在下面的示例代码中,我想获取函数worker的返回值。我该怎么做呢?这个值存储在哪里?
示例代码:
import multiprocessing
def worker(procnum):
'''worker function'''
print str(procnum) + ' represent!'
return procnum
if __name__ == '__main__':
jobs = []
for i in range(5):
p = multiprocessing.Process(target=worker, args=(i,))
jobs.append(p)
p.start()
for proc in jobs:
proc.join()
print jobs
输出:
0 represent!
1 represent!
2 represent!
3 represent!
4 represent!
[<Process(Process-1, stopped)>, <Process(Process-2, stopped)>, <Process(Process-3, stopped)>, <Process(Process-4, stopped)>, <Process(Process-5, stopped)>]
我似乎无法在存储在作业中的对象中找到相关属性。
这个例子展示了如何使用multiprocessing的列表。管道实例从任意数量的进程中返回字符串:
import multiprocessing
def worker(procnum, send_end):
'''worker function'''
result = str(procnum) + ' represent!'
print result
send_end.send(result)
def main():
jobs = []
pipe_list = []
for i in range(5):
recv_end, send_end = multiprocessing.Pipe(False)
p = multiprocessing.Process(target=worker, args=(i, send_end))
jobs.append(p)
pipe_list.append(recv_end)
p.start()
for proc in jobs:
proc.join()
result_list = [x.recv() for x in pipe_list]
print result_list
if __name__ == '__main__':
main()
输出:
0 represent!
1 represent!
2 represent!
3 represent!
4 represent!
['0 represent!', '1 represent!', '2 represent!', '3 represent!', '4 represent!']
这种解决方案使用的资源比多进程少。使用
一个管道
至少一个锁
一个缓冲区
一个线程
或者多处理。SimpleQueue使用
一个管道
至少一个锁
查看这些类型的源代码是非常有指导意义的。
我修改了vartec的答案,因为我需要从函数中获得错误代码。(由于vertec ! !这是一个很棒的技巧)
这也可以通过经理来实现。列表,但我认为最好是在字典中,并在其中存储一个列表。这样,我们就可以保留函数和结果,因为我们不能确定列表将被填充的顺序。
from multiprocessing import Process
import time
import datetime
import multiprocessing
def func1(fn, m_list):
print 'func1: starting'
time.sleep(1)
m_list[fn] = "this is the first function"
print 'func1: finishing'
# return "func1" # no need for return since Multiprocess doesnt return it =(
def func2(fn, m_list):
print 'func2: starting'
time.sleep(3)
m_list[fn] = "this is function 2"
print 'func2: finishing'
# return "func2"
def func3(fn, m_list):
print 'func3: starting'
time.sleep(9)
# if fail wont join the rest because it never populate the dict
# or do a try/except to get something in return.
raise ValueError("failed here")
# if we want to get the error in the manager dict we can catch the error
try:
raise ValueError("failed here")
m_list[fn] = "this is third"
except:
m_list[fn] = "this is third and it fail horrible"
# print 'func3: finishing'
# return "func3"
def runInParallel(*fns): # * is to accept any input in list
start_time = datetime.datetime.now()
proc = []
manager = multiprocessing.Manager()
m_list = manager.dict()
for fn in fns:
# print fn
# print dir(fn)
p = Process(target=fn, name=fn.func_name, args=(fn, m_list))
p.start()
proc.append(p)
for p in proc:
p.join() # 5 is the time out
print datetime.datetime.now() - start_time
return m_list, proc
if __name__ == '__main__':
manager, proc = runInParallel(func1, func2, func3)
# print dir(proc[0])
# print proc[0]._name
# print proc[0].name
# print proc[0].exitcode
# here you can check what did fail
for i in proc:
print i.name, i.exitcode # name was set up in the Process line 53
# here will only show the function that worked and where able to populate the
# manager dict
for i, j in manager.items():
print dir(i) # things you can do to the function
print i, j
我认为@sega_sai建议的方法更好。但它确实需要一个代码示例,所以如下:
import multiprocessing
from os import getpid
def worker(procnum):
print('I am number %d in process %d' % (procnum, getpid()))
return getpid()
if __name__ == '__main__':
pool = multiprocessing.Pool(processes = 3)
print(pool.map(worker, range(5)))
它将打印返回值:
I am number 0 in process 19139
I am number 1 in process 19138
I am number 2 in process 19140
I am number 3 in process 19139
I am number 4 in process 19140
[19139, 19138, 19140, 19139, 19140]
如果你熟悉map (Python 2内置的),这应该不是太有挑战性。否则,请查看sega_Sai的链接。
注意,只需要很少的代码。(还要注意如何重用流程)。
出于某种原因,我找不到一个通用的例子,如何做到这一点与队列在任何地方(甚至Python的文档示例不会产生多个进程),所以这是我得到的工作后,10次尝试:
from multiprocessing import Process, Queue
def add_helper(queue, arg1, arg2): # the func called in child processes
ret = arg1 + arg2
queue.put(ret)
def multi_add(): # spawns child processes
q = Queue()
processes = []
rets = []
for _ in range(0, 100):
p = Process(target=add_helper, args=(q, 1, 2))
processes.append(p)
p.start()
for p in processes:
ret = q.get() # will block
rets.append(ret)
for p in processes:
p.join()
return rets
Queue是一个阻塞的、线程安全的队列,您可以使用它来存储来自子进程的返回值。因此,您必须将队列传递给每个进程。这里不太明显的一点是,您必须在加入进程之前从队列中获取(),否则队列将被填满并阻塞所有内容。
面向对象的更新(在Python 3.4中测试):
from multiprocessing import Process, Queue
class Multiprocessor():
def __init__(self):
self.processes = []
self.queue = Queue()
@staticmethod
def _wrapper(func, queue, args, kwargs):
ret = func(*args, **kwargs)
queue.put(ret)
def run(self, func, *args, **kwargs):
args2 = [func, self.queue, args, kwargs]
p = Process(target=self._wrapper, args=args2)
self.processes.append(p)
p.start()
def wait(self):
rets = []
for p in self.processes:
ret = self.queue.get()
rets.append(ret)
for p in self.processes:
p.join()
return rets
# tester
if __name__ == "__main__":
mp = Multiprocessor()
num_proc = 64
for _ in range(num_proc): # queue up multiple tasks running `sum`
mp.run(sum, [1, 2, 3, 4, 5])
ret = mp.wait() # get all results
print(ret)
assert len(ret) == num_proc and all(r == 15 for r in ret)