hello i have 53*3 matrix but include rows that three column of the row are zeros i want to delete the entire row i tried this but since every delete shift the location of the other rows i had problem with the for loop so what can i do to overcome this
[c u]=size(test_ary); %dimensions
for i=1:c
for b=1:u
if(test_ary(i,1)==(o)));
test_ary(i,:)=[];
end
end
end

 Réponse acceptée

James Tursa
James Tursa le 24 Mai 2016
Modifié(e) : James Tursa le 24 Mai 2016

0 votes

Replace your for-loop with this vectorized code:
x = all(test_ary==0,2); % Which rows are all 0's
test_ary(x,:) = []; % Delete those rows

Plus de réponses (3)

Jos (10584)
Jos (10584) le 26 Mai 2016
Using the statement "test_ary(i,:) = []" you will change the size of it, which will cause problems!
tf = all(test_ary==0,2) % true for rows with only 0's
test_ary(tf,:) = [] % remove those rows using logical indexing
Azzi Abdelmalek
Azzi Abdelmalek le 24 Mai 2016

0 votes

out=test_ary(~ismember(test_ary,[0 0 0],'rows'),:)

2 commentaires

m. muner
m. muner le 26 Mai 2016
just create new matrix called out which is same as the original test_ary
Azzi Abdelmalek
Azzi Abdelmalek le 26 Mai 2016
No, this is not true

Connectez-vous pour commenter.

Anoire BEN JDIDIA
Anoire BEN JDIDIA le 14 Oct 2016

0 votes

I have a big matrix 599794x2 i want to delete rows which contains values which repeats for exemple if A=[1,1;2,1;3,1;4,1;5,2;6,2;7,2]; i want to have A=[1,1;5,2]

2 commentaires

In the future open up a new Question for this rather than piggyback on an existing Question. But I will answer this here this time:
A = A(logical([1;diff(A(:,2))]),:);
Anoire BEN JDIDIA
Anoire BEN JDIDIA le 17 Oct 2016
Hi, Thank you

Connectez-vous pour commenter.

Catégories

En savoir plus sur Loops and Conditional Statements dans Centre d'aide et File Exchange

Community Treasure Hunt

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

Start Hunting!

Translated by