substitute every n elements of a vector

Hi, I have a vector lets say 24 elements (zeros), and I want to substitute a vector of ones (lets say in this example a vector of two ones) n times (lets say 4 times in this example) starting from an index of the original vector decided by me, examples:

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 (original vector)

1 1 0 0 0 0 1 1 0 0 0 0 1 1 0 0 0 0 1 1 0 0 0 0 (here substituted 4 times a vector of 2 ones starting form index 1)

0 0 1 1 0 0 0 0 1 1 0 0 0 0 1 1 0 0 0 0 1 1 0 0 (here substituted 4 times a vector of 2 ones starting form index 3)

Is it possible to do simply something like this (possibly without a for)? just having the vector and deciding how many times substitute it, and the index to start? Thank you so much.

Answer 1

[Stephen23]

It may be that using a loop is actually faster than other solutions, and less obfuscated:

num = 3; % start index
stp = 6; % step size
vec = [1,2,3]; % vector
out = zeros(1,24)
for k = 1:numel(vec)
  out(num+k-1:stp:end) = vec(k);
end
out

Displays this in the command window:

>> temp
out =
   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0
out =
   0   0   1   2   3   0   0   0   1   2   3   0   0   0   1   2   3   0   0   0   1   2   3   0

[simone clochiatti]

Thank you everyone for the help, but this is the solution that works for me! Thank you very very much Stephen!!

Answer 2

[Szilárd Ecsenyi]

more faster:

v = 0:1:23;

v(rem(fix(v/3),2) == 0) = 0;

v(v~=0) = 1;