Help with updating for loop

3 Ansichten (letzte 30 Tage)
Neda
Neda am 23 Apr. 2024
Kommentiert: Voss am 24 Apr. 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

Akzeptierte Antwort

Voss
Voss am 23 Apr. 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 Kommentare
Neda
Neda am 24 Apr. 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 am 24 Apr. 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.

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Kategorien

Mehr zu App Building finden Sie in Help Center und File Exchange

Tags

Community Treasure Hunt

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

Start Hunting!

Translated by