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Counting every other number in a list

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Howdy.
I have a matrix
mylist = [1:10]
I need to count every other number in this matrix
I can't use the odd or even distinguishing tests such as...
for i = 1:length(mylist)
if rem(mylist(i),2) ~= 0;
eo_total = eo_total + 1
end
end
Because this only counts the odd numbers in my for loop. So if I had a matrix
mylist(2) = [ 2 4 6 8 10]
eo_total = 0 instead of 2
any ideas?

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Accepted Answer

the cyclist
the cyclist on 12 Sep 2020
I'm not sure what you mean by "count" them, but the following vector will store every other element from mylist, starting with the first one:
out = mylist(1:2:end)

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Nathaniel Wolff
Nathaniel Wolff on 12 Sep 2020
by count I mean that litterally. I want to count every other other digit in my matrix.
for instance
[1 2 3 4 5]
Ideally i would have a function that would output 2. Becuause if I start at 1, not counting itself, the next other number would be 3 then the last is 5. But I can't have it be odd or even because this list might be something disproportionalty even or odd.
thank you for the code, I can just use a length(out) to count the number of digits in this new matrix.
that's actually much simpler then what I was trying to do with for loops
the cyclist
the cyclist on 12 Sep 2020
Oh. Then it's probably more efficient to do
ceil(numel(mylist)/2)
Then you don't need to create a second vector.

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