inserting headers into a vector

I have numbers 127 or 129 and I need insert some headers two numbers 27425 and 1216.

Header, data(1216 numbers), header...

I solved this problem using for cycle, but it was very slow (I have very long vector).

n=0;
for j=1:(length(input)+ceil(length(data)/1216)*2)
    if j==1 || mod(j-1,1218)==0
        output(j) = 27425;
    elseif j==2 || mod(j-2,1218)==0
        output(j) = 1216;
    else
        n=n+1;
        output(j) = data(n);
    end
end

Are there some functions to insert columns?

[Sean de Wolski]

Can you please reformat your code so it's readable using the markup help?

Answer 1

[Paulo Silva]

v is your vector, val is just a number

insert column in the first index

v=[val v]

insert column in the last index

v=[v val]

insert values (headers), val1 and val2 are just numbers

v=[val1 v val2]

Answer 2

[Matt Fig]

Boris, please format your code and fix the errors. When I tried to copy and paste your code, it looks like you are missing some operations.

n=0; 
for j=1:(length(input)+ceil(length(data)/1216)*2) 
    if j==1 mod(j-1,1218)==0 
        output(j) = 27425; 
    elseif j==2 mod(j-2,1218)==0 
        output(j) = 1216; 
    else
        n=n+1;
        output(j) = data(n); 
    end
end

For example, what is the MOD function doing there? Did you mean to have an operator between the 1 and mod(j-1,1218)==0, and between the 2 and mod(j-2,1218)==0? If so, what operator?

[Boris]

I use function mod to find position where must be header.

sorry I made a mistake

n=0;

for j=1:(length(data)+ceil(length(data)/1216)*2)

if j==1 || mod(j-1,1218)==0

output(j) = 27425;

elseif j==2 || mod(j-2,1218)==0

output(j) = 1216;

else

n=n+1;

output(j) = data(n);

end

end

I tried to split vector data to smaller vectors (1216 numbers) and between each smaller vector I wanted to insert header.

[Matt Fig]

But the mod function call is not doing anything here. It simply returns true or false when j==1 or j==2. It will not even execute after the first two passes through the loop, so I don't understand why it is even in there.

Answer 3

[Oleg Komarov]

My solution, based on the assumption that headers don't have to be repmatted:

% Example input, stepsize and one header (multiple headers may require repmatting)
A    = rand(9,1);
step = 4;
head = 1000;

% Create index of new positions
numA = numel(A);
idx  = true(numA + fix(numA/step)+1, 1);
idx(1:step:end) = false;

% Reassign values to new positions and add header in between
A(idx)  = A;
A(~idx) = head;