Add a known column to a matrix 'n' times without a loop?

Hi Guys!

Is there a way of qriting the script below without the loop? I'm sure there's a way but I can't find it in the help section or the internet.

I want to add a known column of numbers to a matrix 'n' times.

Thank you!

Ken

   clear all
  clc

a=zeros(5,6); % matrix a

[m,n]=size(a);

b=zeros(m,n); % matrix b

b(1:10)=[8 8 8 8 8 10 10 10 10 10];
for i=3:n;

      b(:,i)=[1 2 3 4 5];

end

[Daniel Shub]

There are lots of ways, but why would you want to?

[ken]

I just wanted to remove the 'for loop' to make the script more efficient/faster. Am I right in thinking 'for loops' are harder to compute?

I have done a 'tic toc' test and the for loop takes about the same time as andrei's first two suggestions. Its just a tiny tiny bit slower.

[Walter Roberson]

These days "for loops" are not necessarily slower, and in some cases are faster.

[ken]

Thanks for letting me know Walter

Answer 1

[Andrei Bobrov]

b(:,k:n) = (1:5)'*ones(1,n-k+1);

more

V = 1:5;
b(:,k:n) = V(ones(1,n-k+1),:).';

more more

b(:,k:n) = repmat(1:5,n-k+1,1).';

[ken]

Thanks Andrei for all your suggestions! I was initially trying to do your first one but it didn't work because I wasn't multiplying by a matrix.

[Jan]

@Andrei: The ONES method is applied inside REPMAT also. Therefore I expect using it directly is faster, because the overhead of calling a function is avoided.

Usually a matrix with repeated rows wastes memory, because of the redundancy. Therefore a BSXFUN method can be faster: Do not create the matrix explicitely, but instruct BSXFUN to perform an operation by "inflating" the vector dynamically.

[Andrei Bobrov]

Thank Jan for your comment

Answer 2

[Fangjun Jiang]

help repmat

[ken]

Thanks Fangjun!