Grouping Key Value pairs with Linq and List
I'm trying to take a long list of items, a key/value pair, and group them by key. Doing this I want to get the count of each key/value pair so I can for a weighted list later on. The code I have for generating the list is similar to this sample:
class Notes
{
public int NoteId { get; set; }
public string NoteName { get; set; }
}
class Program
{
static void Main(string[] args)
{
List<Notes> _popularNotes = new List<Notes> {
new Notes { NoteId = 1, NoteName = "Title 1" },
new Notes { NoteId = 1, NoteName = "Title 1" },
new Notes { NoteId = 2, NoteName = "Title 2" },
new Notes { NoteId = 4, NoteName = "Title 4" },
new Notes { NoteId = 4, NoteName = "Title 4" } };
开发者_JAVA百科 foreach (var _note in _popularNotes)
Console.WriteLine(_note.NoteId + ": " + _note.NoteName);
IEnumerable<IGrouping<int, string>> _query = _popularNotes.GroupBy(x => x.NoteId, x => x.NoteName);
foreach (var _noteGroup in _query)
{
Console.WriteLine(_noteGroup.Key + ": " + _noteGroup.Count());
}
Console.ReadKey();
}
}
This build the list and groups them and I can get the count of each object, I just cant get the value. I can only seem to get just the key.
With I'm sure a million ways to do this I'm really trying to pick one I understand. And well I'm just not understanding it I guess.
So should I go back and get the name from the _popularNotes
list with a lookup? Or is there another way of actually building and outputting the list with the key/value pair plus the count?
You can write _noteGroup.First()
IGrouping<TKey, TElement>
is an IEnumerable<TElement>
which means you can enumerate over it.
As per the documentation of IGrouping<TKey, TElement>
:
public interface IGrouping<out TKey, out TElement> : IEnumerable<TElement>,
IEnumerable
In other words, to spit out the key + count, and then all the elements (names in your case) in that group, you can do:
foreach (var _noteGroup in _query)
{
Console.WriteLine(_noteGroup.Key + ": " + _noteGroup.Count());
foreach (var name in _noteGroup)
Console.WriteLine(" " + name);
}
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