Help with jumping one position using circshift function in a for loop

Hi there,

I am trying to create a transformation matrix for the matrix stiffness method, and to save time I am using the circshift function in a for loop.

So, I am starting off with this

n = 12;
T = zeros(2,n);

T(1,1) = 1;
T(2,2) = 1

for i = 1:9
  Tn(:,:,i) = circshift(T,[0,i])
end

However, every 3rd interation I want the values (the 1's) to shift one extra place to the right. So I want the Tn matrix to follow this pattern:

Tn = [1 1 0 1 1 0 1 1 0 1 1 0;
      0 1 1 0 1 1 0 1 1 0 1 1]

I guessed at trying this:

for i = 1:9
  Tn(:,:,i) = circshift(T,[0,i])
  if i == 3,6,9
    Tn(:,:,i) = circshift(T,[0,i+1])
  endif
end

But it didn't work.

I hope I have explained this clearly. Can somebody help please?

Many thanks.

[Chhayank Srivastava]

Could you clarify what do you mean when you say you want Tn to follow that pattern, does that mean after running the loop Tn should look like that?

Answer 1

[Star Strider]

The circshift function does not duplicate any values, so I don’t understand how you expect to get the ‘Tn’ matrix at the end.

I’m not certain what result you want otherwise.

Try this —

n = 12;
T = zeros(2,n);

T(1,1) = 1;
T(2,2) = 1

T = 2×12

1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0

si = 0;
for i = 1:9
    sia = rem(i,3) == 0;
    si = [si+1+sia]                                             % Shift Increments
    Tn(:,:,i) = circshift(T,[0,si]);
end

si = 1

si = 2

si = 4

si = 5

si = 6

si = 8

si = 9

si = 10

si = 12

Tn

Tn = 
Tn(:,:,1) =

     0     1     0     0     0     0     0     0     0     0     0     0
     0     0     1     0     0     0     0     0     0     0     0     0


Tn(:,:,2) =

     0     0     1     0     0     0     0     0     0     0     0     0
     0     0     0     1     0     0     0     0     0     0     0     0


Tn(:,:,3) =

     0     0     0     0     1     0     0     0     0     0     0     0
     0     0     0     0     0     1     0     0     0     0     0     0


Tn(:,:,4) =

     0     0     0     0     0     1     0     0     0     0     0     0
     0     0     0     0     0     0     1     0     0     0     0     0


Tn(:,:,5) =

     0     0     0     0     0     0     1     0     0     0     0     0
     0     0     0     0     0     0     0     1     0     0     0     0


Tn(:,:,6) =

     0     0     0     0     0     0     0     0     1     0     0     0
     0     0     0     0     0     0     0     0     0     1     0     0


Tn(:,:,7) =

     0     0     0     0     0     0     0     0     0     1     0     0
     0     0     0     0     0     0     0     0     0     0     1     0


Tn(:,:,8) =

     0     0     0     0     0     0     0     0     0     0     1     0
     0     0     0     0     0     0     0     0     0     0     0     1


Tn(:,:,9) =

     1     0     0     0     0     0     0     0     0     0     0     0
     0     1     0     0     0     0     0     0     0     0     0     0

Ths ‘si’ values are the increments provided to circshift so that you can keep track of them. (Suppress that line’s output when its display is no longer necessary.)

.

Answer 2

[Voss]

n = 12;
T = zeros(2,n);

T(1,1) = 1;
T(2,2) = 1

T = 2×12

1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0

Here's one way to generate that Tn from this T using circshift in a loop:

Tn = zeros(2,n,8);
shift = 0;
for i = 1:size(Tn,3)
    if mod(i,2) == 1
        shift = shift+1;
    end
    Tn(:,:,i) = circshift(T,[0,shift-1]);
    shift = shift+1;
end
disp(Tn);

(:,:,1) =

     1     0     0     0     0     0     0     0     0     0     0     0
     0     1     0     0     0     0     0     0     0     0     0     0


(:,:,2) =

     0     1     0     0     0     0     0     0     0     0     0     0
     0     0     1     0     0     0     0     0     0     0     0     0


(:,:,3) =

     0     0     0     1     0     0     0     0     0     0     0     0
     0     0     0     0     1     0     0     0     0     0     0     0


(:,:,4) =

     0     0     0     0     1     0     0     0     0     0     0     0
     0     0     0     0     0     1     0     0     0     0     0     0


(:,:,5) =

     0     0     0     0     0     0     1     0     0     0     0     0
     0     0     0     0     0     0     0     1     0     0     0     0


(:,:,6) =

     0     0     0     0     0     0     0     1     0     0     0     0
     0     0     0     0     0     0     0     0     1     0     0     0


(:,:,7) =

     0     0     0     0     0     0     0     0     0     1     0     0
     0     0     0     0     0     0     0     0     0     0     1     0


(:,:,8) =

     0     0     0     0     0     0     0     0     0     0     1     0
     0     0     0     0     0     0     0     0     0     0     0     1

Answer 3

[Chhayank Srivastava]

Hi,

I see some issue with the if statement mentioned above.

