Help with updating for loop

2 vues (au cours des 30 derniers jours)
Neda
Neda le 23 Avr 2024
Commenté : Voss le 24 Avr 2024
Hi all,I have a problem with the following sample code:
M=[1 2 3 4 5 6 7 8 9 10] for i=1:length(M) M(i)=[] end
My code is not complete, but my problem is that when one row is omitted from the matrix, the size reduces to 9, so the code gets stock with an error to find M(10). Actually, I want to update the cell size in for loop. In my code, I must make some rows empty, I can not keep them. Thanks

Réponse acceptée

Voss
Voss le 23 Avr 2024
You can iterate backwards, so that subsequent indexing is never off the end of M.
Example:
M=[1 2 3 4 5 6 7 8 9 10];
for i=length(M):-1:1
M(i)=[];
end
M
M = 1x0 empty double row vector
Or you can store the indices to delete, and delete them all after the loop.
Example, using numeric indices:
M=[1 2 3 4 5 6 7 8 9 10];
to_delete = [];
for i=1:length(M)
to_delete(end+1) = i;
end
M(to_delete)=[];
M
M = 1x0 empty double row vector
Example, using logical indices:
M=[1 2 3 4 5 6 7 8 9 10];
to_delete = false(1,numel(M));
for i=1:length(M)
to_delete(i) = true;
end
M(to_delete)=[];
M
M = 1x0 empty double row vector
  4 commentaires
Neda
Neda le 24 Avr 2024
Dear Voss;
Is there any way that I can send you my full code to check, since I still have problem in my code? my code has a main function and some functions !!!
Thanks
Voss
Voss le 24 Avr 2024
Neda,
Yes, please post a new question, upload your code file(s) there using the paperclip button, and specify what problems you are having.

Connectez-vous pour commenter.

Plus de réponses (0)

Catégories

En savoir plus sur Loops and Conditional Statements dans Help Center et File Exchange

Tags

Community Treasure Hunt

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

Start Hunting!

Translated by