我一直在使用以下函数来制作一个“更具可读性”(据称)的格式,用于从Oracle获取数据。这是功能:
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
然后我会使用这样的函数:
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相当好,但我想知道是否有更高效或“pythonic”的方式来做到这一点。
答案 0 :(得分:25)
还有其他一些改进,但这真的让我跳了起来:
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
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。请改用enumerate
:
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]
但那是小土豆,真的。这是这个函数的更紧凑版本:
def rows_to_dict_list(cursor):
columns = [i[0] for i in cursor.description]
return [dict(zip(columns, row)) for row in cursor]
如果有效,请告诉我。
答案 1 :(得分:10)
为了避免在预先列出列表中的所有内容的内存使用的干净方法,您可以将光标包装在生成器函数中:
def rows_as_dicts(cursor):
""" returns cx_Oracle rows as dicts """
colnames = [i[0] for i in cursor.description]
for row in cursor:
yield dict(zip(colnames, row))
然后使用如下 - 迭代时将光标中的行转换为dicts:
for row in rows_as_dicts(cursor):
item1 = row["col1"]
item2 = row["col2"]
答案 2 :(得分:4)
您不应该将dict用于大结果集,因为内存使用量会很大。我经常使用cx_Oracle而没有一个好的字典光标让我感到困扰,为它编写了一个模块。我还必须将Python连接到许多不同的数据库,因此我以一种可以与任何DB API 2连接器一起使用的方式进行连接。
这取决于PyPi DBMS - DataBases Made Simpler
>>> import dbms
>>> db = dbms.OraConnect('myUser', 'myPass', 'myInstance')
>>> cur = db.cursor()
>>> cur.execute('SELECT * FROM people WHERE id = :id', {'id': 1123})
>>> row = cur.fetchone()
>>> row['last_name']
Bailey
>>> row.last_name
Bailey
>>> row[3]
Bailey
>>> row[0:4]
[1123, 'Scott', 'R', 'Bailey']
答案 3 :(得分:0)
假设游标“Cursor”已经定义并且raring to go:
byCol = {cl:i for i,(cl,type, a, b, c,d,e) in enumerate(Cursor.description)}
然后你可以去:
for row in Cursor:
column_of_interest = row[byCol["COLUMN_NAME_OF_INTEREST"]]
不像系统本身处理它那样干净和光滑,但并不可怕。
答案 4 :(得分:0)
创建一个词典
cols=dict()
for col, desc in enumerate(cur.description):
cols[desc[0]] = col
访问:
for result in cur
print (result[cols['COL_NAME']])
答案 5 :(得分:0)
我有一个更好的:
import cx_Oracle
def makedict(cursor):
"""Convert cx_oracle query result to be a dictionary
"""
cols = [d[0] for d in cursor.description]
def createrow(*args):
return dict(zip(cols, args))
return createrow
db = cx_Oracle.connect('user', 'pw', 'host')
cursor = db.cursor()
rs = cursor.execute('SELECT * FROM Tablename')
cursor.rowfactory = makedict(cursor)