How to rotate rows of a matrix?

I have a matrix A where each row of A has only one value of 1 and the rest are some other number.

A = [9 8 7 1; 9 1 8 7; 9 8 1 7; 9 1 8 7; 9 8 7 1; 1 9 8 7]

How can I rotate all of the rows individually such that the 1 value is in the first column of each row in the resultant matrix? How about the second column? I'd like to do this without a for-loop because I am working with large matrices.

%%% Result:

% First column.

B = [1 9 8 7; 1 8 7 9; 1 7 9 8; 1 8 7 9; 1 9 8 7; 1 9 8 7];

% Second column.

C = [7 1 9 8; 9 1 8 7; 8 1 7 9; 9 1 8 7; 7 1 9 8; 7 1 9 8];

Answer 1

[Roger Stafford]

I think a for-loop is your best bet:

for k = 1:size(A,1)
  f = find(A(k,:)==1,1);
  A(k,:) = circshift(A(k,:),1-f);
end

Answer 2

[Image Analyst]

Try this:

A = [1 0 0 0; 0 1 0 0; 0 0 1 0; 0 1 0 0; 0 0 0 1; 0 0 1 0];
desiredColumn = 1; % or 2 or whatever
output = zeros(size(A)); % Initialize
output(:, desiredColumn) = 1 % Assign desired column to all ones.

[Dominik Mattioli]

Eek, I edited this just after I posted it, I'm sorry. I forgot to add that the order of each row is important. Do you know how to do that too?

Answer 3

[Guillaume]

A version without loop. Not sure it'd be faster than Roger's answer:

destcol = 1;  %column where the 1 must be located
[c, r] = find(A' == 1);
A(sub2ind(size(A), repmat(r, 1, size(A, 2)), mod((1:size(A, 2)) + c - 1 - destcol, size(A, 2)) + 1))