CODEWITHSUNDEEP
c#linq
Robert J.
Robert J.

Reputation: 2711

c# LINQ Distinct() on multi level list

I have a List<List<int>> with the following values:

[1] = 1,2
[2] = 4,5,6
[3] = 1,3

What I want to do is to retrieve simple List with unique values => 1,2,3,4,5,6 (number one was duplicated in original list).

If it was 1D List, I would use 

variable.Select(n=>n).Distinct();

however when I tried to use

variable.Select(n=>n.Select(x=>x)).Distinct();

I got 2D list (with unique values I guess). How can I solve this? Thanks

Upvotes: 1

Views: 509

Answers (3)

saj
saj

Reputation: 4816

You could do SelectMany if you want a distinct of your overall integers, but if you want distinc lists, you could do something like this;

void Main()
{
    List<List<int>> xss = new List<List<int>>()
    {
        new List<int>() {1, 2},
        new List<int>() {1, 2},
        new List<int>() {1, 2},
        new List<int>() {4, 5, 6},
        new List<int>() {1, 3}
    };

    xss.Select (x => string.Join(",", x)).Distinct();

}

Upvotes: 1

Rezo Megrelidze
Rezo Megrelidze

Reputation: 3060

private static void Main()
{
    List<List<int>> xss = new List<List<int>>()
    {
        new List<int>() {1, 2},
        new List<int>() {4, 5, 6},
        new List<int>() {1, 3}
    };
    var xs = xss.SelectMany(x => x).Distinct();
    Console.WriteLine(string.Join(" ",xs));
}

Upvotes: 1

Selman Gen&#231;
Selman Gen&#231;

Reputation: 101732

You could just use SelectMany with Distinct

var uniqueList = list.SelectMany(x => x).Distinct().ToList();

SelectMany flattents the List<List<int>> into a IEnumerable<int> and Distinct eliminates the duplicates.

Upvotes: 8

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