Python将字典转成scipy sparse matrix求方法

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))