我知道BCL中没有一个，但有人能指出我的开源好吗？
By Multi我指的是2把钥匙。 ; - ）
答案 0 :(得分：64)
我也使用jason in his answer作为元组。但是，我建议您只需将元组定义为结构：
public struct Tuple<T1, T2> {
public readonly T1 Item1;
public readonly T2 Item2;
public Tuple(T1 item1, T2 item2) { Item1 = item1; Item2 = item2;}
}
public static class Tuple { // for type-inference goodness.
public static Tuple<T1,T2> Create<T1,T2>(T1 item1, T2 item2) {
return new Tuple<T1,T2>(item1, item2);
}
}
你可以免费获得免疫力，.GetHashcode和.Equals，这（当你等待C＃4.0时）非常简单......
一个警告但是：默认GetHashcode实施（有时）only considers the first field因此请务必使第一个字段最具辨别力或自己实施GetHashcode（例如，使用ValueUtils中的FieldwiseHasher.Hash(this)，否则您可能会遇到可伸缩性问题。
此外，您可以避免使用往往使问题复杂化的空值（如果您确实需要空值，则只需使Tuple<>可以为空）。稍微偏离主题，我是唯一一个对框架级别缺乏对非空引用的支持而烦恼的人吗？我在大型项目上工作，偶尔会在某个地方出现一个无法实现的小问题 - 而且嘿嘿，你得到一个空引用异常 - 但是有一个堆栈跟踪指向引用的第一个用法，而不是实际有缺陷的代码
当然，.NET 4.0现在还很老了;我们大多数人都可以使用.NET 4.0的元组。
编辑： 解决.NET为我编写的ValueUtils结构提供的不良GetHashCode实现，这也允许您使用真实姓名作为您的多字段键;这意味着你可以写一些类似的东西：
sealed class MyValueObject : ValueObject<MyValueObject> {
public DayOfWeek day;
public string NamedPart;
//properties work fine too
}
...希望能够更容易地为具有值语义的数据提供人类可读的名称，至少在some future version of C# implements proper tuples with named members之前;希望有合适的哈希码; - ）。
答案 1 :(得分：54)
我使用Tuple作为Dictionary中的键。
public class Tuple<T1, T2> {
public T1 Item1 { get; private set; }
public T2 Item2 { get; private set; }
// implementation details
}
请务必覆盖Equals和GetHashCode，并根据需要定义operator!=和operator==。您可以展开Tuple以根据需要保留更多项目。 .NET 4.0将包含内置的Tuple。
答案 2 :(得分：31)
元组将在.Net 4.0中。在那之前，你也可以使用
Dictionary<key1, Dictionary<key2, TypeObject>>
或者，创建一个自定义集合类来表示这个......
public class TwoKeyDictionary<K1, K2, T>:
Dictionary<K1, Dictionary<K2, T>> { }
或者，有三把钥匙......
public class ThreeKeyDictionary<K1, K2, K3, T> :
Dictionary<K1, Dictionary<K2, Dictionary<K3, T>>> { }
答案 3 :(得分：18)
这里有很多好的解决方案，
我在这里缺少的是一个基于Tuple类型构建的实现，所以我自己写了一个。
由于它只是继承自Dictionary<Tuple<T1,T2>, T>，因此您可以始终使用这两种方式。
var dict = new Dictionary<int, int, Row>();
var row = new Row();
dict.Add(1, 2, row);
dict.Add(Tuple.Create(1, 2, row));
dict.Add(new Tuple<int, int>(1, 2));
这是代码。
public class Dictionary<TKey1,TKey2,TValue> : Dictionary<Tuple<TKey1, TKey2>, TValue>, IDictionary<Tuple<TKey1, TKey2>, TValue>
{
public TValue this[TKey1 key1, TKey2 key2]
{
get { return base[Tuple.Create(key1, key2)]; }
set { base[Tuple.Create(key1, key2)] = value; }
}
public void Add(TKey1 key1, TKey2 key2, TValue value)
{
base.Add(Tuple.Create(key1, key2), value);
}
public bool ContainsKey(TKey1 key1, TKey2 key2)
{
return base.ContainsKey(Tuple.Create(key1, key2));
}
}
请注意，此实现取决于Tuple.Equals（）实现本身：
http://msdn.microsoft.com/en-us/library/dd270346(v=vs.110).aspx
在以下条件下，obj参数被视为等于当前实例：
答案 4 :(得分：11)
我写了并成功地使用了它。
public class MultiKeyDictionary<K1, K2, V> : Dictionary<K1, Dictionary<K2, V>> {
public V this[K1 key1, K2 key2] {
get {
