Hashtable with MultiDimensional Key in C#

C#DictionaryHashtable

C# Problem Overview


I'm basically looking for a way to access a hashtable value using a two-dimensional typed key in c#.

Eventually I would be able to do something like this

HashTable[1][false] = 5;
int a = HashTable[1][false];
//a = 5

This is what I've been trying...hasn't worked

Hashtable test = new Hashtable();
test.Add(new Dictionary<int, bool>() { { 1, true } }, 555);
Dictionary<int, bool> temp = new Dictionary<int, bool>() {{1, true}};
string testz = test[temp].ToString(); 

C# Solutions


Solution 1 - C#

I think a better approach is to encapsulate the many fields of your multi-dimensional key into a class / struct. For example

struct Key {
  public readonly int Dimension1;
  public readonly bool Dimension2;
  public Key(int p1, bool p2) {
    Dimension1 = p1;
    Dimension2 = p2;
  }
  // Equals and GetHashCode ommitted
}

Now you can create and use a normal HashTable and use this wrapper as a Key.

Solution 2 - C#

You can do this in C# 7.0 now with the new tuples:

// Declare
var test = new Dictionary<(int, bool), int>();

// Add
test.Add((1, false), 5);

// Get
int a = test[(1, false)];

Solution 3 - C#

How about using a regular Dictionary with some kind of Tuple structure as a key?

public class TwoKeyDictionary<K1,K2,V>
{
    private readonly Dictionary<Pair<K1,K2>, V> _dict;

    public V this[K1 k1, K2 k2]
    {
        get { return _dict[new Pair(k1,k2)]; }
    }

    private struct Pair
    {
        public K1 First;
        public K2 Second;

        public override Int32 GetHashCode()
        {
            return First.GetHashCode() ^ Second.GetHashCode();
        }

        // ... Equals, ctor, etc...
    }
}

Solution 4 - C#

I think this might be closer to what you're looking for...

var data = new Dictionary<int, Dictionary<bool, int>>();

Solution 5 - C#

Just in case anyone is here recently, an example of how to do this the quick and dirty way in .Net 4.0, as described by one of the commenters.

class Program
{
  static void Main(string[] args)
  {
     var twoDic = new Dictionary<Tuple<int, bool>, String>();
     twoDic.Add(new Tuple<int, bool>(3, true), "3 and true." );
     twoDic.Add(new Tuple<int, bool>(4, true), "4 and true." );
     twoDic.Add(new Tuple<int, bool>(3, false), "3 and false.");

     // Will throw exception. Item with the same key already exists.
     // twoDic.Add(new Tuple<int, bool>(3, true), "3 and true." );

     Console.WriteLine(twoDic[new Tuple<int, bool>(3,false)]);
     Console.WriteLine(twoDic[new Tuple<int, bool>(4,true)]);
     // Outputs "3 and false." and "4 and true."
  }
}

Solution 6 - C#

I'd suggest a slight variation on jachymko's solution which will allow you to avoid creating a class for key pairs. Instead, wrap a private dictionary of dictionaries, as so:

public class MultiDictionary<K1, K2, V>
{
    private Dictionary<K1, Dictionary<K2, V>> dict = 
        new Dictionary<K1, Dictionary<K2, V>>();

    public V this[K1 key1, K2 key2]
    {
        get
        {
            return dict[key1][key2];
        }

        set
        {
            if (!dict.ContainsKey(key1))
            {
                dict[key1] = new Dictionary<K2, V>();
            }
            dict[key1][key2] = value;
        }
    }
}

Solution 7 - C#

You need a key class for the Dictonary that implements GetHashCode correctly. And you can extend Dictonary to let you access it in a friendly way.

