Select every second element from array using lambda

Question

C# 4.0. How can the following be done using lambda expressions?

int[] a = new int[8] { 0, 1, 2, 3, 4, 5, 6, 7 };
// Now fetch every second element so that we get { 0, 2, 4, 6 }

score 28 · Accepted Answer · edited Nov 25 '15 at 12:06

28

int[] list = a.Where((value, index) => index % 2 == 0)
              .ToArray();

It will only select even indexes, as calculate by the % (mod) operator .

5 % 2 // returns 1
4 % 2 // returns 0

According to MSDN:

% Operator

edited Nov 25 '15 at 12:06

Loofer

answered Oct 26 '12 at 07:52

Asif Mushtaq

1

+1, I forgot, that `where` also has overload with index :) – Sergey Berezovskiy Oct 26 '12 at 07:56

cuongle · Answer 2 · 2012-10-26T08:18:17.537

10

Another approach using Enumerable.Range

var result = Enumerable.Range(0, a.Length/2)
                       .Select(i => a[2*i])
                       .ToArray();

Or use bitwise for more efficient to check even:

var result = a.Where((i, index) => (index & 1) == 0)
              .ToArray();

edited Oct 26 '12 at 08:18

answered Oct 26 '12 at 07:57

cuongle

2

+1 for bitwise odd/even test which is by far more efficient than '%' operator. So classy :-) – Askolein Oct 26 '12 at 08:15
1

http://stackoverflow.com/questions/2229107/what-is-the-fastest-way-to-find-if-a-number-is-even-or-odd – Cihan Yakar Oct 26 '12 at 08:33

score 8 · Answer 3 · answered Oct 26 '12 at 07:54

The remainder operator is your friend.

int[] everySecond = a.Where((i, ind) => ind % 2 == 0).ToArray();

The % operator computes the remainder after dividing its first operand by its second. All numeric types have predefined remainder operators.

3 Answers3