if (!ContainsKey(key1) || !this[key1].ContainsKey(key2))
throw new ArgumentOutOfRangeException();
return base[key1][key2];
}
set {
if (!ContainsKey(key1))
this[key1] = new Dictionary<K2, V>();
this[key1][key2] = value;
}
}
public void Add(K1 key1, K2 key2, V value) {
if (!ContainsKey(key1))
this[key1] = new Dictionary<K2, V>();
this[key1][key2] = value;
}
public bool ContainsKey(K1 key1, K2 key2) {
return base.ContainsKey(key1) && this[key1].ContainsKey(key2);
}
public new IEnumerable<V> Values {
get {
return from baseDict in base.Values
from baseKey in baseDict.Keys
select baseDict[baseKey];
}
}
}
public class MultiKeyDictionary<K1, K2, K3, V> : Dictionary<K1, MultiKeyDictionary<K2, K3, V>> {
public V this[K1 key1, K2 key2, K3 key3] {
get {
return ContainsKey(key1) ? this[key1][key2, key3] : default(V);
}
set {
if (!ContainsKey(key1))
this[key1] = new MultiKeyDictionary<K2, K3, V>();
this[key1][key2, key3] = value;
}
}
public bool ContainsKey(K1 key1, K2 key2, K3 key3) {
return base.ContainsKey(key1) && this[key1].ContainsKey(key2, key3);
}
}
public class MultiKeyDictionary<K1, K2, K3, K4, V> : Dictionary<K1, MultiKeyDictionary<K2, K3, K4, V>> {
public V this[K1 key1, K2 key2, K3 key3, K4 key4] {
get {
return ContainsKey(key1) ? this[key1][key2, key3, key4] : default(V);
}
set {
if (!ContainsKey(key1))
this[key1] = new MultiKeyDictionary<K2, K3, K4, V>();
this[key1][key2, key3, key4] = value;
}
}
public bool ContainsKey(K1 key1, K2 key2, K3 key3, K4 key4) {
return base.ContainsKey(key1) && this[key1].ContainsKey(key2, key3, key4);
}
}
public class MultiKeyDictionary<K1, K2, K3, K4, K5, V> : Dictionary<K1, MultiKeyDictionary<K2, K3, K4, K5, V>> {
public V this[K1 key1, K2 key2, K3 key3, K4 key4, K5 key5] {
get {
return ContainsKey(key1) ? this[key1][key2, key3, key4, key5] : default(V);
}
set {
if (!ContainsKey(key1))
this[key1] = new MultiKeyDictionary<K2, K3, K4, K5, V>();
this[key1][key2, key3, key4, key5] = value;
}
}
public bool ContainsKey(K1 key1, K2 key2, K3 key3, K4 key4, K5 key5) {
return base.ContainsKey(key1) && this[key1].ContainsKey(key2, key3, key4, key5);
}
}
public class MultiKeyDictionary<K1, K2, K3, K4, K5, K6, V> : Dictionary<K1, MultiKeyDictionary<K2, K3, K4, K5, K6, V>> {
public V this[K1 key1, K2 key2, K3 key3, K4 key4, K5 key5, K6 key6] {
get {
return ContainsKey(key1) ? this[key1][key2, key3, key4, key5, key6] : default(V);
}
set {
if (!ContainsKey(key1))
this[key1] = new MultiKeyDictionary<K2, K3, K4, K5, K6, V>();
this[key1][key2, key3, key4, key5, key6] = value;
}
}
public bool ContainsKey(K1 key1, K2 key2, K3 key3, K4 key4, K5 key5, K6 key6) {
return base.ContainsKey(key1) && this[key1].ContainsKey(key2, key3, key4, key5, key6);
}
}
public class MultiKeyDictionary<K1, K2, K3, K4, K5, K6, K7, V> : Dictionary<K1, MultiKeyDictionary<K2, K3, K4, K5, K6, K7, V>> {
