Why do I receive error for reshape function?

1 Ansicht (letzte 30 Tage)
Ryan
Ryan am 7 Jun. 2021
Kommentiert: Ryan am 7 Jun. 2021
I am using the following code below, however, I keep getting geting the error message for reshape function. How would the code or reshape function be changed to stop the error?
>> s = 1/sqrt(2);
A = [1 1 1 0 0 0 0 0 0; 0 0 0 0 0 0 1 1 1; 1 0 0 1 0 0 1 0 0; 0 0 1 0 0 1 0 0 1; s s 0 s 0 0 0 0 0; 0 0 0 0 0 s 0 s s; 0 0 0 s 0 0 s s 0; 0 s s 0 0 s 0 0 0];
b = [6; 24; 12; 18; 4.9497; 16.2634; 13.4350; 7.7781];
k = 1:20;
X = kaczmarz(A,b,k);
figure(1)
for i=1:kmax, subplot(4,5,i), imagesc(reshape(X(:,i),N,N)), end
Error using reshape
Number of elements must not change. Use [] as one of the size inputs
to automatically calculate the appropriate size for that dimension.
  3 Kommentare
Scott MacKenzie
Scott MacKenzie am 7 Jun. 2021
I suspect the issue is that X(:,i) does not have exactly NxN elements.
Ryan
Ryan am 7 Jun. 2021
I've checked the size of X:
size(X)
ans =
9 7
>> N*N
ans =
4096
What would I need to change X size to be to get figure(1)?

Melden Sie sich an, um zu kommentieren.

Akzeptierte Antwort

KSSV
KSSV am 7 Jun. 2021
The number of elements in X(:,j) is not equal to N*N, thats why you are getting error.
N = 3 ;
X = rand(10) ;
x = reshape(X(:,1),N,N) ; % this will lead to error
Error using reshape
Number of elements must not change. Use [] as one of the size inputs to automatically calculate the appropriate size for that dimension.
See to it that the number of elements should not change. You may use imresize to achieve this.
[m,n] = size(X) ;
X = imresize(X,[N*N,n]) ;
imagesc(reshape(X(:,1),N,N)) ; % this will work
  3 Kommentare
KSSV
KSSV am 7 Jun. 2021
It would be because the dimensions (number of rows) in x and X are different. You can change the dimensions of x or use same X as previous.
Ryan
Ryan am 7 Jun. 2021
Great! All working now, thanks for the help!

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Produkte


Version

R2020b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by