Compressed Sparse Row Transpose

Question

As you know we can write sparse matrices in compressed row storage (CRS) (or alternatively, compressed sparse row (CSR)). Let A be an m n matrix. The transpose of A is an n x m matrix A' such that for all 0 <= i < n and 0 <= j < m, A'(i; j) = A(j; i).

I need to write the algorithm for transposing a matrix in CRS representation. How can i approach this problem?

Answer 1

I was looking for something like that. Here is my algorithm. I don't know if it is the fastest, but I think it's quite good.

EDIT: Essentially the same algorithm is implemented in C++ module for scipy.

Suppose matrix is represent by this struct:

struct CRSMatrix
{
    int n; // number of rows
    int m; // number of columns
    int nz; // number of non-zero elements
    std::vector<double> val; // non-zero elements
    std::vector<int> colIndex; // column indices
    std::vector<int> rowPtr; // row ptr
};

This function does it:

CRSMatrix sparse_transpose(const CRSMatrix& input) {
    CRSMatrix res{
        input.m,
        input.n,
        input.nz,
        std::vector<double>(input.nz, 0.0),
        std::vector<int>(input.nz, 0),
        std::vector<int>(input.m + 2, 0) // one extra
    };

    // count per column
    for (int i = 0; i < input.nz; ++i) {
        ++res.rowPtr[input.colIndex[i] + 2];
    }

    // from count per column generate new rowPtr (but shifted)
    for (int i = 2; i < res.rowPtr.size(); ++i) {
        // create incremental sum
        res.rowPtr[i] += res.rowPtr[i - 1];
    }

    // perform the main part
    for (int i = 0; i < input.n; ++i) {
        for (int j = input.rowPtr[i]; j < input.rowPtr[i + 1]; ++j) {
            // calculate index to transposed matrix at which we should place current element, and at the same time build final rowPtr
            const int new_index = res.rowPtr[input.colIndex[j] + 1]++;
            res.val[new_index] = input.val[j];
            res.colIndex[new_index] = i;
        }
    }
    res.rowPtr.pop_back(); // pop that one extra

    return res;
}