Help with jumping one position using circshift function in a for loop
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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.
1 comentario
Chhayank Srivastava
el 15 de Ag. de 2023
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?
Respuesta aceptada
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
si = 0;
for i = 1:9
sia = rem(i,3) == 0;
si = [si+1+sia] % Shift Increments
Tn(:,:,i) = circshift(T,[0,si]);
end
Tn
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.)
.
Más respuestas (3)
n = 12;
T = zeros(2,n);
T(1,1) = 1;
T(2,2) = 1
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);
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
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)
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)
Scott Banks
el 15 de Ag. de 2023
5 comentarios
Star Strider
el 15 de Ag. de 2023
@Scott Banks — Thank you!
Scott Banks
el 17 de Ag. de 2023
Star Strider
el 17 de Ag. de 2023
@Scott Banks — Thank you!
Voss
el 17 de Ag. de 2023
@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?
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