值不一定是唯一的，所以必须进行查找。你可以这样做:

var myKey = types.FirstOrDefault(x => x.Value == "one").Key;

如果值是唯一的，并且插入的频率低于读取，那么创建一个反向字典，其中值是键，键是值。

Dictionary<K,V>扩展。我已经用了很长时间了::

public static bool TryGetKey<K, V>(this IDictionary<K, V> instance, V value, out K key)
{
    foreach (var entry in instance)
    {
        if (!entry.Value.Equals(value))
        {
            continue;
        }
        key = entry.Key;
        return true;
    }
    key = default(K);
    return false;
}

并使用as:

public static void Main()
{
    Dictionary<string, string> dict = new Dictionary<string, string>()
    {
        {"1", "one"},
        {"2", "two"},
        {"3", "three"}
    };
     
     string value="two"; 
     if (dict.TryGetKey(value, out var returnedKey))
         Console.WriteLine($"Found Key {returnedKey}");
     else
         Console.WriteLine($"No key found for value {value}");
}

我创建了一个双重查找类:

/// <summary>
/// dictionary with double key lookup
/// </summary>
/// <typeparam name="T1">primary key</typeparam>
/// <typeparam name="T2">secondary key</typeparam>
/// <typeparam name="TValue">value type</typeparam>
public class cDoubleKeyDictionary<T1, T2, TValue> {
    private struct Key2ValuePair {
        internal T2 key2;
        internal TValue value;
    }
    private Dictionary<T1, Key2ValuePair> d1 = new Dictionary<T1, Key2ValuePair>();
    private Dictionary<T2, T1> d2 = new Dictionary<T2, T1>();

    /// <summary>
    /// add item
    /// not exacly like add, mote like Dictionary[] = overwriting existing values
    /// </summary>
    /// <param name="key1"></param>
    /// <param name="key2"></param>
    public void Add(T1 key1, T2 key2, TValue value) {
        lock (d1) {
            d1[key1] = new Key2ValuePair {
                key2 = key2,
                value = value,
            };
            d2[key2] = key1;
        }
    }

    /// <summary>
    /// get key2 by key1
    /// </summary>
    /// <param name="key1"></param>
    /// <param name="key2"></param>
    /// <returns></returns>
    public bool TryGetValue(T1 key1, out TValue value) {
        if (d1.TryGetValue(key1, out Key2ValuePair kvp)) {
            value = kvp.value;
            return true;
        } else {
            value = default;
            return false;
        }
    }

    /// <summary>
    /// get key1 by key2
    /// </summary>
    /// <param name="key2"></param>
    /// <param name="key1"></param>
    /// <remarks>
    /// 2x O(1) operation
    /// </remarks>
    /// <returns></returns>
    public bool TryGetValue2(T2 key2, out TValue value) {
        if (d2.TryGetValue(key2, out T1 key1)) {
            return TryGetValue(key1, out value);
        } else {
            value = default;
            return false;
        }
    }

    /// <summary>
    /// get key1 by key2
    /// </summary>
    /// <param name="key2"></param>
    /// <param name="key1"></param>
    /// <remarks>
    /// 2x O(1) operation
    /// </remarks>
    /// <returns></returns>
    public bool TryGetKey1(T2 key2, out T1 key1) {
        return d2.TryGetValue(key2, out key1);
    }

    /// <summary>
    /// get key1 by key2
    /// </summary>
    /// <param name="key2"></param>
    /// <param name="key1"></param>
    /// <remarks>
    /// 2x O(1) operation
    /// </remarks>
    /// <returns></returns>
    public bool TryGetKey2(T1 key1, out T2 key2) {
        if (d1.TryGetValue(key1, out Key2ValuePair kvp1)) {
            key2 = kvp1.key2;
            return true;
        } else {
            key2 = default;
            return false;
        }
    }

    /// <summary>
    /// remove item by key 1
    /// </summary>
    /// <param name="key1"></param>
    public void Remove(T1 key1) {
        lock (d1) {
            if (d1.TryGetValue(key1, out Key2ValuePair kvp)) {
                d1.Remove(key1);
                d2.Remove(kvp.key2);
            }
        }
    }

    /// <summary>
    /// remove item by key 2
    /// </summary>
    /// <param name="key2"></param>
    public void Remove2(T2 key2) {
        lock (d1) {
            if (d2.TryGetValue(key2, out T1 key1)) {
                d1.Remove(key1);
                d2.Remove(key2);
            }
        }
    }

    /// <summary>
    /// clear all items
    /// </summary>
    public void Clear() {
        lock (d1) {
            d1.Clear();
            d2.Clear();
        }
    }

    /// <summary>
    /// enumerator on key1, so we can replace Dictionary by cDoubleKeyDictionary
    /// </summary>
    /// <param name="key1"></param>
    /// <returns></returns>
    public TValue this[T1 key1] {
        get => d1[key1].value;
    }

    /// <summary>
    /// enumerator on key1, so we can replace Dictionary by cDoubleKeyDictionary
    /// </summary>
    /// <param name="key1"></param>
    /// <returns></returns>
    public TValue this[T1 key1, T2 key2] {
        set {
            lock (d1) {
                d1[key1] = new Key2ValuePair {
                    key2 = key2,
                    value = value,
                };
                d2[key2] = key1;
            }
        }
    }

也许是这样的:

foreach (var keyvaluepair in dict)
{
    if(Object.ReferenceEquals(keyvaluepair.Value, searchedObject))
    {
        //dict.Remove(keyvaluepair.Key);
        break;
    }
}

types.Values.ToList().IndexOf("one");

values . tolist()将您的字典值转换为对象列表。 IndexOf("one")搜索你的新列表，寻找"one"，并返回与字典中Key/Value对的索引匹配的Index。

这个方法并不关心字典中的键，它只是返回您正在寻找的值的索引。

请记住，在你的字典中可能有不止一个“1”值。这就是没有“get key”方法的原因。

keys中的键的顺序是未指定的，但它与values中的相关值相同(来自c#文档)。

因此，(在某些情况下)对值的集合执行此操作的有效方法如下:

    /// <summary>
    /// Gets the 1st key matching each value
    /// </summary>
    public static IEnumerable<TKey> GetKeys<TKey,TValue>(this Dictionary<TKey, TValue> dic, IEnumerable<TValue> values) where TKey : notnull
    {
        //The order of the keys in Keys is unspecified, but it is the same as the associated values in Values
        var dicKeys = dic.Keys.ToList();
        var dicValues = dic.Values.ToList();
        foreach (var value in values)
        {
            var i = dicValues.IndexOf(value); //Will return the index of the 1st found value, even when multiple values are present
            //we could test if i==-1 there.
            yield return dicKeys[i];
        }
    }