[英]Working around lambda limitation in python
为了更熟悉Python,我正在尝试编写许多与sql相关的函数。 其中之一应该执行查询并将结果转换为命名元组。 相关代码段。
from collections import namedtuple, Iterable
import psycopg2.extras
import psycopg2
def tuple_to_named_tuple(tuple, cursor_description) -> list:
rdef = namedtuple('row', ' '.join([x[0] for x in cursor_description]))
return rdef._make(tuple);
def run_query(connection_string, callback_function):
with psycopg2.connect(connection_string) as conn:
with conn.cursor() as cur:
return callback_function(cur);
该代码有效,并且我可以使用嵌套函数传递回调:
def test_param(value):
def nested_function(cur):
cur.execute("SELECT val from table1 where id =%", (value,));
return tuple_to_named_tuple(cur.fetchone(), cur.description);
val = run_query(CONNECTION_STRING, nested_function);
print(val);
或具有多个lambda,例如
def test_param(value):
cb = lambda cur: (cur.execute("SELECT val from table1 where id =%", (value,)), cur);
cb1 = lambda ignore, cur: tuple_to_named_tuple(cur.fetchone(), cur.description);
cb3 = lambda c1: cb1(*cb(c1));
val = run_query(CONNECTION_STRING, cb3);
print(val);
但是,我不知道如何将后者包装成一系列包装好的lambda。 我想要类似的东西:
# not working code
lambda_callback = lambda cur :
lambda ignore, cur :
*(cur.execute("SELECT val from table1 where id =%", (value,))[0],
*(cur.execute("SELECT val from table1 where id =%", (value,))[1];
val = run_query(CONNECTION_STRING, lambdas_callback );
我想知道是否有可能
谢谢。
如果您正在编写的函数足够重要以具有多个语句,那么将其定义为适当的函数也足够重要。
您可以这样定义它:
def my_important_multistatement_lambda_func(*args, **kwargs):
# TODO
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