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##### A.1.6.2 Creating Sparse Matrices in Oct-Files

You have several alternatives for creating a sparse matrix. You can first create the data as three vectors representing the row and column indexes and the data, and from those create the matrix. Or alternatively, you can create a sparse matrix with the appropriate amount of space and then fill in the values. Both techniques have their advantages and disadvantages.

Here is an example of how to create a small sparse matrix with the first technique

         int nz = 4, nr = 3, nc = 4;

         ColumnVector ridx (nz);
         ColumnVector cidx (nz);
         ColumnVector data (nz);

         ridx(0) = 0; ridx(1) = 0; ridx(2) = 1; ridx(3) = 2;
         cidx(0) = 0; cidx(1) = 1; cidx(2) = 3; cidx(3) = 3;
         data(0) = 1; data(1) = 2; data(2) = 3; data(3) = 4;

         SparseMatrix sm (data, ridx, cidx, nr, nc);

which creates the matrix given in section Storage of Sparse Matrices. Note that the compressed matrix format is not used at the time of the creation of the matrix itself, however it is used internally.

As previously mentioned, the values of the sparse matrix are stored in increasing column-major ordering. Although the data passed by the user does not need to respect this requirement, the pre-sorting the data significantly speeds up the creation of the sparse matrix.

The disadvantage of this technique of creating a sparse matrix is that there is a brief time where two copies of the data exists. Therefore for extremely memory constrained problems this might not be the right technique to create the sparse matrix.

The alternative is to first create the sparse matrix with the desired number of non-zero elements and then later fill those elements in. The easiest way to do this is

         int nz = 4, nr = 3, nc = 4;
         SparseMatrix sm (nr, nc, nz);
         sm(0,0) = 1; sm(0,1) = 2; sm(1,3) = 3; sm(2,3) = 4;

That creates the same matrix as previously. Again, although it is not strictly necessary, it is significantly faster if the sparse matrix is created in this manner that the elements are added in column-major ordering. The reason for this is that if the elements are inserted at the end of the current list of known elements then no element in the matrix needs to be moved to allow the new element to be inserted. Only the column indexes need to be updated.

There are a few further points to note about this technique of creating a sparse matrix. Firstly, it is possible to create a sparse matrix with fewer elements than are actually inserted in the matrix. Therefore

         int nz = 4, nr = 3, nc = 4;
         SparseMatrix sm (nr, nc, 0);
         sm(0,0) = 1; sm(0,1) = 2; sm(1,3) = 3; sm(2,3) = 4;

is perfectly valid. However it is a very bad idea. The reason is that as each new element is added to the sparse matrix the space allocated to it is increased by reallocating the memory. This is an expensive operation, that will significantly slow this means of creating a sparse matrix. Furthermore, it is possible to create a sparse matrix with too much storage, so having nz above equaling 6 is also valid. The disadvantage is that the matrix occupies more memory than strictly needed.

It is not always easy to know the number of non-zero elements prior to filling a matrix. For this reason the additional storage for the sparse matrix can be removed after its creation with the maybe_compress function. Furthermore, the maybe_compress can deallocate the unused storage, but it can equally remove zero elements from the matrix. The removal of zero elements from the matrix is controlled by setting the argument of the maybe_compress function to be ‘`true`’. However, the cost of removing the zeros is high because it implies resorting the elements. Therefore, if possible it is better is the user doesn't add the zeros in the first place. An example of the use of maybe_compress is

           int nz = 6, nr = 3, nc = 4;

           SparseMatrix sm1 (nr, nc, nz);
           sm1(0,0) = 1; sm1(0,1) = 2; sm1(1,3) = 3; sm1(2,3) = 4;
           sm1.maybe_compress ();  // No zero elements were added

           SparseMatrix sm2 (nr, nc, nz);
           sm2(0,0) = 1; sm2(0,1) = 2; sm(0,2) = 0; sm(1,2) = 0;
           sm1(1,3) = 3; sm1(2,3) = 4;
           sm2.maybe_compress (true);  // Zero elements were added

The use of the maybe_compress function should be avoided if possible, as it will slow the creation of the matrices.

A third means of creating a sparse matrix is to work directly with the data in compressed row format. An example of this technique might be

         octave_value arg;
         ...
         int nz = 6, nr = 3, nc = 4;   // Assume we know the max no nz
         SparseMatrix sm (nr, nc, nz);
         Matrix m = arg.matrix_value ();

         int ii = 0;
         sm.cidx (0) = 0;
         for (int j = 1; j < nc; j++)
           {
             for (int i = 0; i < nr; i++)
               {
                 double tmp = foo (m(i,j));
                 if (tmp != 0.)
                   {
                     sm.data(ii) = tmp;
                     sm.ridx(ii) = i;
                     ii++;
                   }
               }
             sm.cidx(j+1) = ii;
          }
         sm.maybe_compress ();  // If don't know a-priori
                                // the final no of nz.

which is probably the most efficient means of creating the sparse matrix.

Finally, it might sometimes arise that the amount of storage initially created is insufficient to completely store the sparse matrix. Therefore, the method `change_capacity` exists to reallocate the sparse memory. The above example would then be modified as

         octave_value arg;
         ...
         int nz = 6, nr = 3, nc = 4;   // Assume we know the max no nz
         SparseMatrix sm (nr, nc, nz);
         Matrix m = arg.matrix_value ();

         int ii = 0;
         sm.cidx (0) = 0;
         for (int j = 1; j < nc; j++)
           {
             for (int i = 0; i < nr; i++)
               {
                 double tmp = foo (m(i,j));
                 if (tmp != 0.)
                   {
                     if (ii == nz)
                       {
                         nz += 2;   // Add 2 more elements
                         sm.change_capacity (nz);
                       }
                     sm.data(ii) = tmp;
                     sm.ridx(ii) = i;
                     ii++;
                   }
               }
             sm.cidx(j+1) = ii;
          }
         sm.maybe_mutate ();  // If don't know a-priori
                              // the final no of nz.

Note that both increasing and decreasing the number of non-zero elements in a sparse matrix is expensive, as it involves memory reallocation. Also as parts of the matrix, though not its entirety, exist as the old and new copy at the same time, additional memory is needed. Therefore if possible this should be avoided.