Get a list of distinct values in List

Question

In C#, say I have a class called Note with three string member variables.

public class Note
{
    public string Title;
    public string Author;
    public string Text;
}

And I have a list of type Note:

List<Note> Notes = new List<Note>();

What would be the cleanest way to get a list of all distinct values in the Author column?

I could iterate through the list and add all values that aren't duplicates to another list of strings, but this seems dirty and inefficient. I have a feeling there's some magical Linq construction that'll do this in one line, but I haven't been able to come up with anything.

Answer 1

Notes.Select(x => x.Author).Distinct();

This will return a sequence (IEnumerable<string>) of Author values -- one per unique value.

Answer 2

Distinct the Note class by Author

var DistinctItems = Notes.GroupBy(x => x.Author).Select(y => y.First());

foreach(var item in DistinctItems)
{
    //Add to other List
}

Answer 3

Jon Skeet has written a library called morelinq which has a DistinctBy() operator. See here for the implementation. Your code would look like

IEnumerable<Note> distinctNotes = Notes.DistinctBy(note => note.Author);

Update: After re-reading your question, Kirk has the correct answer if you're just looking for a distinct set of Authors.

Added sample, several fields in DistinctBy:

res = res.DistinctBy(i => i.Name).DistinctBy(i => i.ProductId).ToList();

Answer 4

public class KeyNote
{
    public long KeyNoteId { get; set; }
    public long CourseId { get; set; }
    public string CourseName { get; set; }
    public string Note { get; set; }
    public DateTime CreatedDate { get; set; }
}

public List<KeyNote> KeyNotes { get; set; }
public List<RefCourse> GetCourses { get; set; }    

List<RefCourse> courses = KeyNotes.Select(x => new RefCourse { CourseId = x.CourseId, Name = x.CourseName }).Distinct().ToList();

By using the above logic, we can get the unique Courses.

Answer 5

mcilist = (from mci in mcilist select mci).Distinct().ToList();