriteLine(string.Join(",", values));
}

List类实际上适用于包含重复项的键/值集合，在此集合上进行迭代。例子:

List<KeyValuePair<string, string>> list = new List<KeyValuePair<string, string>>();

// add some values to the collection here

for (int i = 0;  i < list.Count;  i++)
{
    Print(list[i].Key, list[i].Value);
}

我认为List<KeyValuePair<object object>>可以完成任务。

你可以创建自己的字典包装器，就像这样，作为奖励，它支持空值作为键:

/// <summary>
/// Dictionary which supports duplicates and null entries
/// </summary>
/// <typeparam name="TKey">Type of key</typeparam>
/// <typeparam name="TValue">Type of items</typeparam>
public class OpenDictionary<TKey, TValue>
{
    private readonly Lazy<List<TValue>> _nullStorage = new Lazy<List<TValue>>(
        () => new List<TValue>());

    private readonly Dictionary<TKey, List<TValue>> _innerDictionary =
        new Dictionary<TKey, List<TValue>>();

    /// <summary>
    /// Get all entries
    /// </summary>
    public IEnumerable<TValue> Values =>
        _innerDictionary.Values
            .SelectMany(x => x)
            .Concat(_nullStorage.Value);

    /// <summary>
    /// Add an item
    /// </summary>
    public OpenDictionary<TKey, TValue> Add(TKey key, TValue item)
    {
        if (ReferenceEquals(key, null))
            _nullStorage.Value.Add(item);
        else
        {
            if (!_innerDictionary.ContainsKey(key))
                _innerDictionary.Add(key, new List<TValue>());

            _innerDictionary[key].Add(item);
        }

        return this;
    }

    /// <summary>
    /// Remove an entry by key
    /// </summary>
    public OpenDictionary<TKey, TValue> RemoveEntryByKey(TKey key, TValue entry)
    {
        if (ReferenceEquals(key, null))
        {
            int targetIdx = _nullStorage.Value.FindIndex(x => x.Equals(entry));
            if (targetIdx < 0)
                return this;

            _nullStorage.Value.RemoveAt(targetIdx);
        }
        else
        {
            if (!_innerDictionary.ContainsKey(key))
                return this;

            List<TValue> targetChain = _innerDictionary[key];
            if (targetChain.Count == 0)
                return this;

            int targetIdx = targetChain.FindIndex(x => x.Equals(entry));
            if (targetIdx < 0)
                return this;

            targetChain.RemoveAt(targetIdx);
        }

        return this;
    }

    /// <summary>
    /// Remove all entries by key
    /// </summary>
    public OpenDictionary<TKey, TValue> RemoveAllEntriesByKey(TKey key)
    {
        if (ReferenceEquals(key, null))
        {
            if (_nullStorage.IsValueCreated)
                _nullStorage.Value.Clear();
        }       
        else
        {
            if (_innerDictionary.ContainsKey(key))
                _innerDictionary[key].Clear();
        }

        return this;
    }

    /// <summary>
    /// Try get entries by key
    /// </summary>
    public bool TryGetEntries(TKey key, out IReadOnlyList<TValue> entries)
    {
        entries = null;

        if (ReferenceEquals(key, null))
        {
            if (_nullStorage.IsValueCreated)
            {
                entries = _nullStorage.Value;
                return true;
            }
            else return false;
        }
        else
        {
            if (_innerDictionary.ContainsKey(key))
            {
                entries = _innerDictionary[key];
                return true;
            }
            else return false;
        }
    }
}

用法示例:

var dictionary = new OpenDictionary<string, int>();
dictionary.Add("1", 1); 
// The next line won't throw an exception; 
dictionary.Add("1", 2);

dictionary.TryGetEntries("1", out List<int> result); 
// result is { 1, 2 }

dictionary.Add(null, 42);
dictionary.Add(null, 24);
dictionary.TryGetEntries(null, out List<int> result); 
// result is { 42, 24 }

我无意中发现了这篇文章，想要寻找相同的答案，但没有找到，所以我使用字典列表构建了一个简单的示例解决方案，当所有其他字典都有给定的键(set)时，重写[]操作符向列表中添加一个新字典，并返回一个值列表(get)。 它既丑陋又低效，它只通过键获取/设置，并且总是返回一个列表，但它是有效的:

 class DKD {
        List<Dictionary<string, string>> dictionaries;
        public DKD(){
            dictionaries = new List<Dictionary<string, string>>();}
        public object this[string key]{
             get{
                string temp;
                List<string> valueList = new List<string>();
                for (int i = 0; i < dictionaries.Count; i++){
                    dictionaries[i].TryGetValue(key, out temp);
                    if (temp == key){
                        valueList.Add(temp);}}
                return valueList;}
            set{
                for (int i = 0; i < dictionaries.Count; i++){
                    if (dictionaries[i].ContainsKey(key)){
                        continue;}
                    else{
                        dictionaries[i].Add(key,(string) value);
                        return;}}
                dictionaries.Add(new Dictionary<string, string>());
                dictionaries.Last()[key] =(string)value;
            }
        }
    }

你可以添加相同的键与不同的情况，如:

key1 Key1 KEY1 KeY1 kEy1 keY1

我知道这是个愚蠢的答案，但对我来说很管用。