2 differenct size arrays, I want to make the shorter array the same size as the longer by adding Zeros between the indices or the missing numbers and make them the same length

A=[1:10,1:10]'
B=[2 4 6 8 ; 3 3 3 3]'

The output I am trying to aim for is a matrix same size as A but column 2 has the values of B(:,2) see below

Output=[1:10 ; 0 3 0 3 0 3 0 3 0 0]'

I have tried various loops but I get indicies not same used vertcat to add to B the missing values this does not work either, tried all sorts for eg

for ii= 1:10
  if ii~=B(ii,1)
    B(ii,1)=vertcat(ii,B(:,1));
    B(ii,2)=vertcat(0,B(:,2)); 
  else
    B(ii,1)=B(ii,1);
    B(ii,2)=B(ii,2);
  end
end

Answer 1

[dpb]

No loops needed (or, in Matlab, wanted... :) )

>> O=[A(:,1) zeros(size(A,1),1)];
>> O(B(:,1),2)=B(:,2)
O =
   1     0
   2     3
   3     0
   4     3
   5     0
   6     3
   7     0
   8     3
   9     0
  10     0
>>

I'd note you're using the complex-conjugate transpose operator where you intend the simple array transpose which is given by the element-wise "dot operator" version, .' This makes no numerical difference with reals but is a bad habit to get into that can bite unexpectedly if/when one has complex numerical values. It's easier to develop the habit of using the right operator for the purpose from the git-go than have to break a bad habit later or debug a case that is wrong (or worse yet, not realize the answer gotten is wrong).

[Ranjit Kaur]

Many thanks this is much an easier way of doing it I thought you had to index each element and do a loop which would never work because the two matrices are different sizes. Great appreciate the help and quick response.

[Stephen23]

Learn to avoid using loops, and make your code neater and faster:

http://www.mathworks.com/help/matlab/matlab_prog/vectorization.html

http://www.mathworks.com/company/newsletters/articles/matrix-indexing-in-matlab.html

[dpb]

"...and do a loop which would never work"

Of course a loop can work, just that the power of Matlab is largely wasted if one uses them exclusively where vectorization is possible. But, to demonstrate, given the initialization step done above, then

for i=1:length(B)
  O(B(i,1),2)=B(i,2);
end

You just have to recognize over what it is that you wish to/need to iterate.