The KeyPair class:

public class KeyPair<Tkey1, Tkey2>
{
    public KeyPair(Tkey1 key1, Tkey2 key2)
    {
        Key1 = key1;
        Key2 = key2;
    }

    public Tkey1 Key1 { get; set; }
    public Tkey2 Key2 { get; set; }

    public override int GetHashCode()
    {
        return Key1.GetHashCode() ^ Key2.GetHashCode();
    }
    public override bool Equals(object obj)
    {
        KeyPair<Tkey1, Tkey2> o = obj as KeyPair<Tkey1, Tkey2>;
        if (o == null)
            return false;
        else
            return Key1.Equals(o.Key1) && Key2.Equals(o.Key2);
    }
}

Extend Dictonary<>:

public class KeyPairDictonary<Tkey1, Tkey2, Tvalue> 
    : Dictionary<KeyPair<Tkey1, Tkey2>, Tvalue>
{
    public Tvalue this[Tkey1 key1, Tkey2 key2]
    {
        get
        {
            return this[new KeyPair<Tkey1, Tkey2>(key1, key2)];
        }
        set
        {
            this[new KeyPair<Tkey1, Tkey2>(key1, key2)] = value;
        }
    }
}

You can use it like this:

KeyPairDictonary<int, bool, string> dict = 
    new KeyPairDictonary<int, bool, string>();
    
dict[1, false] = "test";
string test = dict[1, false];

Solution 8 - C#

I would suggest that you create a small custom class exposing the bool and int properties, and override its GetHashCode and Equals methods, then use this as the key.

Solution 9 - C#

Essentially you need to use an embedded hashtable. If you think about your question, a hashtable with two keys is a function with two independent variables, and f(x,y) is 2-dimensional by definition.

But you want to use it like it were one hashtable, and not embedded hashes. So what you need to do is create an object that wraps around that embedded hashtable idea and operates like a single hash.

A couple of snags:

  • You want to iterate over it, so you need to overwrite the GetEnumerator() method. And you need your own Iterator that will iterate correctly in 2 dimensions.
  • You need to do more checking to be sure that there are no duplicates.

I have included my code to do it:

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Collections;
using System.Windows.Forms;

namespace YourProjectNameHere
{
    public class Hashtable2D
    {
        /// <summary>
        /// This is a hashtable of hashtables
        /// The X dim is the root key, and the y is the internal hashes key
        /// </summary>
        /// 
        private Hashtable root = new Hashtable();
        public bool overwriteDuplicates = false;
        public bool alertOnDuplicates = true;

        public void Add(object key_x, object key_y, object toStore)
        {
            if(root[key_x]!=null)//If key_x has already been entered 
            {
                Hashtable tempHT = (Hashtable)root[key_x];//IF the hash table does not exist then focus will skip to the catch statement
                if (tempHT[key_y] == null)  tempHT.Add(key_y, toStore);
                else handleDuplicate(tempHT, key_y, toStore);
            }else{//Making a new hashtable 
                Hashtable tempHT = new Hashtable();
                tempHT.Add(key_y, toStore);
                root.Add(key_x, tempHT);
            }

        }

        public void Remove(object key_x, object key_y)
        {
            try{
                ((Hashtable)root[key_x]).Remove(key_y);
            }catch(Exception e){
                MessageBox.Show("That item does not exist");
            }

        }
        
        public void handleDuplicate (Hashtable tempHT, object key_y, object toStore)
        {
            if (alertOnDuplicates) MessageBox.Show("This Item already Exists in the collection");

            if (overwriteDuplicates)
            {
                tempHT.Remove(key_y);
                tempHT.Add(key_y,toStore);
            }
        }

        public object getItem(object key_x, object key_y)
        {
            Hashtable tempHT = (Hashtable)root[key_x];
            return tempHT[key_y];
        }

        public ClassEnumerator GetEnumerator()
        {
            return new ClassEnumerator(root);
        }

        public class ClassEnumerator : IEnumerator
        {
            private Hashtable ht;
            private IEnumerator iEnumRoot;
            private Hashtable innerHt;
            private IEnumerator iEnumInner;

            public ClassEnumerator(Hashtable _ht)
            {
                ht = _ht;
                iEnumRoot = ht.GetEnumerator();

                iEnumRoot.MoveNext();//THIS ASSUMES THAT THERE IS AT LEAST ONE ITEM

                innerHt = (Hashtable)((DictionaryEntry)iEnumRoot.Current).Value;
                iEnumInner = innerHt.GetEnumerator();
            }

            #region IEnumerator Members

            public void Reset()
            {
                iEnumRoot = ht.GetEnumerator();
            }

            public object Current
            {
                get
                {
                    return iEnumInner.Current; 
                }
            }

            public bool MoveNext()
            {
                if(!iEnumInner.MoveNext())
                {
                    if (!iEnumRoot.MoveNext()) return false;
                    innerHt = (Hashtable)((DictionaryEntry)iEnumRoot.Current).Value;
                    iEnumInner = innerHt.GetEnumerator();
                    iEnumInner.MoveNext();
                }
                return true;
            }

            #endregion
        }
        
    }
}

Solution 10 - C#

You might be able to "double-nest" your hashtables - in other words, your main Dictionary is of type Dictionary<int, Dictionary<bool, my_return_type>>.

