C＃字典：每个值有多个KEYS [英] C# Dictionary: Multiple KEYS per Value

问题描述

我正在寻找一种通过单个值获取多个键的方法。是的，我已经使用过搜索功能，但是大多数答案都是相反的方向（每个键有多个值），但我想相反。

其背后的原因是，我想为每个主 ID保留多个Item-ID（用于Bot），并将这些多个ID放入一个ID中的值太慢了进行修改（寻找一个值=>遍历所有主要ID并获取每个值，然后检查该ID是否存在）。

示例

 键1 =>值
键2 =>值
键3 =>值
键4 =>值
键5 =>价值2 
  

寻找价值应返回：键1-4，而不是5

所以我正在寻找一种更轻松的方法-就像我上面所说的。

任何人都知道这是否可行以及如何会吗
预先感谢。

解决方案

用另一种方法来做字典，并将值设为项目列表。 / p>

例如，如果 Value 是字符串，并且键1-4 是您的字典的整数，可能看起来像这样：

  var theDictionary = new Dictionary< string，List< int>>（ ）; 
  

通过值 > theDictionary [ Value] 然后将返回包含1、2、3和4的整数的列表。

编辑-添加了示例：

  var theDictionary = new字典< string，List< string>> 
 {
 {值，新列表< string> {键1，键2，键3，键4，键5，}}，
 { Value2，新列表< string> {键5，键2}} 
}; 
 
 var oneToFour = theDictionary [ Value]; 
  

I'm looking for a way to get multiple keys with a single value. Yes, I've already used the search function, but most answers are for the opposite direction (multiple values per key), but I want the opposite.

The reasoning behind this is that I want keep multiple Item-IDs (it's for a Bot) per "main" ID, and throwing those multiple IDs into a value of the one is too slow to modify (looking for one value => looping trough all main IDs & getting each value, then checking if that ID exists).

Example

Key 1 => Value
Key 2 => Value
Key 3 => Value
Key 4 => Value
Key 5 => Value 2

Looking for Value should return: Key 1-4, not 5

So I'm looking for a way to do that easier - like I said above.

Anyone knows if that's possible and how to do it? Thanks in advance.

解决方案

Do the dictionary the other way around and make the value a list of items.

if for example Value is a string and Key 1-4 are ints your dictionary could look something like:

var theDictionary = new Dictionary<string, List<int>>();

retrieving Value by theDictionary["Value"] would then return a list of ints containing 1, 2, 3 and 4.

Edit - Added example:

var theDictionary = new Dictionary<string, List<string>>
    {
        {"Value", new List<string> {"Key 1", "Key 2", "Key 3", "Key 4", "Key 5",}},
        {"Value2", new List<string> {"Key 5", "Key 2"}}
    };

var oneToFour = theDictionary["Value"];

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