public V this[K1 key1, K2 key2, K3 key3, K4 key4, K5 key5, K6 key6, K7 key7] {
get {
return ContainsKey(key1) ? this[key1][key2, key3, key4, key5, key6, key7] : default(V);
}
set {
if (!ContainsKey(key1))
this[key1] = new MultiKeyDictionary<K2, K3, K4, K5, K6, K7, V>();
this[key1][key2, key3, key4, key5, key6, key7] = value;
}
}
public bool ContainsKey(K1 key1, K2 key2, K3 key3, K4 key4, K5 key5, K6 key6, K7 key7) {
return base.ContainsKey(key1) && this[key1].ContainsKey(key2, key3, key4, key5, key6, key7);
}
}
public class MultiKeyDictionary<K1, K2, K3, K4, K5, K6, K7, K8, V> : Dictionary<K1, MultiKeyDictionary<K2, K3, K4, K5, K6, K7, K8, V>> {
public V this[K1 key1, K2 key2, K3 key3, K4 key4, K5 key5, K6 key6, K7 key7, K8 key8] {
get {
return ContainsKey(key1) ? this[key1][key2, key3, key4, key5, key6, key7, key8] : default(V);
}
set {
if (!ContainsKey(key1))
this[key1] = new MultiKeyDictionary<K2, K3, K4, K5, K6, K7, K8, V>();
this[key1][key2, key3, key4, key5, key6, key7, key8] = value;
}
}
public bool ContainsKey(K1 key1, K2 key2, K3 key3, K4 key4, K5 key5, K6 key6, K7 key7, K8 key8) {
return base.ContainsKey(key1) && this[key1].ContainsKey(key2, key3, key4, key5, key6, key7, key8);
}
}
public class MultiKeyDictionary<K1, K2, K3, K4, K5, K6, K7, K8, K9, V> : Dictionary<K1, MultiKeyDictionary<K2, K3, K4, K5, K6, K7, K8, K9, V>> {
public V this[K1 key1, K2 key2, K3 key3, K4 key4, K5 key5, K6 key6, K7 key7, K8 key8, K9 key9] {
get {
return ContainsKey(key1) ? this[key1][key2, key3, key4, key5, key6, key7, key8, key9] : default(V);
}
set {
if (!ContainsKey(key1))
this[key1] = new MultiKeyDictionary<K2, K3, K4, K5, K6, K7, K8, K9, V>();
this[key1][key2, key3, key4, key5, key6, key7, key8, key9] = value;
}
}
public bool ContainsKey(K1 key1, K2 key2, K3 key3, K4 key4, K5 key5, K6 key6, K7 key7, K8 key8, K9 key9) {
return base.ContainsKey(key1) && this[key1].ContainsKey(key2, key3, key4, key5, key6, key7, key8, key9);
}
}
public class MultiKeyDictionary<K1, K2, K3, K4, K5, K6, K7, K8, K9, K10, V> : Dictionary<K1, MultiKeyDictionary<K2, K3, K4, K5, K6, K7, K8, K9, K10, V>> {
public V this[K1 key1, K2 key2, K3 key3, K4 key4, K5 key5, K6 key6, K7 key7, K8 key8, K9 key9, K10 key10] {
get {
return ContainsKey(key1) ? this[key1][key2, key3, key4, key5, key6, key7, key8, key9, key10] : default(V);
}
set {
if (!ContainsKey(key1))
this[key1] = new MultiKeyDictionary<K2, K3, K4, K5, K6, K7, K8, K9, K10, V>();
this[key1][key2, key3, key4, key5, key6, key7, key8, key9, key10] = value;
}
}
public bool ContainsKey(K1 key1, K2 key2, K3 key3, K4 key4, K5 key5, K6 key6, K7 key7, K8 key8, K9 key9, K10 key10) {
return base.ContainsKey(key1) && this[key1].ContainsKey(key2, key3, key4, key5, key6, key7, key8, key9, key10);
}
}
