Deleting rows in a matrix
4 visualizaciones (últimos 30 días)
Mostrar comentarios más antiguos
Bas
el 11 de Ag. de 2015
Comentada: Brendan Hamm
el 11 de Ag. de 2015
Hi,
I have two matrices which are generated in a loop for n years:
A (m-rows x n-columns) B (k-rows)
Matrix B includes values which represent row numbers of matrix A that should be deleted. So, I want for every value in matrix B, the corresponding row in matrix A to be deleted.
For example,
A=[1 2; 2 2; 3 4], B= [2]
The result should be a matrix [1 2;3 4]. I.e. the middle row is deleted as B includes value 2.
For some reason I'm not able to find the correct coding for this. I tried several things, but inside the loop it doesn't work. This is what I tried, but this does't work:
for j= 1:n
B;
for i= 1:length(B)
indices = find(A(:,1)==B(i));
A(indices,:) = [];
end
end
Can anyone help me with this?
2 comentarios
James Tursa
el 11 de Ag. de 2015
Does B really contain the row numbers, or does it contain values that must match values in the first column of A? Your words say the former but the code looks like the latter.
Respuesta aceptada
Brendan Hamm
el 11 de Ag. de 2015
If you want to delete the second row because B has the value 2 you can use this vector to index into A as such:
A(B,:) = [];
That is how I interpret the question you pose, but the code seems to approach a different problem, outlined below.
If you want to delete the second row because the first element of A is 2, then you would find if each element of the first column of A is in the vector B and delete those locations:
locA = ismember(A(:,1),B); % Logical vector (true if A(i,1) is in B, false otherwise)
% use the logical vector to index into A and delete
A(locA,:) = [];
2 comentarios
Brendan Hamm
el 11 de Ag. de 2015
I am a bit confused with your question here still, what is not working specifically? Does this work fine on the first iteration of the loop? Is it that you are overwriting the excel file at every iteration of the loop? Are you receiving any errors?
Taking a stab at it:
If UitDienst is a matrix and not a vector as in the example you gave, then you likely do not want to perform a linear indexing as you do with
UitDienst(j)
Matlab is a column major language which means that elements of a matrix are stored column by column:
A = [1 2 3; 4 5 6]
is stored in memory as [1 4 2 5 3 6] (column-by-column) This means that A(2) will return the value 4, A(3) will return the value 2, etc. It seems that maybe you need to change that to read:
UitDienst(:,j)
Más respuestas (2)
James Tursa
el 11 de Ag. de 2015
Editada: James Tursa
el 11 de Ag. de 2015
"Matrix B includes values which represent row numbers of matrix A that should be deleted"
Assuming this is exactly what you want, then this will do the job:
A(B,:) = [];
0 comentarios
Jon
el 11 de Ag. de 2015
Editada: Jon
el 11 de Ag. de 2015
Your code produces the result you expect. You'll need to better specify your problem. You can lose the loop completely, though, with a simple indexing:
A(B,:) = [];
This is assuming that B is unique and contains no elements larger than the length of A.
0 comentarios
Ver también
Categorías
Más información sobre Creating and Concatenating Matrices 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!