circshift() one number ,in each iteration?

can I am do shifting the number A(2,1) of matrix (A) ,one times in each iteration,('iter=0' ,iter=iter+1)? how can do that?
A=[1 2 3;4 5 6;7 8 9]
% Shift A(2,1)
for k=1:size(A,2)-1
A(2,:)=circshift(A(2,:),[0 1])
end

2 个评论

That's what the code is doing
It shifts the entire second row one element at each iteration of the loop. Is that not what you wanted? Did you want to shift only the (2,1) element and not all of the elements in the second row? Either way, why? Seems like a funny thing to do unless it's homework or something. But it can't be homework because you would have tagged it as homework, so why do you want to do this?

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 采纳的回答

Maybe you want to save all the results
clear
A=[1 2 3;4 5 6;7 8 9]
% Shift A(2,1)
for k=1:size(A,2)-1
A(2,:)=circshift(A(2,:),[0 1])
out{k}=A
end
celldisp(out)

6 个评论

I am have this error!!!
Undefined function or variable "out".
Error in ==> sumaei at 26 celldisp(out)
(by using matlab R14)
copy and paste the code, there is no error
yes, there is no error, but when I am applied on my code Iam have error!!!
my code is
d=2:1:33;
E_0=(1/(36*pi))*1e-9;
E_r=1;
E=E_0*E_r*ones(111,34);
E(2:5,d)=10;
E(7:17,2:33)=4;
E(20,1)=100; % Shift E(20,1)=100
for k=1:size(E,20)-1
E(20,:)=circshift(E(20,:),[0 1])
out{k}=E
end
celldisp(out)
E doesn't have a 20th dimension so size(E,20)-1 = 0-1 = -1 which means your loop never gets entered. So when it comes time to do celldisp(), there is no out at all to display.
isnt it? that mean E(20,1) at row number 20,column number 1,
I am want shifting it (2:33),with save all the result not last result only
size(array, n) means the length of array in the nth dimension. It DOES NOT mean array(n) which is the value of the array at index n. They're totally different things. By the way, you never answered my question in the comment at the very top.

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