我有两个用于查询数据库的函数。假设有两个独立的查询，如何并行运行这些查询以查询同一数据库，并在继续执行其余代码之前，等待两个结果返回？
def query1(param1, param2):
result = None
logging.info("Connecting to database...")
try:
conn = connect(host=host, port=port, database=db)
curs = conn.cursor()
curs.execute(query)
result = curs
curs.close()
conn.close()
except Exception as e:
logging.error("Unable to access database %s" % str(e))
return result
def query2(param1, param2):
result = None
logging.info("Connecting to database...")
try:
conn = connect(host=host, port=port, database=db)
curs = conn.cursor()
curs.execute(query)
result = curs
curs.close()
conn.close()
except Exception as e:
logging.error("Unable to access database %s" % str(e))
return result
据我所知,两个功能可以并行使用运行multiprocessing或threading模块,例如使在同一时间运行两个功能和Python的多重并行进程.
但以上示例仅使用打印功能.是否有可能在python中运行并行返回列表的函数,如果是,如何？
我尝试过使用线程:
from threading import Thread
def func1(x):
return [i*i for i in x]
def func2(x):
return [i*i*i for i in x]
nums = [1,2,3,4,5]
p1 = Thread(target = func1(nums)).start()
p2 = Thread(target = func2(nums)).start()
print p1
print p2
但我得到了以下错误:
Exception in thread Thread-1:
Traceback (most recent call last):
File "/usr/lib/python2.7/threading.py", line 808, in __bootstrap_inner
self.run()
File "/usr/lib/python2.7/threading.py", line 761, in run
self.__target(*self.__args, **self.__kwargs)
TypeError: 'list' object is not callable …我想创建一个函数的多线程版本。我发现t.start()返回None，所以我必须使用队列。我搜索了文档，但我不明白如何在我的示例中使用它。
这是函数：
def derivative(lst, var): # Example of lst = [1 + [3 * x]]
if len(lst) == 1:
return derive_solver(lst[0], var)
if lst[1] == '+':
return [derivative(lst[0], var), '+', derivative(lst[2], var)]
if lst[1] == '*':
return [[derivative(lst[0], var), '*', lst[2]], '+', [lst[0], '*', derivative(lst[2], var)]]
这是我尝试多线程该功能：
def derivative(lst, var): # Example of lst = [1 + [3 * x]]
if len(lst) == 1:
return derive_solver(lst[0], var)
if lst[1] == '+':
t1 = threading.Thread(target = derivative, …通常我一直在浏览这个问答网站,使用多线程和处理的答案告诉我使用这样的格式:
(target=foo, args=(bar, baz))
我的问题是,什么target意思,有人可以解释它是如何使用的？
我无法在文档或其他地方找到好的解释.
我有以下代码:
def upload_to_s3(filepath, unique_id):
# do something
print s3_url # <-- Confirming that this `s3_url` variable is not None
return s3_url
threads = []
for num, list_of_paths in enumerate(chunked_paths_as_list):
for filepath in list_of_paths:
t = threading.Thread(target=upload_to_s3, args=(filepath, self.unique_id))
t.start()
threads.append(t)
results = map(lambda t: t.join(), threads)
print results
不幸的是,这是None为每个项目返回:
[None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None]
>>>>> TIME: 13.9884989262
我需要做些什么才能获得上述return声明 …