That accomplishes your goal of being able to use the double bracket notation in your first code snippet.

Of course, the management side is a little trickier. Every time you add an entry, you need to test if the main dictionary contains a dictionary for the primary key, and add a new dictionary if not, then add the secondary key and value to the inner Dictionary.

Solution 11 - C#

Could you use a Dictionary<KeyValuePair<int,bool>,int>?

Solution 12 - C#

Wrap your two-dimensional key in a separate type and use that type as a key. Also consider overriding GetHashCode() and Equals() methods. Preferably use Dictionary<> instead of HashTable since apparently you can use that.

Solution 13 - C#

A quick and dirty way would be to create a composite key from the two pieces of information, e.g.

IDictionary<string, int> values = new Dictionary<string, int>();
int i = ...;
bool b = ...;
string key = string.Concat(i, '\0', b);
values[key] = 555;

To encapsulate this a bit better you could wrap the dictionary:

public class MyDict
{
    private readonly IDictionary<string, int> values = new Dictionary<string, int>();

    public int this[int i, bool b]
    {
        get
        {
            string key = BuildKey(i, b);
            return values[key];
        }

        set
        {
            string key = BuildKey(i, b);
            values[key] = value;
        }
    }

    private static string BuildKey(int i, bool b)
    {
        return string.Concat(i, '\0', b);
    }
}

To make this more robust, encapsulate the composite key as a type, e.g. a class that contains the two fields, ensuring you override the Equals() and GetHashCode() methods correctly.

Solution 14 - C#

Look, this code works just fine:

    public Form1()
    {
            InitializeComponent();
    }

    private void Form1_Load(object sender, EventArgs e)
    {

        this.Services = new Dictionary<object, Hashtable>();
        this.Services.Add("array1", new Hashtable());
        
        this.Services["array1"]["qwe"] = "123";
        this.Services["array1"][22] = 223;

        object zz = null;
        zz = this.Services["array1"]["qwe"];
        MessageBox.Show(zz.ToString()); // shows qwe

        zz = this.Services["array1"][22];
        MessageBox.Show(zz.ToString()); // shows 22
    }

Now we just need a wrapper to avoid manually doing this.Services.Add("array1", new Hashtable());

Solution 15 - C#

I think the easiest way to do it now is to use Tupple.Create and ValueTuple.Create:

> var k1 =  Tuple.Create("test", int.MinValue, DateTime.MinValue, double.MinValue);
> var k2 = Tuple.Create("test", int.MinValue, DateTime.MinValue, double.MinValue);
> var dict = new Dictionary<object, object>();
> dict.Add(k1, "item");
> dict.Add(k2, "item");
An item with the same key has already been added....
> dict[k1] == dict[k2]
true

or use new c#7 tuple syntax to create tuple-keys:

var k = (item1: "value1", item2: 123);

Solution 16 - C#

This is my nested Dictionary implementation:

public class TwoKeysDictionary<K1, K2, T>:
		Dictionary<K1, Dictionary<K2, T>>
{
	public T this[K1 key1, K2 key2]
	{
		get => base.ContainsKey(key1) && base[key1].ContainsKey(key2) ? base[key1][key2] : default;
		set
		{
			if (ContainsKey(key1) && base[key1].ContainsKey(key2))
				base[key1][key2] = value;
			else
				Add(key1, key2, value);
		}
	}

	public void Add(K1 key1, K2 key2, T value)
	{
		if (ContainsKey(key1))
		{
			if (base[key1].ContainsKey(key2))
				throw new Exception("Couple " + key1 + "/" + key2 + " already exists!");
			base[key1].Add(key2, value);
		}
		else
			Add(key1, new Dictionary<K2, T>() { { key2, value } });
	}

	public bool ContainsKey(K1 key1, K2 key2) => ContainsKey(key1) && base[key1].ContainsKey(key2);
}

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