Creating Sparse Matrices in Oct-Files (GNU Octave (version 9.1.0))

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

There are two useful strategies for creating a sparse matrix. The first is to create three vectors representing the row index, column index, and data values, and from these create the matrix. The second alternative is to create a sparse matrix with the appropriate amount of space, and then fill in the values. Both techniques have their advantages and disadvantages.

Below is an example of creating a small sparse matrix using the first technique

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

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

    ridx(0) = 1; cidx(0) = 1; data(0) = 1;
    ridx(1) = 2; cidx(1) = 2; data(1) = 2;
    ridx(2) = 2; cidx(2) = 4; data(2) = 3;
    ridx(3) = 3; cidx(3) = 4; 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, but is used internally.

As discussed in the chapter on Sparse Matrices, the values of the sparse matrix are stored in increasing column-major ordering. Although the data passed by the user need not respect this requirement, pre-sorting the data will significantly speed up creation of the sparse matrix.

The disadvantage of this technique for creating a sparse matrix is that there is a brief time when two copies of the data exist. For extremely memory constrained problems this may not be the best technique for creating a sparse matrix.

The alternative is to first create a sparse matrix with the desired number of nonzero elements and then later fill those elements in. Sample code:

    int nz, nr, nc;
    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;

This creates the same matrix as previously. Again, although not strictly necessary, it is significantly faster if the sparse matrix is created and the elements are added in column-major ordering. The reason for this is that when 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 indices need to be updated.

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

    int nr, nc;
    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 because as each new element is added to the sparse matrix the matrix needs to request more space and reallocate memory. This is an expensive operation that will significantly slow this means of creating a sparse matrix. It is possible to create a sparse matrix with excess storage, so having nz greater than 4 in this example is also valid. The disadvantage is that the matrix occupies more memory than strictly needed.

Of course, it is not always possible to know the number of nonzero elements prior to filling a matrix. For this reason the additional unused storage of a sparse matrix can be removed after its creation with the `maybe_compress` function. In addition to deallocating unused storage, `maybe_compress` can also 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 re-sorting the elements. If possible, it is better for the user to avoid adding the unnecessary zeros in the first place. An example of the use of `maybe_compress` is

    int nz, nr, nc;
    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 matrix.

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

    octave_value arg;
    ...
    int nz, nr, nc;
    nz = 6, nr = 3, nc = 4;   // Assume we know the max # 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 = 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 # of nz.

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

Finally, it may 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, nr, nc;
    nz = 6, nr = 3, nc = 4;   // Guess the number of nz elements
    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 = 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_compress ();  // If don't know a priori the final # of nz.

Note that both increasing and decreasing the number of nonzero elements in a sparse matrix is expensive as it involves memory reallocation. Also because parts of the matrix, though not its entirety, exist as old and new copies at the same time, additional memory is needed. Therefore, if possible avoid changing capacity.