Two-way / bidirectional Dictionary in C#?

C#.NetDictionary

C# Problem Overview


I want to store words in a dictionary in following way:

I can get word code by word: dict["SomeWord"] -> 123 and get word by word code: dict[123] -> "SomeWord"

Is it real? Of course one way to do it is two dictionaries: Dictionary<string,int> and Dictionary<int,string> but is there another way?

C# Solutions


Solution 1 - C#

I wrote a quick couple of classes that lets you do what you want. You'd probably need to extend it with more features, but it is a good starting point.

The use of the code looks like this:

var map = new Map<int, string>();

map.Add(42, "Hello");

Console.WriteLine(map.Forward[42]);
// Outputs "Hello"

Console.WriteLine(map.Reverse["Hello"]);
//Outputs 42

Here's the definition:

public class Map<T1, T2>
{
	private Dictionary<T1, T2> _forward = new Dictionary<T1, T2>();
	private Dictionary<T2, T1> _reverse = new Dictionary<T2, T1>();
	
	public Map()
	{
		this.Forward = new Indexer<T1, T2>(_forward);
		this.Reverse = new Indexer<T2, T1>(_reverse);
	}
	
	public class Indexer<T3, T4>
	{
		private Dictionary<T3, T4> _dictionary;
		public Indexer(Dictionary<T3, T4> dictionary)
		{
			_dictionary = dictionary;
		}
		public T4 this[T3 index]
		{
			get { return _dictionary[index]; }
			set { _dictionary[index] = value; }
		}
	}
	
	public void Add(T1 t1, T2 t2)
	{
		_forward.Add(t1, t2);
		_reverse.Add(t2, t1);
	}
	
	public Indexer<T1, T2> Forward { get; private set; }
	public Indexer<T2, T1> Reverse { get; private set; }
}

Solution 2 - C#

Regrettably, you need two dictionaries, one for each direction. However, you can easily get the inverse dictionary using LINQ:

Dictionary<T1, T2> dict = new Dictionary<T1, T2>();
Dictionary<T2, T1> dictInverse = dict.ToDictionary((i) => i.Value, (i) => i.Key);

Solution 3 - C#

Expanded on Enigmativity code by adding initializes and Contains method.

public class Map<T1, T2> : IEnumerable<KeyValuePair<T1, T2>>
{
    private readonly Dictionary<T1, T2> _forward = new Dictionary<T1, T2>();
    private readonly Dictionary<T2, T1> _reverse = new Dictionary<T2, T1>();

    public Map()
    {
        Forward = new Indexer<T1, T2>(_forward);
        Reverse = new Indexer<T2, T1>(_reverse);
    }

    public Indexer<T1, T2> Forward { get; private set; }
    public Indexer<T2, T1> Reverse { get; private set; }

    public void Add(T1 t1, T2 t2)
    {
        _forward.Add(t1, t2);
        _reverse.Add(t2, t1);
    }

	public void Remove(T1 t1)
	{
		T2 revKey = Forward[t1];
		_forward.Remove(t1);
		_reverse.Remove(revKey);
	}
	
	public void Remove(T2 t2)
	{
		T1 forwardKey = Reverse[t2];
		_reverse.Remove(t2);
		_forward.Remove(forwardKey);
	}

    IEnumerator IEnumerable.GetEnumerator()
    {
        return GetEnumerator();
    }

    public IEnumerator<KeyValuePair<T1, T2>> GetEnumerator()
    {
        return _forward.GetEnumerator();
    }

    public class Indexer<T3, T4>
    {
        private readonly Dictionary<T3, T4> _dictionary;

        public Indexer(Dictionary<T3, T4> dictionary)
        {
            _dictionary = dictionary;
        }

        public T4 this[T3 index]
        {
            get { return _dictionary[index]; }
            set { _dictionary[index] = value; }
        }

        public bool Contains(T3 key)
        {
            return _dictionary.ContainsKey(key);
        }
    }
}

Here is a use case, check valid parentheses

public static class ValidParenthesisExt
{
    private static readonly Map<char, char>
        _parenthesis = new Map<char, char>
        {
            {'(', ')'},
            {'{', '}'},
            {'[', ']'}
        };

    public static bool IsValidParenthesis(this string input)
    {
        var stack = new Stack<char>();
        foreach (var c in input)
        {
            if (_parenthesis.Forward.Contains(c))
                stack.Push(c);
            else
            {
                if (stack.Count == 0) return false;
                if (_parenthesis.Reverse[c] != stack.Pop())
                    return false;
            }
        }
        return stack.Count == 0;
    }
}

Solution 4 - C#

You could use two dictionaries, as others have said, but note also that if both TKey and TValue are the of same type (and their runtime value domains are known to be disjoint) then you can just use the same dictionary by creating two entries for each key/value pairing:

dict["SomeWord"]= "123" and dict["123"]="SomeWord"

This way a single dictionary can be used for either type of lookup.

Solution 5 - C#

What the heck, I'll throw my version into the mix:

public class BijectiveDictionary<TKey, TValue> 
{
	private EqualityComparer<TKey> _keyComparer;
	private Dictionary<TKey, ISet<TValue>> _forwardLookup;
	private EqualityComparer<TValue> _valueComparer;
	private Dictionary<TValue, ISet<TKey>> _reverseLookup;             

	public BijectiveDictionary()
		: this(EqualityComparer<TKey>.Default, EqualityComparer<TValue>.Default)
	{
	}

	public BijectiveDictionary(EqualityComparer<TKey> keyComparer, EqualityComparer<TValue> valueComparer)
		: this(0, EqualityComparer<TKey>.Default, EqualityComparer<TValue>.Default)
	{
	}

	public BijectiveDictionary(int capacity, EqualityComparer<TKey> keyComparer, EqualityComparer<TValue> valueComparer)
	{
		_keyComparer = keyComparer;
		_forwardLookup = new Dictionary<TKey, ISet<TValue>>(capacity, keyComparer);            
		_valueComparer = valueComparer;
		_reverseLookup = new Dictionary<TValue, ISet<TKey>>(capacity, valueComparer);            
	}

	public void Add(TKey key, TValue value)
	{
		AddForward(key, value);
		AddReverse(key, value);
	}

	public void AddForward(TKey key, TValue value)
	{
		ISet<TValue> values;
		if (!_forwardLookup.TryGetValue(key, out values))
		{
			values = new HashSet<TValue>(_valueComparer);
			_forwardLookup.Add(key, values);
		}
		values.Add(value);
	}

	public void AddReverse(TKey key, TValue value) 
	{
		ISet<TKey> keys;
		if (!_reverseLookup.TryGetValue(value, out keys))
		{
			keys = new HashSet<TKey>(_keyComparer);
			_reverseLookup.Add(value, keys);
		}
		keys.Add(key);
	}

	public bool TryGetReverse(TValue value, out ISet<TKey> keys)
	{
		return _reverseLookup.TryGetValue(value, out keys);
	}

	public ISet<TKey> GetReverse(TValue value)
	{
		ISet<TKey> keys;
		TryGetReverse(value, out keys);
		return keys;
	}

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

	public bool TryGetForward(TKey key, out ISet<TValue> values)
	{
		return _forwardLookup.TryGetValue(key, out values);
	}

	public ISet<TValue> GetForward(TKey key)
	{
		ISet<TValue> values;
		TryGetForward(key, out values);
		return values;
	}

	public bool ContainsReverse(TValue value)
	{
		return _reverseLookup.ContainsKey(value);
	}

	public void Clear()
	{
		_forwardLookup.Clear();
		_reverseLookup.Clear();
	}
}

Add some data to it:

var lookup = new BijectiveDictionary<int, int>();

lookup.Add(1, 2);
lookup.Add(1, 3);
lookup.Add(1, 4);
lookup.Add(1, 5);

lookup.Add(6, 2);
lookup.Add(6, 8);
lookup.Add(6, 9);
lookup.Add(6, 10);

And then do the lookup:

lookup[2] --> 1, 6
lookup[3] --> 1
lookup[8] --> 6

Solution 6 - C#

You can use this extension method, although it uses enumeration, and thus may not be as performant for large data sets. If you are worried about efficiency, then you need two dictionaries. If you want to wrap the two dictionaries into one class, see the accepted answer for this question: https://stackoverflow.com/questions/268321/bidirectional-1-to-1-dictionary-in-c-sharp

public static class IDictionaryExtensions
{
    public static TKey FindKeyByValue<TKey, TValue>(this IDictionary<TKey, TValue> dictionary, TValue value)
    {
        if (dictionary == null)
            throw new ArgumentNullException("dictionary");

        foreach (KeyValuePair<TKey, TValue> pair in dictionary)
            if (value.Equals(pair.Value)) return pair.Key;

        throw new Exception("the value is not found in the dictionary");
    }
}

Solution 7 - C#

There is an expanded version of Enigmativity's answer available as a nuget package https://www.nuget.org/packages/BidirectionalMap/

It is open sourced here

Solution 8 - C#

A modified version of Xavier John's answer, with an additional constructor to take forward and reverse Comparers. This would support case-insensitive keys, for example. Further constructors could be added, if needed, to pass further arguments to the forward and reverse Dictionary constructors.

public class Map<T1, T2> : IEnumerable<KeyValuePair<T1, T2>>
{
    private readonly Dictionary<T1, T2> _forward;
    private readonly Dictionary<T2, T1> _reverse;

    /// <summary>
    /// Constructor that uses the default comparers for the keys in each direction.
    /// </summary>
    public Map()
        : this(null, null)
    {
    }

    /// <summary>
    /// Constructor that defines the comparers to use when comparing keys in each direction.
    /// </summary>
    /// <param name="t1Comparer">Comparer for the keys of type T1.</param>
    /// <param name="t2Comparer">Comparer for the keys of type T2.</param>
    /// <remarks>Pass null to use the default comparer.</remarks>
    public Map(IEqualityComparer<T1> t1Comparer, IEqualityComparer<T2> t2Comparer)
    {
        _forward = new Dictionary<T1, T2>(t1Comparer);
        _reverse = new Dictionary<T2, T1>(t2Comparer);
        Forward = new Indexer<T1, T2>(_forward);
        Reverse = new Indexer<T2, T1>(_reverse);
    }

    // Remainder is the same as Xavier John's answer:
    // https://stackoverflow.com/a/41907561/216440
    ...
}

Usage example, with a case-insensitive key:

Map<int, string> categories = 
new Map<int, string>(null, StringComparer.CurrentCultureIgnoreCase)
{
	{ 1, "Bedroom Furniture" },
	{ 2, "Dining Furniture" },
	{ 3, "Outdoor Furniture" }, 
	{ 4, "Kitchen Appliances" }
};

int categoryId = 3;
Console.WriteLine("Description for category ID {0}: '{1}'", 
    categoryId, categories.Forward[categoryId]);

string categoryDescription = "DINING FURNITURE";
Console.WriteLine("Category ID for description '{0}': {1}", 
    categoryDescription, categories.Reverse[categoryDescription]);

categoryDescription = "outdoor furniture";
Console.WriteLine("Category ID for description '{0}': {1}", 
    categoryDescription, categories.Reverse[categoryDescription]);

// Results:
/*
Description for category ID 3: 'Outdoor Furniture'
Category ID for description 'DINING FURNITURE': 2
Category ID for description 'outdoor furniture': 3
*/

Solution 9 - C#

Here's my code. Everything is O(1) except for the seeded constructors.

using System.Collections.Generic;
using System.Linq;

public class TwoWayDictionary<T1, T2>
{
    Dictionary<T1, T2> _Forwards = new Dictionary<T1, T2>();
    Dictionary<T2, T1> _Backwards = new Dictionary<T2, T1>();
    
    public IReadOnlyDictionary<T1, T2> Forwards => _Forwards;
    public IReadOnlyDictionary<T2, T1> Backwards => _Backwards;
    
    public IEnumerable<T1> Set1 => Forwards.Keys;
    public IEnumerable<T2> Set2 => Backwards.Keys;
    
    
    public TwoWayDictionary()
    {
        _Forwards = new Dictionary<T1, T2>();
        _Backwards = new Dictionary<T2, T1>();
    }
    
    public TwoWayDictionary(int capacity)
    {
        _Forwards = new Dictionary<T1, T2>(capacity);
        _Backwards = new Dictionary<T2, T1>(capacity);
    }
    
    public TwoWayDictionary(Dictionary<T1, T2> initial)
    {
        _Forwards = initial;
        _Backwards = initial.ToDictionary(kvp => kvp.Value, kvp => kvp.Key);
    }
    
    public TwoWayDictionary(Dictionary<T2, T1> initial)
    {
        _Backwards = initial;
        _Forwards = initial.ToDictionary(kvp => kvp.Value, kvp => kvp.Key);
    }
    
    
    public T1 this[T2 index]
    {
        get => _Backwards[index];
        set
        {
            if (_Backwards.TryGetValue(index, out var removeThis))
                _Forwards.Remove(removeThis);
    
            _Backwards[index] = value;
            _Forwards[value] = index;
        }
    }
    
    public T2 this[T1 index]
    {
        get => _Forwards[index];
        set
        {
            if (_Forwards.TryGetValue(index, out var removeThis))
                _Backwards.Remove(removeThis);
    
            _Forwards[index] = value;
            _Backwards[value] = index;
        }
    }
    
    public int Count => _Forwards.Count;
    
    public bool Contains(T1 item) => _Forwards.ContainsKey(item);
    public bool Contains(T2 item) => _Backwards.ContainsKey(item);
    
    public bool Remove(T1 item)
    {
        if (!this.Contains(item))
            return false;
    
        var t2 = _Forwards[item];
    
