我想通过以下方式将单词存储在字典中：
我可以逐字获取代码：dict["SomeWord"]->123并逐字获取代码：dict[123]->"SomeWord"
这是真的吗？当然，一种方法是使用两本词典：但Dictionary<string,int>还有Dictionary<int,string>另一种方法吗？
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55736 次
15 回答
125
我写了几个快速的类，可以让你做你想做的事。您可能需要使用更多功能对其进行扩展，但这是一个很好的起点。
代码的使用是这样的：
var map = new Map<int, string>();
map.Add(42, "Hello");
Console.WriteLine(map.Forward[42]);
// Outputs "Hello"
Console.WriteLine(map.Reverse["Hello"]);
//Outputs 42
这是定义：
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; }
}
于 2012-06-10T06:02:55.210 回答
32
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);
于 2018-05-17T08:50:05.757 回答
14
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;
}
}
于 2017-01-28T07:52:54.957 回答
7
正如其他人所说，您可以使用两个字典，但还要注意，如果两者TKey和TValue都是相同类型（并且它们的运行时值域已知是不相交的），那么您可以通过为每个键创建两个条目来使用相同的字典/值配对：
dict["SomeWord"]= "123"和dict["123"]="SomeWord"
这样，单个字典可以用于任何一种类型的查找。
于 2012-06-10T04:18:12.630 回答
7
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
于 2019-03-04T20:34:19.730 回答
6
您可以使用此扩展方法，尽管它使用枚举，因此对于大型数据集的性能可能不那么好。如果您担心效率，那么您需要两个字典。如果要将两个字典包装到一个类中，请参阅此问题的公认答案：Bidirectional 1 to 1 Dictionary in C#
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");
}
}
于 2012-06-10T04:47:42.250 回答
3
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
于 2020-04-24T14:17:53.700 回答
2
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
*/
于 2019-03-04T06:07:42.493 回答
1
于 2015-09-19T17:55:40.960 回答
1
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;
}
于 2017-09-26T19:28:15.663 回答
1
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();
}
}
于 2019-05-12T07:21:56.103 回答
1
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();
}
}
于 2019-05-15T17:02:41.733 回答
0
这使用索引器进行反向查找。
反向查找是 O(n) 但它也不使用两个字典
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;
}
}
}
于 2012-10-02T13:27:39.087 回答
0
以下封装类在 1 个字典实例上使用 linq（IEnumerable Extensions）。
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"];
}
}
于 2015-09-18T22:05:35.420 回答
0
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.
于 2020-04-05T17:31:02.907 回答