Build pairs out of list - without repetition

Question

to do it in linqI have an integer List and want to group these to a list of integer pairs.

var input = new[] {1, 24, 3, 2, 26, 11, 18, 13};

result should be: {{1, 24}, {3, 2}, {26, 11}, {18, 13}}

I tried:

List<int> src = new List<int> { 1, 24, 3, 2, 26, 11, 18, 13 };
var agr = src.Select((n, i) => new Tuple<int, int>(i++ % 2, n))
             .GroupBy(t => t.Item1)
             .ToList();

var wanted = agr[0].Zip(agr[1], (d, s) => new Tuple<int, int>(d.Item2, s.Item2));

Is there a better way to do it in linq?

Of course I can do it with a simple for-loop.

Edit: I think I give MoreLinq a try. I also mark this as the answer even if it's an extension and not pure linq.

By the way - I think doing it with a for-loop is much more understandable.

Possible duplicate of [Getting pair-set using LINQ](https://stackoverflow.com/questions/1624341/getting-pair-set-using-linq) — Sinatr, Aug 04 '17 at 07:53
@Sinatr wait.. no it's not - the other question repeats the values. This question does not. — default, Aug 04 '17 at 08:00
@MongZhu yes, I would like to have a linq solution, as simple as possible. — GreenEyedAndy, Aug 04 '17 at 08:09
"I think doing it with a for-loop is much more understandable" I guess it is also more performant, depends of course strongly on the size of your collection — Mong Zhu, Aug 04 '17 at 08:25

score 4 · Accepted Answer · answered Aug 04 '17 at 07:57

4

You can use MoreLINQ Batch to split your input into a list of "length 2" lists. Or any other length you want.

List<int> src = new List<int> { 1, 24, 3, 2, 26, 11, 18, 13 };
List<IEnumerable<int>> wanted = src.Batch(2).ToList();

answered Aug 04 '17 at 07:57

Hans Keﬆing

38,117
9
79
111

score 1 · Answer 2 · answered Aug 04 '17 at 08:31

1

No need for MoreLINQ; Enumerate even- and odd-indexed values, and Zip

int[] input = new int[8] { 1, 24, 3, 2, 26, 11, 18, 13 };

var evenPositioned = input.Where((o, i) => i % 2 == 0);
var oddPositioned = input.Where((o, i) => i % 2 != 0);
var wanted = evenPositioned.Zip(oddPositioned, (even, odd) => new { even, odd }).ToList();

answered Aug 04 '17 at 08:31

Fredrik

2,247
18
21

this is nearly what I posted - or? Anyway, yours is looking much better. – GreenEyedAndy Aug 04 '17 at 09:58
i believe our thinking was the same yes. – Fredrik Aug 04 '17 at 11:17

score 0 · Answer 3 · answered Aug 04 '17 at 07:54

If you can garantee, that the length of the source can always be devided by 2:

List<int> src = new List<int> { 1, 24, 3, 2, 26, 11, 18, 13 };
var Tuple<int, int>[] wanted = new Tuple<int, int>[src.Count /2];
for(var i = 0; i < src.Count; i = i + 2)
    wanted[i/2] = new Tuple<int, int>(src[i], src[i+1]);

score 0 · Answer 4 · answered Aug 04 '17 at 08:01

a simple for loop is enough for this.Just start with 1 and increment it by 2

List<int> src = new List<int> { 1, 24, 3, 2, 26, 11, 18, 13 };
var list = new List<Tuple<int, int>>();
for(int i =1;i<src.Count;i=i+2)
{
    list.Add(new Tuple<int, int>(src[i-1],src[i]));
}

In case of odd count last item will be skipped

score 0 · Answer 5 · answered Aug 04 '17 at 08:08

Another simple loop solution featuring C# 7 tuples.

var input = new List<int> { 1, 24, 3, 2, 26, 11, 18, 13 };
var wanted = new List<(int, int)>();
for (var index = 0; index < input.Count; index += 2) {
    wanted.Add((input[index], input[index + 1]));
}

Build pairs out of list - without repetition

5 Answers5