How Zero padding inside a matrix ?

Hello, how can i zero pad inside a matrix ? For example :

if true
  % code
end
A = [ 1 2 3 4
    5 6 7 8
    9 10 11 12
    13 14 15 16]
And i want 
A = [ 1 2 3 4
    5 6 7 8
    0 0 0 0
    0 0 0 0
    ......
    0 0 0 0 
    0 0 0 0
    9 10 11 12
    13 14 15 16]

I know padarray can zero pad but only outside not in the inside.. Thank you.

Answer 1

[Star Strider]

Create a second matrix of zeros, then assign the appropriate rows of your first matrix to the rows you want in the second matrix:

A = [ 1 2 3 4
    5 6 7 8
    9 10 11 12
    13 14 15 16];

Desired_Rows = 10;                                          % Pick A Number
B = zeros(Desired_Rows, size(A,2));
B([1:2 end-1:end],:) = A

B =

       1     2     3     4
       5     6     7     8
       0     0     0     0
       0     0     0     0
       0     0     0     0
       0     0     0     0
       0     0     0     0
       0     0     0     0
       9    10    11    12
      13    14    15    16

[So]

Thank you for your answer. I don't really understand that line : B([1:2 end-1:end],:) = A

For example, if my initial matrix has 20 rows and I want

if true
  % code
end
myFirst10Rows
40RowsOfZeros
myLast10Rows

Thank you

[Star Strider]

My pleasure.

The assignment:

B([1:2 end-1:end],:) = A

creates a vector of row indices ‘[1:2 end-1:end]’ indicating the first two rows (the ‘1:2’) and the last two rows (the ‘end-1:end’), using the colon operator to create the continuity.

For your (20xN) matrix, my slightly revised code becomes:

B = zeros(size(A,1)+40, size(A,2));
B([1:10 end-9:end],:) = A

This creates a (60xN) matrix of zeros (in your example), then fills the first 10 rows of it with the first 10 rows of ‘A’ and the last 10 rows of it with the last 10 rows of ‘A’. Note that ‘end-9’ is the tenth row from the end.

Experiment with the indexing to see how it works.

[So]

I understand now, and it works perfectly with my code. Thank you very much.

[Star Strider]

My pleasure.

If my Answer solved your problem, please Accept it.

[Guillaume]

In my opinion, it'd be simpler to split the original matrix and insert a zero matrix in between: B = [A(1:10, :); zeros(40, size(A, 2); A(11:20, :)]

[Star Strider]

I thought about that, but considered that this could be a ‘proxy problem’ and the rows could end up being inserted anywhere in the target matrix, not just at the ends. The approach I took allows for that more easily.