public class MultiKeyDictionary<K1, K2, K3, K4, K5, K6, K7, K8, K9, K10, K11, V> : Dictionary<K1, MultiKeyDictionary<K2, K3, K4, K5, K6, K7, K8, K9, K10, K11, V>> {
public V this[K1 key1, K2 key2, K3 key3, K4 key4, K5 key5, K6 key6, K7 key7, K8 key8, K9 key9, K10 key10, K11 key11] {
get {
return ContainsKey(key1) ? this[key1][key2, key3, key4, key5, key6, key7, key8, key9, key10, key11] : default(V);
}
set {
if (!ContainsKey(key1))
this[key1] = new MultiKeyDictionary<K2, K3, K4, K5, K6, K7, K8, K9, K10, K11, V>();
this[key1][key2, key3, key4, key5, key6, key7, key8, key9, key10, key11] = value;
}
}
public bool ContainsKey(K1 key1, K2 key2, K3 key3, K4 key4, K5 key5, K6 key6, K7 key7, K8 key8, K9 key9, K10 key10, K11 key11) {
return base.ContainsKey(key1) && this[key1].ContainsKey(key2, key3, key4, key5, key6, key7, key8, key9, key10, key11);
}
}
答案 5 :(得分：7)
我目前只是简单地将密钥连接成一个字符串作为解决方法。当然，这不适用于非字符串键。我也很想知道答案。
答案 6 :(得分：6)
看看Wintellect的PowerCollections（CodePlex download）。我认为他们的MultiDictionary就是这样的。
这是一本字典词典，所以你有2个键可以访问每个对象，主词典的键可以获得所需的子词典，然后是子词典的第二个键，可以获得所需的项目。这是你的意思吗？
答案 7 :(得分：6)
<{p>
new Dictionary<KeyValuePair<object, object>, object>是否有问题？
答案 8 :(得分：6)
我经常使用它，因为它很短并且提供了我需要的语法糖......
public class MultiKeyDictionary<T1, T2, T3> : Dictionary<T1, Dictionary<T2, T3>>
{
new public Dictionary<T2, T3> this[T1 key]
{
get
{
if (!ContainsKey(key))
Add(key, new Dictionary<T2, T3>());
Dictionary<T2, T3> returnObj;
TryGetValue(key, out returnObj);
return returnObj;
}
}
}
使用它：
dict[cat][fish] = 9000;
其中“Cat”键也不一定存在。
答案 9 :(得分：4)
我用Google搜索了这个：http://www.codeproject.com/KB/recipes/multikey-dictionary.aspx。我想这与使用struct在常规字典中包含2个键相比的主要特征是，您可以稍后通过其中一个键引用，而不必提供2个键。
答案 10 :(得分：3)
如果有人正在寻找ToMultiKeyDictionary()这里的实施应该适用于大多数答案（基于Herman's）：
public static class Extensions_MultiKeyDictionary {
public static MultiKeyDictionary<K1, K2, V> ToMultiKeyDictionary<S, K1, K2, V>(this IEnumerable<S> items, Func<S, K1> key1, Func<S, K2> key2, Func<S, V> value) {
var dict = new MultiKeyDictionary<K1, K2, V>();
foreach (S i in items) {
dict.Add(key1(i), key2(i), value(i));
}
return dict;
}
public static MultiKeyDictionary<K1, K2, K3, V> ToMultiKeyDictionary<S, K1, K2, K3, V>(this IEnumerable<S> items, Func<S, K1> key1, Func<S, K2> key2, Func<S, K3> key3, Func<S, V> value) {
var dict = new MultiKeyDictionary<K1, K2, K3, V>();
foreach (S i in items) {
dict.Add(key1(i), key2(i), key3(i), value(i));
}
return dict;
}
}
答案 11 :(得分：2)
我认为你需要一个类似Tuple2的课程。确保它是GetHashCode（）和Equals（）基于两个包含的元素。
请参阅Tuples in C#
答案 12 :(得分：2)
您可以使用Dictionary<TKey1,Dictionary<TKey2,TValue>>吗？
你甚至可以将其子类化：
public class DualKeyDictionary<TKey1,TKey2,TValue> : Dictionary<TKey1,Dictionary<TKey2,TValue>>
编辑：现在这是一个重复的答案。它的实用性也受到限制。虽然它确实“有效”并且提供了代码dict[key1][key2]的能力，但是有很多“变通办法”可以让它“正常工作”。
HOWEVER：只是为了踢，但仍然可以实现Dictionary，但此时它有点冗长：
