Question

我将一个python代码传递给C ++，我发现这样的python表达式：

J11 = dS_dVa[array([pvpq]).T, pvpq].real

此处，J11和dS_dVa是稀疏矩阵，pvpq是一系列索引，可以按任何增长顺序排列（即1,2,5,7,9 ）

查看文档here我推断了以下内容：

arma::Row<int> pvpq(calc->pqpv);

arma::sp_mat J11 = arma::real(dS_dVa.submat(pvpq, pvpq));

其中calc->pqpv的类型为std::vector<int>。

然而GCC编译器说：

engine.h:2436: error: no matching function for call to ‘arma::SpMat<std::complex<double> >::submat(arma::Row<int>&, arma::Row<int>&)’
         arma::sp_mat J11 = arma::real(dS_dVa.submat(pvpq, pvpq));
                                                               ^

我该如何解决这个问题？

它是否告诉我稀疏矩阵没有submat方法？

Answer 1

Armadillo仅支持连续表单的子矩阵视图。请参阅sp_mat doc。

中的注意事项部分

Answer 2

过了一会儿，我做了自己的功能。它使用内部CSC结构。

    /**
    * @brief sp_submatrix Function to extract columns and rows from a sparse matrix
    * @param A Sparse matrix pointer
    * @param rows vector of the rown indices to keep (must be sorted)
    * @param cols vector of the clumn indices to keep (must be sorted)
    * @return Sparse matrix of the indicated indices
    */
    arma::sp_mat sp_submatrix(arma::sp_mat *A, std::vector<std::size_t> *rows, std::vector<std::size_t> *cols) {

        std::size_t n_rows = rows->size();
        std::size_t n_cols = cols->size();

        bool found = false;
        std::size_t n = 0;
        std::size_t p = 0;
        std::size_t found_idx = 0;

        arma::vec new_val(A->n_nonzero);
        arma::uvec new_row_ind(A->n_nonzero);
        arma::uvec new_col_ptr(n_cols + 1);

        new_col_ptr(p) = 0;

        for (auto const& j: *cols) { // for every column in the cols vector

            for (std::size_t k = A->col_ptrs[j]; k < A->col_ptrs[j + 1]; k++) {  // k is the index of the "values" and "row_indices" that corresponds to the column j

                // search row_ind[k] in rows
                found = false;
                found_idx = 0;
                while (!found && found_idx < n_rows) {
                    if (A->row_indices[k] == rows->at(found_idx))
                        found = true;
                    found_idx++;
                }

                // store the values if the row was found in rows
                if (found) { // if the row index is in the designated rows...                   
                    new_val(n) = A->values[k]; // store the value
                    new_row_ind(n) = found_idx - 1;  // store the index where the original index was found inside "rows"
                    n++;
                }
            }

            p++;
            new_col_ptr(p) = n;
        }
        new_col_ptr(p) = n ;

        // reshape the vectors to the actual number of elements
        new_val.reshape(n, 1);
        new_row_ind.reshape(n, 1);

        return arma::sp_mat(new_row_ind, new_col_ptr, new_val, n_rows, n_cols);
    }

选择具有非连续索引的Armadillo子矩阵

2 个答案: