I am trying to put zero on top of each generated column.

Hello,

I am trying to put zero on top of each generated column. Assuming that "A", "ds" and "ws" differ in each of my data files, I need your help to make c2 generic for every file with known "ws", "ds" and "A". In below sample, data are not that much; I do not know how to expand it for big files.

I hope someone can help in this :)

Thanks

ws = 6                            %row
ds = 4                            %column
A = [1;3;5;5;6;7;8;5;1;6;9;2;6;7;4;5;6;7;8;5;5;3;2;1]            
for i = 1:ds
c2 = [0;A(1:ws*1);0;A(ws+1:ws*2);0;A(ws*2+1:ws*3);0;A(ws*3+1:end)]
end
g= reshape (A,ws,ds)
g2 = reshape (c2 , ws+1, ds) 

Answer 1

[BobH]

Your first reshape is the right approach, to arrange the values into a matrix with the right number of columns

ws = 6;                            %row
ds = 4;                            %column
A = [1;3;5;5;6;7;8;5;1;6;9;2;6;7;4;5;6;7;8;5;5;3;2;1];
g= reshape (A,ws,ds)
g =
     1     8     6     8
     3     5     7     5
     5     1     4     5
     5     6     5     3
     6     9     6     2
     7     2     7     1

Flip the columns, append a row of 0 at the new bottom, flip again

fg = flipud(g);
fg(end+1,:) = 0; 
g0 = flipud(fg)
g0 =
     0     0     0     0
     1     8     6     8
     3     5     7     5
     5     1     4     5
     5     6     5     3
     6     9     6     2
     7     2     7     1

[Masoud Taleb]

Thanks for the response. It works great :)

[BobH]

David's answer is far superior, and he rightly mentions that a proper reshape depends on your original array being neatly divisible by the number of columns or rows.

Please use his answer.

Answer 2

[David Hill]

Assuming you know that mod(length(A),ws)==0

g=reshape(A,ws,[]);
g2=[zeros(1,size(g,2));g];

[Masoud Taleb]

Thanks David. very nice idea :)