public class DualKeyDictionary<TKey1, TKey2, TValue> : Dictionary<TKey1, Dictionary<TKey2, TValue>> , IDictionary< object[], TValue >
{
#region IDictionary<object[],TValue> Members
void IDictionary<object[], TValue>.Add( object[] key, TValue value )
{
if ( key == null || key.Length != 2 )
throw new ArgumentException( "Invalid Key" );
TKey1 key1 = key[0] as TKey1;
TKey2 key2 = key[1] as TKey2;
if ( !ContainsKey( key1 ) )
Add( key1, new Dictionary<TKey2, TValue>() );
this[key1][key2] = value;
}
bool IDictionary<object[], TValue>.ContainsKey( object[] key )
{
if ( key == null || key.Length != 2 )
throw new ArgumentException( "Invalid Key" );
TKey1 key1 = key[0] as TKey1;
TKey2 key2 = key[1] as TKey2;
if ( !ContainsKey( key1 ) )
return false;
if ( !this[key1].ContainsKey( key2 ) )
return false;
return true;
}
答案 13 :(得分：2)
这是一个充实的对类示例，它可以用作词典的关键。
public class Pair<T1, T2> {
public T1 Left { get; private set; }
public T2 Right { get; private set; }
public Pair(T1 t1, T2 t2) {
Left=t1;
Right=t2;
}
public override bool Equals(object obj) {
if(ReferenceEquals(null, obj)) return false;
if(ReferenceEquals(this, obj)) return true;
if(obj.GetType()!=typeof(Pair<T1, T2>)) return false;
return Equals((Pair<T1, T2>)obj);
}
public bool Equals(Pair<T1, T2> obj) {
if(ReferenceEquals(null, obj)) return false;
if(ReferenceEquals(this, obj)) return true;
return Equals(obj.Left, Left) && Equals(obj.Right, Right);
}
public override int GetHashCode() {
unchecked {
return (Left.GetHashCode()*397)^Right.GetHashCode();
}
}
}
答案 14 :(得分：1)
这是我的实施。我想要隐藏Tuple概念的实现。
public class TwoKeyDictionary<TKey1, TKey2, TValue> : Dictionary<TwoKey<TKey1, TKey2>, TValue>
{
public static TwoKey<TKey1, TKey2> Key(TKey1 key1, TKey2 key2)
{
return new TwoKey<TKey1, TKey2>(key1, key2);
}
public TValue this[TKey1 key1, TKey2 key2]
{
get { return this[Key(key1, key2)]; }
set { this[Key(key1, key2)] = value; }
}
public void Add(TKey1 key1, TKey2 key2, TValue value)
{
Add(Key(key1, key2), value);
}
public bool ContainsKey(TKey1 key1, TKey2 key2)
{
return ContainsKey(Key(key1, key2));
}
}
public class TwoKey<TKey1, TKey2> : Tuple<TKey1, TKey2>
{
public TwoKey(TKey1 item1, TKey2 item2) : base(item1, item2) { }
public override string ToString()
{
return string.Format("({0},{1})", Item1, Item2);
}
}
它有助于保持使用看起来像字典
item.Add(1, "D", 5.6);
value = item[1, "D"];
答案 15 :(得分：0)
这是使用Tuple类和Dictionary的另一个例子。
// Setup Dictionary
Dictionary<Tuple<string, string>, string> testDictionary = new Dictionary<Tuple<string, string>, string>
{
{new Tuple<string, string>("key1","key2"), "value1"},
{new Tuple<string, string>("key1","key3"), "value2"},
{new Tuple<string, string>("key2","key3"), "value3"}
};
//Query Dictionary
public string FindValue(string stuff1, string stuff2)
{
return testDictionary[Tuple.Create(stuff1, stuff2)];
}