Effacer les filtres
Effacer les filtres

How know the size of matrix after delete raws

2 vues (au cours des 30 derniers jours)
asma khaled
asma khaled le 22 Juin 2018
Commenté : dpb le 23 Juin 2018
What function do you use to know the size of the array after one or more rows are removed after a condition is achieved within a loop

Réponses (4)

dpb
dpb le 22 Juin 2018
Modifié(e) : dpb le 22 Juin 2018
"I want to delete all rows that start with an element greater than 5 in the matrix"
In Matlab, use "logical addressing" for such problems...
>> s(s(:,1)>5,:)=[]
s =
2 98 52 41 63
4 5 2 6 9
>>

Shweta Singh
Shweta Singh le 22 Juin 2018
You can use the 'size' function as follows:
matrix_example = rand(7,8); %matrix of size 7x8
[matrix_x, matrix_y] = size(matrix_example);
% Delete a row
matrix_example(3,:) = [];
% Get the updated dimensions
[matrix_x, matrix_y] = size(matrix_example);
Hope this helps!

asma khaled
asma khaled le 22 Juin 2018
But if I place this code inside a loop for 1 to the number of rows of matrix, a error message will appear because the size of the array has changed.
  1 commentaire
Shweta Singh
Shweta Singh le 22 Juin 2018
That should be a warning and not an error. Can you share your code?

Connectez-vous pour commenter.


asma khaled
asma khaled le 22 Juin 2018
s=[10 20 30 50 60;80 90 40 10 60;2 98 52 41 63;4 5 2 6 9];
[count,count1]=size(s);
for i=1:count
if (s(i,1)>5)
s(i,:)=[];
[count,count1]=size(s);
end
end
error:Index exceeds matrix dimensions.
  4 commentaires
dpb
dpb le 22 Juin 2018
Modifié(e) : dpb le 23 Juin 2018
Because when you got to the loop index with i = 4, you've already removed more than that number of rows so when you write
s(i,1)
where i==4 and there are only three rows left, the array addressing is outside the array dimensions and that causes the error.
To do something like this in a loop, you would want to start at the end of the array and work backwards so that the loop index is decreasing at at least the same rate as the size of the array--
nr=size(s,1); % initial number rows in s
for i=nr:-1:1 % start from end, work to beginning...
if s(i,1)>5, s(i,:)=[]; end % eliminate if desired.
end
By going backwards, you always have at least as many rows left in the array as the index value because you've removed only those of higher initial position previously for the next iteration.
BTW, don't feel badly; this is a general precept that every beginning programmer has had to learn the hard way first... :)
dpb
dpb le 23 Juin 2018
Again, move to Comment...-dpb
Thank you so much for your help

Connectez-vous pour commenter.

Catégories

En savoir plus sur Matrix Indexing dans Help Center et File Exchange

Community Treasure Hunt

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

Start Hunting!

Translated by