0

I have a List of integer tuple

List<Tuple<int, int>> list =  new  List<Tuple<int, int>>();
list.Add(new Tuple<int, int>(1,12));
list.Add(new Tuple<int, int>(1,2));
list.Add(new Tuple<int, int>(1,18));
list.Add(new Tuple<int, int>(1,12));

I want to remove the redundant value of item2 which is 12 in this case

The updated list should also be List>. (same type) but with distinct values.The new tuple of type List> should contain only following: 

 list.Add(new Tuple<int, int>(1,2));
 list.Add(new Tuple<int, int>(1,18));
 list.Add(new Tuple<int, int>(1,12));

Any help?

Rimmy
  • 21
  • 7
  • 1
    Do you only care about the second tuple value? Meaning, thats the only one you need to distinct on? Or only duplicate matches of item1 and item2? – maccettura Oct 11 '17 at 16:51
  • so thats the problem i need both values i.e i want the tuple of the same type with both the values – Rimmy Oct 11 '17 at 16:55
  • that comment makes no sense. What happens if you have a Tuple with (2,12), should that be part of your output since the first value is unique or should it be excluded because the second value is a duplicate? – maccettura Oct 11 '17 at 16:58
  • i want to retain both the values item1 and item2 in the new tuple but i have to apply distinct on the basis of item2. Acc. to the above example one 2,12 will be removed because item2 12 is 2 times. So my new tue will contain 3 values instead of 4 – Rimmy Oct 11 '17 at 17:06
  • Then the second part of my answer is what you are looking for. – maccettura Oct 11 '17 at 17:07

1 Answers1

4

If you want only the unique pairs of numbers this can easily be done by using Distinct():

list = list.Distinct().ToList();

If you only care about removing duplicates of only Item2, then use GroupBy() and Select() the First() of each group:

list = list.GroupBy(x => x.Item2).Select(x => x.First());

I made a fiddle here that demonstrates both methods

EDIT

From your latest comment, it sounds like you want to use the second method I have proposed (GroupBy Item2 and Select the First of each Group).

maccettura
  • 10,514
  • 3
  • 28
  • 35