I have this problem

I have a column vector

k =[ 0 0 0 0 0 0 2 5 6 8 7 2 1 0 0 0 0 3 5 4 1 8 6 8 ];

I want to keep the shape of the vector while replacing every 1st and 3rd non zero element with 100 so I end up with

k = [ 0 0 0 0 0 0 100 5 6 8 100 100 1 2 5 100 0 100 5 4 1 100 6 8];

[Rik]

How are you counting? I can't reproduce your vector by any way I can think of

%               1 2 3 1 2 3 1         1 2 3 1 2 3 1
%               1 2 3 1 2 3 1         2 3 1 2 3 1 2
%  (1 2 3 1 2 3)1 2 3 1 2 3 1(2 3 1 2)3 1 2 3 1 2 3
k=[ 0 0 0 0 0 0 2 5 6 8 7 2 1 0 0 0 0 3 5 4 1 8 6 8 ];
%               x   x x   x x         x   x x   x x
%               x   x x   x x           x x   x x  
%               x   x x   x x         x x   x x   x

None of these patterns result in your second vector. Also, there seem to be some extra values. Please explain what your pattern is, then we can try to help you find a way to implement it.

[Ev]

 %                 1 2 3 1 2 3 1         1 2 3 1 2 3 1
 k = [ 0 0 0 0 0 0 2 5 6 8 7 2 1 0 0 0 0 3 5 4 1 8 6 8 ];
 % Counting only the non-zero elements

original vector. I want

Answer 1

[Rik]

The easiest way to do this is with a loop. A second method is less easy to follow, but removes the need for a loop. The final statement shows that the two methods are indeed interchangeable.

%                 1 2 3 1 2 3 1         1 2 3 1 2 3 1
k = [ 0 0 0 0 0 0 2 5 6 8 7 2 1 0 0 0 0 3 5 4 1 8 6 8 ];
k2=k;
counter=0;
for n=1:numel(k)
    if n>1 && k(n-1)==0
        counter=0;%restart for next run
    end
    counter=counter+1;
    if k(n)~=0 && ...
            mod(counter,3)~=2 %"not 2" is equivalent to "1 or 3"
        k2(n)=100;
    end
end

run_start_inds=find([k(1)~=0 diff(k~=0)>0]);
base=ones(size(k));
base(run_start_inds)=base(run_start_inds)-mod(run_start_inds,3)+1;
countlist=cumsum(base);
countlist(k==0)=NaN;
countlist=mod(countlist,3);
k3=k;
k3(countlist==mod(1,3))=100;
k3(countlist==mod(3,3))=100;

isequal(k2,k3)