deleting a row based on index.

9 visualizaciones (últimos 30 días)
CalebJones
CalebJones el 3 de Sept. de 2019
Comentada: CalebJones el 3 de Sept. de 2019
HbO_index = find(xyz1 > 7.5);%link the index with the HbO table and eliminate the bad channels
HbO_good_channel = HbO;
% for i = 1 : size(HbO,2)
for l = 1 : length(HbO_index)
k = HbO_index(l);
HbO_good_channel(:,k) = [];
end
% end
HbO_index is matrix containing values which are basically index of HbO_good_channels which i want to remove.
Only the first column from HbO_good_channels it removing properly rest its removing one index prior to what is actually mentioned, it something to do with my logic. help me out please.
thank you
  2 comentarios
madhan ravi
madhan ravi el 3 de Sept. de 2019
Attach your data as .mat file.
CalebJones
CalebJones el 3 de Sept. de 2019
Editada: CalebJones el 3 de Sept. de 2019
I have attached the files below madhan ravi

Iniciar sesión para comentar.

Respuesta aceptada

David
David el 3 de Sept. de 2019
The problem is that in your for-loop you delete the l-th column. Thereby all the indecies of the following columns change (are reduced by one). This happens in each iteration of the for-loop. If you want to remove columns "a, b, c, d" with your code you will actually remove columns "a, b+1, c+2, d+3".
You can solve your problem by starting the for loop on the end of "HBO_index" or without a for-loop:
HbO_good_channel = HbO
HbO_good_channel(:,find(xyz1 > 7.5)) = [];
  3 comentarios
David
David el 3 de Sept. de 2019
You could also remove the "find" and use logical indexing for better performance:
HbO_good_channel = HbO
HbO_good_channel(:,xyz1 > 7.5) = [];
CalebJones
CalebJones el 3 de Sept. de 2019
Sure thank you David

Iniciar sesión para comentar.

Más respuestas (0)

Categorías

Más información sobre Matrix Indexing en Help Center y File Exchange.

Community Treasure Hunt

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

Start Hunting!

Translated by