在字典中查找对

Question

For my project i made a dictionary that has a random double and a string that belongs to that double: Dictionary<double, string> myDict = new Dictionary<double, string>();

For this project i know that the double is a random value, and within the dictionaries all strings are unique, with the exception that about 80% one of them is twice in the dictionary. So what i want to do, is find the 2 strings that are a pair (the same string) and find the 2 double values that belong to these 2 string.

Basically my idea of doing this is by using IEnumerator counter = myDict.GetEnumerator(); and use the while (counter.MoveNext() == true) to start another IEnumerator that loops again through all the entries of the dictionary and compares by string, so if will find the pairs this way. So for each entry in the dictionary, it will loop through the whole dictionary again to find pairs.

Now i get the feeling this might not be the best solution to handle this. Are there alternatives to find the pairs in the dictionary, or is this looping through the only real way of doing this?

Answer 1

I believe, you are looking to get Keys for those items where there is a pair of string available in Values .

var result  = myDict.GroupBy(r => r.Value)
                    .Where(grp => grp.Count() == 2)
                    .SelectMany(grp => grp.Select(subItem => subItem.Key))
                    .ToList();

If you want to get keys for those items which have multiple string values, (more than two) then modify the condition to:

.Where(grp => grp.Count() >= 2)

Another thing to add, you are adding keys as Random values in the dictionary. Remember, Random doesn't mean Unique . You could end up with an exception since Dictionary keys are unique.

If your dictionary is defined as:

Dictionary<double, string> myDict = new Dictionary<double, string>
{
    {1, "ABC"},
    {2, "ABC"},
    {3,"DEF"},
    {4,"DEF"},
    {5,"DEF2"},
    {6,"XYZ"}
};

For output after the LINQ expression:

foreach (var d in result)
{
    Console.WriteLine(d);
}

Output:

1
2
3
4