python - 使用 cx_Oracle 制作字典列表

Question

我一直在使用以下函数来制作一种“更易读”(据说)格式来从 Oracle 获取数据。这是函数:

def rows_to_dict_list(cursor):
    """ 
    Create a list, each item contains a dictionary outlined like so:
    { "col1_name" : col1_data }
    Each item in the list is technically one row of data with named columns,
    represented as a dictionary object
    For example:
    list = [
        {"col1":1234567, "col2":1234, "col3":123456, "col4":BLAH},
        {"col1":7654321, "col2":1234, "col3":123456, "col4":BLAH}
    ]
    """

    # Get all the column names of the query.
    # Each column name corresponds to the row index
    # 
    # cursor.description returns a list of tuples, 
    # with the 0th item in the tuple being the actual column name.
    # everything after i[0] is just misc Oracle info (e.g. datatype, size)
    columns = [i[0] for i in cursor.description]

    new_list = []
    for row in cursor:
        row_dict = dict()
        for col in columns:
            # Create a new dictionary with field names as the key, 
            # row data as the value.
            #
            # Then add this dictionary to the new_list
            row_dict[col] = row[columns.index(col)]

        new_list.append(row_dict)
    return new_list

然后我会像这样使用函数:

sql = "Some kind of SQL statement"
curs.execute(sql)
data = rows_to_dict_list(curs)
#
for row in data:
    item1 = row["col1"]
    item2 = row["col2"]
    # Do stuff with item1, item2, etc...
    # You don't necessarily have to assign them to variables,
    # but you get the idea.

虽然这似乎在不同程度的压力下表现得相当好，但我想知道是否有更有效或“pythonic”的方式来做到这一点。

Answer 1

还有其他改进需要做，但这真的让我大吃一惊:

    for col in columns:
        # Create a new dictionary with field names as the key, 
        # row data as the value.
        #
        # Then add this dictionary to the new_list
        row_dict[col] = row[columns.index(col)]

除了效率低下之外，在这种情况下使用 index 还容易出错，至少在同一项目可能在列表中出现两次的情况下是这样。使用 enumerate 代替:

    for i, col in enumerate(columns):
        # Create a new dictionary with field names as the key, 
        # row data as the value.
        #
        # Then add this dictionary to the new_list
        row_dict[col] = row[i]

但那只是小土 bean ，真的。这是这个函数的一个更紧凑的版本:

def rows_to_dict_list(cursor):
    columns = [i[0] for i in cursor.description]
    return [dict(zip(columns, row)) for row in cursor]

让我知道这是否有效。