Select all (non-zero) columns of sparse matrix
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Quinten Rensen
il 24 Ago 2018
Commentato: Quinten Rensen
il 27 Ago 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 Commenti
Jan
il 26 Ago 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.
Risposta accettata
Matt J
il 26 Ago 2018
Modificato: Matt J
il 26 Ago 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 Commenti
Matt J
il 26 Ago 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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