Get a list of distinct items and their count

Question

I have an object, that has many properties but the only two to worry about are:

myobject.ID which is an int
myobject.Names which is a HashSet

Then I have a List of those objects that looks something similar to this:

List<myobject>

I use Linq to get a some of the data into a repeater, but I'm not sure how to get the list of Names and how often they show up.

Want to use Linq to avoid having to loop through the data.

As my tags should show, this is an ASP.NET solution using C#.

Some clarification:
Lets say I have just three items in my list: Item 1 has John, Fred, Jack in its names. Item 2 has John, Fred, Joe in its names. Item 3 has John in its names.

I am trying to return the following: John - 3 Fred - 2 Jack - 1 Joe - 1

Also, as a note I am familiar with having to write my own comparer for my object, I'm just missing the 'how to' for the overall solution in my thoughts.

Answer 1

Here is how I understand the problem that you are trying to solve. You have a list of myobjects. Each myobject has a property called Names which is a HashSet of strings (i.e., HashSet<string>). You want to count the number of times each string that appears in some myobject.Names appears in all the myobject.Names. That is, you have

"Alice", "Bob", "Charlie"
"Alice", "Bob", "Donald"
"Alice", "Donald", "Ernie"

as three myobject.Names and you want to see

"Alice", 3
"Bob", 2
"Charlie", 1
"Donald", 2
"Ernie", 1

If so:

var query = list.SelectMany(x => x.Names)
                .GroupBy(s => s)
                .Select(g => new { Name = g.Key, Count = g.Count() });

foreach(var result in query) {
    Console.WriteLine("Name: {0}, Count: {1}", result.Name, result.Count);
}

I do not see what role myobject.ID plays here. Please qualify.

Answer 2

var query = yourList
            .SelectMany(x => x.Names)
            .GroupBy(x => x, (y, z) => new { Name = y, Count = z.Count() });

// and to test...
foreach (var item in query)
{
    Console.WriteLine("{0} - {1}", item.Name, item.Count);
}