Create Dictionary(of Dictionary) using Grouping in Linq

Question

I know I could do this with loops (as a matter of fact, I currently am, but I am trying to learn / imporve my Linq skills and i also hope it will provide a more efficent sol

Answer 1

It's not great that you have separate lists with the discrete pieces of information instead of a single list with Student instances. If you want to use group by here you'll probably need to zip the collections together first into a single enumerable with triplets and start using the query operations on that enumerable.

Answer 2

This more or less Spender's solution, but this works.

In C#

var dict = lstTeachers.Zip(lstStudentsSex, (teacher, sex) => new { teacher, sex })
    .Zip(lstStudentName, (x, student) => new { x.teacher, x.sex, student })
    .GroupBy(x => x.teacher)
    .ToDictionary(g => g.Key, g => g.GroupBy(x => x.sex)
        .ToDictionary(p => p.Key, p => p.Select(x => x.student)));

Answer 3

The three lists make things a bit difficult, but not impossible. Either you first zip the lists together as suggested by @fsimonazzi, or you base your query on the list indices rather than the list entries themselves. The result could look like this:

_(C#)

var result = Enumerable
    .Range(0, lstTeachers.Count)
    .GroupBy(i => lstTeachers[i])
    .ToDictionary(g => g.Key, g => g
        .GroupBy(i => lstStudentsSex[i])
        .ToDictionary(h => h.Key, h => h
            .Select(i => lstStudentName[i])
            .ToList()));

// result is Dictionary<string, Dictionary<string, List<string>>>

Answer 4

I only C#, not VB. Apologies in advance.

Let's start by making a single list of anonymous objects with properties teacher, sex, name:

var singleList = lstTeachers
  .Zip(lstStudentsSex, (teacher, sex) => new {teacher, sex})
  .Zip(lstStudentName, (x,name) => new {x.teacher, x.sex, name})

Now we can turn this into a Dictionary of Dictionaries:

singleList
  .GroupBy(x => x.teacher)
  .ToDictionary(
       g => g.Key, 
       g => g.ToDictionary(x => x.sex, x => x.name))

See @nawfal's for a corrected version of my code.