how can make the xor return integer values instead of 1's or 0's?

In the following code, I am xor-ing two different sizes matrices, to get rid of the element overlap in any row of x and y (overlap happens when I have an integer element in x and an integer element in y in same position) example: x=[1 0]; y=[5 0]; 1 & 5 are consider overlapping. so it get rid of these two rows I am xor-ing the matrices.

The problem is that, when you xor you do a logical operation, whcih means the output is 1 or 0 but I want to reserve the element values, Ex: x=[1 0]; y=[0 5]; z=xor(x,y); z=[1 1]; while I want it to be z=[1 5];

Any hint? Is there any function can do xor-ing and reserving the value and not returning only 1 or 0? which means if c & v are any integers: xor(0,0)=xor(c,v)=0 xor(c,0)=c, xor(0,v)=v;

here is the code,

x=[0 1;1 0]; 
y=[2 0;0 2;2 2];
r=size(x,1); 
z = cell(r,1); % intialize z
for n=1:r
  % identify rows where there is no 1's overlap
  k = ~any(bsxfun(@and,x(n,:),y),2);  
  % xor 'n' x row with the 'k' rows of y
  z{n} = bsxfun(@xor,x(n,:),y(k,:)); 
end
S = vertcat(z{:});

Answer 1

[Star Strider]

See if bitxor does what you want:

z{n} = bsxfun(@bitxor,x(n,:),y(k,:));

[JacobM]

thank you agian Star, I am really grateful for your help.

Just maybe an extension question if you don't mind, in S matrix, the last matrix, suppose I want to have a new matrix S_new that return all the rows of S that has minimum element transition, I meant by that if row is [1 0] and another row is [2 0] it is considered one transition as 1 changed to 2. if a row [1 0] and another row is [0 2] is considered 2 transitions. 1 changed to 0, 0 changed to 2.

any hint how can I do that? note that I am comparing all rows together, so there is no reference, the rows with min element transitions will be returned in S_new.

[Star Strider]

As always, my pleasure.

The sincerest expression of appreciation here on MATLAB Answers is to Accept the answer that most closely solves your problem.

I’m not certain exactly what you want with respect to your transition matrix, but if your ‘S’ matrix always contains only integers, experiment with the diff function. It may do what you want without any modification:

S_new = diff(S);

or perhaps using abs as well:

S_new = abs(diff(S));

depending on what you want as the result.

[JacobM]

diff is just taking the difference,

for my case, Ex1: S=[1 0 2;0 1 2;1 1 2]

row1 [1 0 2] and row3[1 1 2] has only transition where 0 in row1 changed to 1 in row2. but if you look to row2, it has two transition compared to row1, 1 changed to 0, and 0 chaned to 1 and so it is deleted. So the new matrix S_new will take row1 and row3 only;

S_new=[1 0 2;1 1 2]

I hope I could explain it well?

[Star Strider]

It took me a bit of time to understand what you want to do. I am still not certain that I completely understand.

I don’t know if this would generalise to other matrices, since I did not test it on any others, but it works here:

S=[1 0 2;0 1 2;1 1 2];
S_new = S(sum(bsxfun(@xor,S,circshift(S,1,1)),2) == 1, :)

S_new =
     1     0     2
     1     1     2

The xor call within the bsxfun call actually ‘counts’ the row-to-row transitions (using circshift to rotate ‘S’ vertically to make the comparisons possible), and then uses sum to sum them row-wise (across columns). If there is only one detected transition, it keeps that row, if more than one it deletes it.

[JacobM]

that's perfect. but the problem is 2 to 1 or 1 to 2 is not consider a transition because of using xor function.

Same example if S=[1 0 2;0 1 1;1 1 2]; between row1 and row2, 2 transition, 1-->0 & 2-->1. while if xor them it will give only one transition as 2 is considered logic1.

what do you think?

[Star Strider]

It would be best to post this as a new Question, and give several examples of initial matrices and your desired results for each. I cannot code a general solution that would work for all your matrices.

[JacobM]

sure and thanks for taking the time to reply to me

[Star Strider]

My pleasure.

Andrei Bobrov can probably come up with a one or two line perfect solution, but it is totally escaping me this afternoon.

I will do what I can to help otherwise.