Python将字典转成scipy sparse matrix求方法，scipysparse,我的数据是从数据库里读出

我搞懂了正确转换方法，比较简单

1.先将dict转换成COO matrix，再转换成CSR matrix

    A[row[k], column[k] = data[k]]    # 创建 COO-matrix    coo = coo_matrix((data,(row,col)))    # Scipy 转换 COO 到 CSR format    return csr_matrix(coo)

代码

    from scipy.sparse import csr_matrix, coo_matrix    def convert(term_dict):        ''' Convert a dictionary with elements of form ('d1', 't1'): 12 to a CSR type         matrix.        The element ('d1', 't1'): 12 becomes entry (0, 0) = 12.        * Conversion from 1-indexed to 0-indexed.        * d is row        * t is column.        '''        # Create the appropriate format for the COO format.        data = []        row = []        col = []        for k, v in term_dict.items():            r = int(k[0][1:])            c = int(k[1][1:])            data.append(v)            row.append(r-1)            col.append(c-1)        # Create the COO-matrix        coo = coo_matrix((data,(row,col)))        # Let Scipy convert COO to CSR format and return        return csr_matrix(coo)    if __name__=='__main__':    doc_term_dict = { ('d1','t1'): 12,             \            ('d2','t3'): 10,             \            ('d3','t2'):  5              \            }       print(convert(doc_term_dict))