How can I get every nth item from a List<T>?

thedorko_68432 0
Viewed 91505

I'm using .NET 3.5 and would like to be able to obtain every *n*th item from a List. I'm not bothered as to whether it's achieved using a lambda expression or LINQ.

Edit

Looks like this question provoked quite a lot of debate (which is a good thing, right?). The main thing I've learnt is that when you think you know every way to do something (even as simple as this), think again!

list lambda linq c#
10 Answers
return list.Where((x, i) => i % nStep == 0);
user_559430

I know it's "old school," but why not just use a for loop with stepping = n?

Michael_Todd_166230

Sounds like

IEnumerator<T> GetNth<T>(List<T> list, int n) {
  for (int i=0; i<list.Count; i+=n)
    yield return list[i]
}

would do the trick. I do not see the need to use Linq or a lambda expressions.

EDIT:

Make it 

public static class MyListExtensions {
  public static IEnumerable<T> GetNth<T>(this List<T> list, int n) {
    for (int i=0; i<list.Count; i+=n)
      yield return list[i];
  }
}

and you write in a LINQish way

from var element in MyList.GetNth(10) select element;

2nd Edit:

To make it even more LINQish

from var i in Range(0, ((myList.Length-1)/n)+1) select list[n*i];
user_814240

You can use the Where overload which passes the index along with the element

var everyFourth = list.Where((x,i) => i % 4 == 0);
JaredPar_232830

For Loop 

for(int i = 0; i < list.Count; i += n)
    //Nth Item..
CQ_127070

I'm not sure if it's possible to do with a LINQ expression, but I know that you can use the Where extension method to do it. For example to get every fifth item:

List<T> list = originalList.Where((t,i) => (i % 5) == 0).ToList();

This will get the first item and every fifth from there. If you want to start at the fifth item instead of the first, you compare with 4 instead of comparing with 0.

Guffa_690830

Imho no answer is right. All solutions begins from 0. But I want to have the real nth element

public static IEnumerable<T> GetNth<T>(this IList<T> list, int n)
{
    for (int i = n - 1; i < list.Count; i += n)
        yield return list[i];
}
user_23401450

@belucha I like this, because the client code is very readable and the Compiler chooses the most efficient Implementation. I would build upon this by reducing the requirements to IReadOnlyList<T> and to save the Division for high-performance LINQ: 

    public static IEnumerable<T> GetNth<T>(this IEnumerable<T> list, int n) {
        if (n <= 0) throw new ArgumentOutOfRangeException(nameof(n), n, null);
        int i = n;
        foreach (var e in list) {
            if (++i < n) { //save Division
                continue;
            }
            i = 0;
            yield return e;
        }
    }

    public static IEnumerable<T> GetNth<T>(this IReadOnlyList<T> list, int n
        , int offset = 0) { //use IReadOnlyList<T>
        if (n <= 0) throw new ArgumentOutOfRangeException(nameof(n), n, null);
        for (var i = offset; i < list.Count; i += n) {
            yield return list[i];
        }
    }
user_37460690
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