Select all (non-zero) columns of sparse matrix
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Quinten Rensen
el 24 de Ag. de 2018
Comentada: Quinten Rensen
el 27 de Ag. de 2018
Hi all,
Since I'm working with very large and sparse matrices I use sparse() to define these matrices. I do this by defining the index of the rows and columns and their corresponding value in separate vectors. For example, for a row i and columns j,j-1,j+1, the values in the matrix are defined according:
s = 1;
row(s) = i; column(s)=j; K(s) = 1; s=s+1;
row(s) = i; column(s)=j-1; K(s) = -1; s=s+1;
row(s) = i; column(s)=j+1; K(s) = -1; s=s+1;
%This is done for all non-zero points
K_matrix = sparse(row,column,K);
This works great when specifying distinct values. However, how can I use this sparse indexing when the value depends on values in other rows of the matrix. For example:
K(i,:) = K(i-1,:) + K(i+1,:);
i.e. I need all the values of another row in order to construct the current row, since I can't do something like this:
row(s) = i; column(s)=:; K(s) = K(i-1,:)+K(i+1,:); s=s+1;
I use this way of defining the sparse matrix since it is recommended for speed. Does anyone know how I can circumvent this problem without Matlab becoming slow when building the matrix?
Kind regards
3 comentarios
Jan
el 26 de Ag. de 2018
@Quinten Rensen: Posting an answer requires to understand, what you want to achieve, so please explain this. The purpose of "column(s)=:" is still not clear and it is no valid Matlab syntax.
Respuesta aceptada
Matt J
el 26 de Ag. de 2018
Editada: Matt J
el 26 de Ag. de 2018
The best way will depend on the specifics of which rows are independent and how the remaining rows are derived from them. However, often the way would be to fill all of the independent rows first and then use vectorized operations to fill the remaining ones simultaneously. For example, starting with the sparse matrix,
>> K=spdiags([1 0 1 0 1].',0,5,5); full(K)
ans =
1 0 0 0 0
0 0 0 0 0
0 0 1 0 0
0 0 0 0 0
0 0 0 0 1
I can fill in the even-numbered rows with the sum of their neighboring rows in a single statement,
>> K(2:2:end,:) = K(1:2:end-1,:)+K(3:2:end,:); full(K)
ans =
1 0 0 0 0
1 0 1 0 0
0 0 1 0 0
0 0 1 0 1
0 0 0 0 1
3 comentarios
Matt J
el 26 de Ag. de 2018
Also, it's probably better to work on columns rather than rows, transposing the final matrix so that columns become rows, if needed.
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