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Home/ Questions/Q 7902585
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Editorial Team
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Editorial Team
Asked: June 3, 20262026-06-03T09:32:20+00:00 2026-06-03T09:32:20+00:00

I’ve been using the following function to make a more readable (supposedly) format for

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I’ve been using the following function to make a “more readable” (supposedly) format for fetching data from Oracle. Here is the function:

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

I would then use the function like this:

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.

While this seems to perform fairly well under varying levels of stress, I’m wondering if there’s a more efficient, or “pythonic” way of doing this.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-03T09:32:21+00:00Added an answer on June 3, 2026 at 9:32 am

    There are other improvements to make, but this really jumped out at me:

        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)]
    

    In addition to being inefficient, using index in situations like this is bug-prone, at least in situations where the same item may occur twice in a list. Use enumerate instead:

        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]
    

    But that’s small potatoes, really. Here’s a much more compact version of this function:

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

    Let me know if that works.

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