        _Backwards.Remove(t2);
        _Forwards.Remove(item);
    
        return true;
    }
    
    public bool Remove(T2 item)
    {
        if (!this.Contains(item))
            return false;
    
        var t1 = _Backwards[item];
    
        _Forwards.Remove(t1);
        _Backwards.Remove(item);
    
        return true;
    }
    
    public void Clear()
    {
        _Forwards.Clear();
        _Backwards.Clear();
    }
}

Solution 10 - C#

Bictionary

Here is a commingling of what I liked in each answer. It implements IEnumerable so it can use collection initializer, as you can see in the example.

Usage Constraint:

  • You are using different datatypes. (i.e., T1T2)

Code:

using System;
using System.Collections.Generic;
using System.Linq;

public class Program
{
	public static void Main()
	{
		Bictionary<string, int> bictionary = 
			new Bictionary<string,int>() {
				{ "a",1 }, 
				{ "b",2 }, 
				{ "c",3 } 
			};
		
		// test forward lookup
		Console.WriteLine(bictionary["b"]);
		// test forward lookup error
		//Console.WriteLine(bictionary["d"]);
		// test reverse lookup
		Console.WriteLine(bictionary[3]); 
		// test reverse lookup error (throws same error as forward lookup does)
		Console.WriteLine(bictionary[4]); 
	}
}

public class Bictionary<T1, T2> : Dictionary<T1, T2>
{
	public T1 this[T2 index]
	{
		get
		{
			if(!this.Any(x => x.Value.Equals(index)))
			   throw new System.Collections.Generic.KeyNotFoundException();
			return this.First(x => x.Value.Equals(index)).Key;
		}
	}
}

Fiddle:

https://dotnetfiddle.net/mTNEuw

Solution 11 - C#

This is an old issue but I wanted to add a two extension methods in case anyone finds it useful. The second is not as useful but it provides a starting point if one to one dictionaries need to be supported.

    public static Dictionary<VALUE,KEY> Inverse<KEY,VALUE>(this Dictionary<KEY,VALUE> dictionary)
    {
        if (dictionary==null || dictionary.Count == 0) { return null; }
        
        var result = new Dictionary<VALUE, KEY>(dictionary.Count);

        foreach(KeyValuePair<KEY,VALUE> entry in dictionary)
        {
            result.Add(entry.Value, entry.Key);
        }

        return result;
    }

    public static Dictionary<VALUE, KEY> SafeInverse<KEY, VALUE>(this Dictionary<KEY, VALUE> dictionary)
    {
        if (dictionary == null || dictionary.Count == 0) { return null; }

        var result = new Dictionary<VALUE, KEY>(dictionary.Count);

        foreach (KeyValuePair<KEY, VALUE> entry in dictionary)
        {
            if (result.ContainsKey(entry.Value)) { continue; }

            result.Add(entry.Value, entry.Key);
        }

        return result;
    }

Solution 12 - C#

Here's an alternative solution to those that were suggested. Removed the inner class and insured the coherence when adding/removing items

using System.Collections;
using System.Collections.Generic;

public class Map<E, F> : IEnumerable<KeyValuePair<E, F>>
{
	private readonly Dictionary<E, F> _left = new Dictionary<E, F>();
	public IReadOnlyDictionary<E, F> left => this._left;
	private readonly Dictionary<F, E> _right = new Dictionary<F, E>();
	public IReadOnlyDictionary<F, E> right => this._right;

	public void RemoveLeft(E e)
	{
		if (!this.left.ContainsKey(e)) return;
		this._right.Remove(this.left[e]);
		this._left.Remove(e);
	}

	public void RemoveRight(F f)
	{
		if (!this.right.ContainsKey(f)) return;
		this._left.Remove(this.right[f]);
		this._right.Remove(f);
	}

	public int Count()
	{
		return this.left.Count;
	}

	public void Set(E left, F right)
	{
		if (this.left.ContainsKey(left))
		{
			this.RemoveLeft(left);
		}
		if (this.right.ContainsKey(right))
		{
			this.RemoveRight(right);
		}
		this._left.Add(left, right);
		this._right.Add(right, left);
	}


	public IEnumerator<KeyValuePair<E, F>> GetEnumerator()
	{
		return this.left.GetEnumerator();
	}

	IEnumerator IEnumerable.GetEnumerator()
	{
		return this.left.GetEnumerator();
	}
}

