我如何得到一个字典键值在c# ?
Dictionary<string, string> types = new Dictionary<string, string>()
{
{"1", "one"},
{"2", "two"},
{"3", "three"}
};
我想要这样的东西:
getByValueKey(string value);
getByValueKey("one")必须返回"1"。
最好的方法是什么?也许是哈希表或排序列表?
我如何得到一个字典键值在c# ?
Dictionary<string, string> types = new Dictionary<string, string>()
{
{"1", "one"},
{"2", "two"},
{"3", "three"}
};
我想要这样的东西:
getByValueKey(string value);
getByValueKey("one")必须返回"1"。
最好的方法是什么?也许是哈希表或排序列表?
当前回答
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;
}
}
}
其他回答
我有一个很简单的方法。这对我来说很完美。
Dictionary<string, string> types = new Dictionary<string, string>();
types.Add("1", "one");
types.Add("2", "two");
types.Add("3", "three");
Console.WriteLine("Please type a key to show its value: ");
string rLine = Console.ReadLine();
if(types.ContainsKey(rLine))
{
string value_For_Key = types[rLine];
Console.WriteLine("Value for " + rLine + " is" + value_For_Key);
}
我遇到的情况是LINQ绑定不可用,必须显式展开lambda。它的结果是一个简单的函数:
public static T KeyByValue<T, W>(this Dictionary<T, W> dict, W val)
{
T key = default;
foreach (KeyValuePair<T, W> pair in dict)
{
if (EqualityComparer<W>.Default.Equals(pair.Value, val))
{
key = pair.Key;
break;
}
}
return key;
}
这样称呼它:
public static void Main()
{
Dictionary<string, string> dict = new Dictionary<string, string>()
{
{"1", "one"},
{"2", "two"},
{"3", "three"}
};
string key = dict.KeyByValue("two");
Console.WriteLine("Key: " + key);
}
它适用于。net 2.0和其他有限的环境。
public static string GetKeyFromValue(string valueVar)
{
foreach (string keyVar in dictionaryVar.Keys)
{
if (dictionaryVar[keyVar] == valueVar)
{
return keyVar;
}
}
return null;
}
其他人可能有更有效的答案,但我个人觉得这个更直观,对我来说很管用。
你可以这样做:
通过循环遍历字典中的所有KeyValuePair<TKey, TValue>(如果字典中有许多条目,这将是一个相当大的性能打击) 使用两个字典,一个用于值到键映射,一个用于键到值映射(这会占用两倍的内存空间)。
如果不考虑性能,请使用方法1;如果不考虑内存,请使用方法2。
此外,所有键都必须是唯一的,但值不要求是唯一的。您可以有多个具有指定值的键。
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。