你可以使用一个队列
将失败反馈到初始化进程
中的循环中,以便将其传递回池
中:
import multiprocessing as mp
import random
def f(x):
if random.getrandbits(1):
f.q.put(x)
return None
return x*x
def f_init(q):
f.q = q
def main(pending):
total_items = len(pending)
successful = []
failure_tracker = []
q = mp.Queue()
p = mp.Pool(None, f_init, [q])
results = p.imap(f, pending)
retry_results = []
while len(successful) < total_items:
successful.extend([r for r in results if not r is None])
successful.extend([r for r in retry_results if not r is None])
failed_items = []
while not q.empty():
failed_items.append(q.get())
if failed_items:
failure_tracker.append(failed_items)
retry_results = p.imap(f, failed_items);
p.close()
p.join()
print "Results: %s" % successful
print "Failures: %s" % failure_tracker
if __name__ == '__main__':
main(range(1, 10))
输出结果如下:
Results: [1, 4, 36, 49, 25, 81, 16, 64, 9]
Failures: [[3, 4, 5, 8, 9], [3, 8, 4], [8, 3], []]
一个Pool
不能在多个进程之间共享。因此采用这种基于Queue
的方法。如果你尝试将一个池作为参数传递给池进程,你会得到这个错误:
NotImplementedError: pool objects cannot be passed between processes or pickled
为了避免同步开销,你可以尝试在函数f
内进行几次立即重试。这主要取决于你的函数需要等待多久才重试,以及如果立即重试成功的可能性有多大。
旧答案:为了完整起见,这是我的旧答案,它不像直接重新提交到池中那样优化,但根据用例可能仍然相关,因为它提供了一种处理/限制n
-级重试的自然方法:
你可以使用一个Queue
来聚合失败并在每次运行结束时重新提交,跨多个运行:
import multiprocessing as mp
import random
def f(x):
if random.getrandbits(1):
f.q.put(x)
return None
return x*x
def f_init(q):
f.q = q
def main(pending):
run_number = 1
while pending:
jobs = pending
pending = []
q = mp.Queue()
p = mp.Pool(None, f_init, [q])
results = p.imap(f, jobs)
p.close()
p.join()
failed_items = []
while not q.empty():
failed_items.append(q.get())
successful = [r for r in results if not r is None]
print "(%d) Succeeded: %s" % (run_number, successful)
print "(%d) Failed: %s" % (run_number, failed_items)
print
pending = failed_items
run_number += 1
if __name__ == '__main__':
main(range(1, 10))
输出结果如下:
(1) Succeeded: [9, 16, 36, 81]
(1) Failed: [2, 1, 5, 7, 8]
(2) Succeeded: [64]
(2) Failed: [2, 1, 5, 7]
(3) Succeeded: [1, 25]
(3) Failed: [2, 7]
(4) Succeeded: [49]
(4) Failed: [2]
(5) Succeeded: [4]
(5) Failed: []
return f(x)
而不是引发ValueError
?只是猜测... - Paulo Freitas