Solution 13 - C#

This uses an indexer for the reverse lookup.
The reverse lookup is O(n) but it also does not use two dictionaries

public sealed class DictionaryDoubleKeyed : Dictionary<UInt32, string>
{   // used UInt32 as the key as it has a perfect hash
    // if most of the lookup is by word then swap
    public void Add(UInt32 ID, string Word)
    {
        if (this.ContainsValue(Word)) throw new ArgumentException();
        base.Add(ID, Word);
    }
    public UInt32 this[string Word]
    {   // this will be O(n)
        get
        {
            return this.FirstOrDefault(x => x.Value == Word).Key;
        }
    } 
}

Solution 14 - C#

The following encapsulating class utilizes linq (IEnumerable Extensions) over 1 dictionary instance.

public class TwoWayDictionary<TKey, TValue>
{
    readonly IDictionary<TKey, TValue> dict;
    readonly Func<TKey, TValue> GetValueWhereKey;
    readonly Func<TValue, TKey> GetKeyWhereValue;
    readonly bool _mustValueBeUnique = true;

    public TwoWayDictionary()
    {
        this.dict = new Dictionary<TKey, TValue>();
        this.GetValueWhereKey = (strValue) => dict.Where(kvp => Object.Equals(kvp.Key, strValue)).Select(kvp => kvp.Value).FirstOrDefault();
        this.GetKeyWhereValue = (intValue) => dict.Where(kvp => Object.Equals(kvp.Value, intValue)).Select(kvp => kvp.Key).FirstOrDefault();
    }

    public TwoWayDictionary(KeyValuePair<TKey, TValue>[] kvps)
        : this()
    {
        this.AddRange(kvps);
    }

    public void AddRange(KeyValuePair<TKey, TValue>[] kvps)
    {
        kvps.ToList().ForEach( kvp => {        
            if (!_mustValueBeUnique || !this.dict.Any(item => Object.Equals(item.Value, kvp.Value)))
            {
                dict.Add(kvp.Key, kvp.Value);
            } else {
                throw new InvalidOperationException("Value must be unique");
            }
        });
    }

    public TValue this[TKey key]
    {
        get { return GetValueWhereKey(key); }
    }

    public TKey this[TValue value]
    {
        get { return GetKeyWhereValue(value); }
    }
}

class Program
{
    static void Main(string[] args)
    {
        var dict = new TwoWayDictionary<string, int>(new KeyValuePair<string, int>[] {
            new KeyValuePair<string, int>(".jpeg",100),
            new KeyValuePair<string, int>(".jpg",101),
            new KeyValuePair<string, int>(".txt",102),
            new KeyValuePair<string, int>(".zip",103)
        });


        var r1 = dict[100];
        var r2 = dict[".jpg"];

    }

}

Solution 15 - C#

There is a BijectionDictionary type available in this open source repo:

https://github.com/ColmBhandal/CsharpExtras.

It isn't qualitatively much different to the other answers given. It uses two dictionaries, like most of those answers.

What is novel, I believe, about this dictionary vs. the other answers so far, is that rather than behaving like a two way dictionary, it just behaves like a one-way, familiar dictionary and then dynamically allows you to flip the dictionary using the Reverse property. The flipped object reference is shallow, so it will still be able to modify the same core object as the original reference. So you can have two references to the same object, except one of them is flipped.

Another thing that is probably unique about this dictionary is that there are some tests written for it in the test project under that repo. It's been used by us in practice and has been pretty stable so far.

Attributions

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionNeir0View Question on Stackoverflow
Solution 1 - C#EnigmativityView Answer on Stackoverflow
Solution 2 - C#Hasan BaidounView Answer on Stackoverflow
Solution 3 - C#Xavier JohnView Answer on Stackoverflow
Solution 4 - C#zmbqView Answer on Stackoverflow
Solution 5 - C#OstatiView Answer on Stackoverflow
Solution 6 - C#moribvndvsView Answer on Stackoverflow
Solution 7 - C#farlee2121View Answer on Stackoverflow
Solution 8 - C#Simon TewsiView Answer on Stackoverflow
Solution 9 - C#IamsodarncoolView Answer on Stackoverflow
Solution 10 - C#toddmoView Answer on Stackoverflow
Solution 11 - C#Felipe RamosView Answer on Stackoverflow
Solution 12 - C#sirpaddowView Answer on Stackoverflow
Solution 13 - C#paparazzoView Answer on Stackoverflow
Solution 14 - C#Brett CaswellView Answer on Stackoverflow
Solution 15 - C#Colm BhandalView Answer on Stackoverflow