So just fixing the if statement

n = 12;
T = zeros(2,n);

T(1,1) = 1;
T(2,2) = 1;

for i = 1:9
  Tn(:,:,i) = circshift(T,[0,i]);
  if mod(i,3) == 0
   Tn(:,:,i) = circshift(T,[0,i+1]);
  end
end
Tn

Tn = 
Tn(:,:,1) =

     0     1     0     0     0     0     0     0     0     0     0     0
     0     0     1     0     0     0     0     0     0     0     0     0


Tn(:,:,2) =

     0     0     1     0     0     0     0     0     0     0     0     0
     0     0     0     1     0     0     0     0     0     0     0     0


Tn(:,:,3) =

     0     0     0     0     1     0     0     0     0     0     0     0
     0     0     0     0     0     1     0     0     0     0     0     0


Tn(:,:,4) =

     0     0     0     0     1     0     0     0     0     0     0     0
     0     0     0     0     0     1     0     0     0     0     0     0


Tn(:,:,5) =

     0     0     0     0     0     1     0     0     0     0     0     0
     0     0     0     0     0     0     1     0     0     0     0     0


Tn(:,:,6) =

     0     0     0     0     0     0     0     1     0     0     0     0
     0     0     0     0     0     0     0     0     1     0     0     0


Tn(:,:,7) =

     0     0     0     0     0     0     0     1     0     0     0     0
     0     0     0     0     0     0     0     0     1     0     0     0


Tn(:,:,8) =

     0     0     0     0     0     0     0     0     1     0     0     0
     0     0     0     0     0     0     0     0     0     1     0     0


Tn(:,:,9) =

     0     0     0     0     0     0     0     0     0     0     1     0
     0     0     0     0     0     0     0     0     0     0     0     1

Moreover, I am assuming after running the program you want the Tn to look like

Tn = [1 1 0 1 1 0 1 1 0 1 1 0;

0 1 1 0 1 1 0 1 1 0 1 1];

But in the above code you are generating a 3D matrix.

A simpler solution would be just to use repmat and transformation matrix

 T = [1,1,0;0,1,1]; 
Tn = repmat(T,1,4)

Tn = 2×12

1 1 0 1 1 0 1 1 0 1 1 0 0 1 1 0 1 1 0 1 1 0 1 1

But, going by your method

clear;clc;
n = 12;
T = zeros(2,n);

T(1,1) = 1;
T(2,2) = 1;

for i = 1:9

  if mod(i,2) == 0
   T = circshift(T,[0,2]);
   Tn(:,:,i) = T;
  else
   
   T = circshift(T,[0,1]);  
   Tn(:,:,i) = T;  
  end
   
 
end
Tn = Tn(:,:,1)+Tn(:,:,2)+Tn(:,:,3)+Tn(:,:,4)+Tn(:,:,5)+Tn(:,:,6)+Tn(:,:,7)+Tn(:,:,8)

Tn = 2×12

1 1 0 1 1 0 1 1 0 1 1 0 0 1 1 0 1 1 0 1 1 0 1 1

Answer 4

[Scott Banks]

Hi guys,

thank you for your repsonses.

Sorry for the confusion for the final Tn matrix. It was a poor way of getting across my idea.

Really, it should look like how Star Strider answered. Which was 9 different matrices in the form of:

Tn = 
Tn(:,:,1) =

     0     1     0     0     0     0     0     0     0     0     0     0
     0     0     1     0     0     0     0     0     0     0     0     0


Tn(:,:,2) =

     0     0     1     0     0     0     0     0     0     0     0     0
     0     0     0     1     0     0     0     0     0     0     0     0


Tn(:,:,3) =

     0     0     0     0     1     0     0     0     0     0     0     0
     0     0     0     0     0     1     0     0     0     0     0     0


Tn(:,:,4) =

     0     0     0     0     1     0     0     0     0     0     0     0
     0     0     0     0     0     1     0     0     0     0     0     0


Tn(:,:,5) =

     0     0     0     0     0     1     0     0     0     0     0     0
     0     0     0     0     0     0     1     0     0     0     0     0


Tn(:,:,6) =

     0     0     0     0     0     0     0     1     0     0     0     0
     0     0     0     0     0     0     0     0     1     0     0     0


Tn(:,:,7) =

     0     0     0     0     0     0     0     1     0     0     0     0
     0     0     0     0     0     0     0     0     1     0     0     0


Tn(:,:,8) =

     0     0     0     0     0     0     0     0     1     0     0     0
     0     0     0     0     0     0     0     0     0     1     0     0


Tn(:,:,9) =

     0     0     0     0     0     0     0     0     0     0     1     0
     0     0     0     0     0     0     0     0     0     0     0     1

[Star Strider]

@Scott Banks — Thank you!

If I provided the correct result, please Accept my Answer.

[Voss]

@Scott Banks: Just to confirm, you want Tn(:,:,3) to be the same as Tn(:,:,4), and Tn(:,:,6) to be the same as Tn(:,:,7). Is that right?

[Scott Banks]

Hi Voss,

Actually no! I want Tn(:,:3) and the others you mentioned to be shifted one increment to the right.

[Star Strider]

@Scott Banks — Thank you!

If I provided the correct result, please Accept my Answer.

(Comment also posted earlier)

[Voss]

@Scott Banks: I'm a little confused because if Tn(:,:,3) and T(n:,:,4) are both shifted to the right, then they are still the same as each other, which contradicts your statement that they should not be the same as each other.

Can you just write down the complete 3D array as it should be?