Getting error while deleting every row of the matrix M that contains the variable x
1 visualizzazione (ultimi 30 giorni)
Mostra commenti meno recenti
I'm trying to write a function N = Deletevar(x,M) that returns a matrix N by deleting every row of the matrix M that contains the variable x.
My code is :
function N = Delete_var(x,M)
N = [];
[a,b] = size(M);
for i = 1:a
for j = 1:b
if M(i,j)== x
M(i,:)=[];
N = M;
end
end
end
end
However when i call Delete_var(3,[1,2,3;3,4,5;7,8,9]), I get:
Index in position 1 exceeds array bounds (must not exceed 2).
Error in Delete_var (line 9)
if M(i,j)== x
Why is that? How to solve the problem? Please help!
2 Commenti
David Fletcher
il 1 Mag 2021
Modificato: David Fletcher
il 1 Mag 2021
The inherent problem you have with the logic of your code is that you are basing the for loops on the starting size of the array (M). Consider what happens when you delete a row in M - the row dimension becomes smaller, but this is not reflected in the for loop which is still based on the starting size of the array. The code will always ultimately throw an error (unless no rows are deleted) since there will be a mismatch between the starting size of M and its size after rows are deleted.
Risposta accettata
per isakson
il 1 Mag 2021
Modificato: per isakson
il 1 Mag 2021
The trick is to iterate in reverse order, i.e a:-1:1 instead of 1:a. Try this
%%
M = magic(5);
x = 13;
N = Delete_var(x,M)
%%
function N = Delete_var(x,M)
N = [];
[a,b] = size(M);
for i = a:-1:1
for j = b:-1:1
if M(i,j)== x
M(i,:)=[];
N = M;
end
end
end
end
Più risposte (1)
Turlough Hughes
il 1 Mag 2021
No need for any loops. Here's an example of data to work with
M = randi(10,[10 3]); % sample data
x = 3; % delete rows with this value
You can delete any rows in M containing x as follows:
M(any(M==x,2),:)=[];
Vedere anche
Categorie
Scopri di più su Loops and Conditional Statements in Help Center e File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!