Fill a zeros matrix with another matrix until it is full

I want to fill the array essai with the value in the array key but my code return zeros

k= 1:length(key);
yr=reshape(y.',1,[]);
essai=zeros(1,length(yr));
essai=uint8(essai);
for n= 1:length(essai)
   if k <length(key)
essai(n)=essai(n)+key(k)
else if k== length(key)

    essai(n)=essai(n)+key(k);
     k=1;
     end
     end
 end

[Stephen23]

@Davidra Fantarina ANDRIAMISAINA: your code is very badly aligned. Badly aligned code is how beginners hide simple bugs and logical errors in their code. You should use the default alignment of the MATLAB editor: select the code and press ctrl+i.

[Sumara]

THANK YOU! I've wanted to do this for when my code became misaligned but didn't know there was a command for it and fixing manually is so tedious !!!

Answer 1

[Stephen23]

MATLAB is not an ugly low-level language like C++ and does not need loops to solve all tasks:

idx = 1+mod(0:numel(y)-1,numel(key));
essai = uint8(key(idx));

And tested on some random data:

>> y = 0:9;
>> key = 2:2:8;
>> idx = 1+mod(0:numel(y)-1,numel(key));
>> essai = uint8(key(idx))
essai =
  2  4  6  8  2  4  6  8  2  4

[Jos (10584)]

C++ is not ugly! ;)

[Davidra Fantarina ANDRIAMISAINA]

Thanks, it works but can you explain this idx thing please

[Stephen23]

"but can you explain this idx thing please"

idx is a vector of indices. Lets look at my example data:

>> y = 0:9 % a vector with ten elements.
y =
   0   1   2   3   4   5   6   7   8   9
>> key = 2:2:8 % a vector with four elements.
key =
   2   4   6   8
>> idx = 1+mod(0:numel(y)-1,numel(key)) % index vector
idx =
   1   2   3   4   1   2   3   4   1   2
>> key(idx) % use the indices
ans =
   2   4   6   8   2   4   6   8   2   4

You can see how the idx values are 1 to 4 repeated up until the vector has the same number of indices as y has elements: these indices determine which elements of key will get selected: so the output vector is equivalent to

[key(1),key(2),key(3),key(4),key(1),key(2),key(3),...]