【发布时间】:2017-09-16 12:34:08
【问题描述】:
可以在as a Gist on GitHub找到代码的清理版本,包括the solution to the problem(感谢@JohanL!)。
以下代码片段 (CPython 3.[4,5,6]) 说明了我的意图(以及我的问题):
from functools import partial
import multiprocessing
from pprint import pprint as pp
NUM_CORES = multiprocessing.cpu_count()
class some_class:
some_dict = {'some_key': None, 'some_other_key': None}
def some_routine(self):
self.some_dict.update({'some_key': 'some_value'})
def some_other_routine(self):
self.some_dict.update({'some_other_key': 77})
def run_routines_on_objects_in_parallel_and_return(in_object_list, routine_list):
func_handle = partial(__run_routines_on_object_and_return__, routine_list)
with multiprocessing.Pool(processes = NUM_CORES) as p:
out_object_list = list(p.imap_unordered(
func_handle,
(in_object for in_object in in_object_list)
))
return out_object_list
def __run_routines_on_object_and_return__(routine_list, in_object):
for routine_name in routine_list:
getattr(in_object, routine_name)()
return in_object
object_list = [some_class() for item in range(20)]
pp([item.some_dict for item in object_list])
new_object_list = run_routines_on_objects_in_parallel_and_return(
object_list,
['some_routine', 'some_other_routine']
)
pp([item.some_dict for item in new_object_list])
verification_object_list = [
__run_routines_on_object_and_return__(
['some_routine', 'some_other_routine'],
item
) for item in object_list
]
pp([item.some_dict for item in verification_object_list])
我正在处理some_class 类型的对象列表。 some_class 有一个属性,一个字典,名为some_dict 和一些方法,可以修改字典(some_routine 和some_other_routine)。有时,我想对列表中的所有对象调用一系列方法。因为这是计算密集型的,所以我打算将对象分布在多个 CPU 内核上(使用 multiprocessing.Pool 和 imap_unordered - 列表顺序无关紧要)。
例程__run_routines_on_object_and_return__ 负责调用单个对象的方法列表。据我所知,这工作得很好。我使用functools.partial 来稍微简化代码结构——因此,多处理池必须仅将对象列表作为输入参数来处理。
问题是......它不起作用。 imap_unordered 返回的列表中包含的对象与我输入的对象相同。对象中的字典看起来就像以前一样。我已经使用类似的机制直接处理字典列表而不会出现故障,所以我怀疑修改恰好是字典的对象属性有问题。
在我的示例中,verification_object_list 包含正确的结果(尽管它是在单个进程/线程中生成的)。 new_object_list 与 object_list 相同,但不应如此。
我做错了什么?
编辑
我找到了以下question,它有一个实际工作且适用的answer。我按照我在每个对象上调用方法列表的想法对其进行了一些修改,它可以工作:
import random
from multiprocessing import Pool, Manager
class Tester(object):
def __init__(self, num=0.0, name='none'):
self.num = num
self.name = name
def modify_me(self):
self.num += random.normalvariate(mu=0, sigma=1)
self.name = 'pla' + str(int(self.num * 100))
def __repr__(self):
return '%s(%r, %r)' % (self.__class__.__name__, self.num, self.name)
def init(L):
global tests
tests = L
def modify(i_t_nn):
i, t, nn = i_t_nn
for method_name in nn:
getattr(t, method_name)()
tests[i] = t # copy back
return i
def main():
num_processes = num = 10 #note: num_processes and num may differ
manager = Manager()
tests = manager.list([Tester(num=i) for i in range(num)])
print(tests[:2])
args = ((i, t, ['modify_me']) for i, t in enumerate(tests))
pool = Pool(processes=num_processes, initializer=init, initargs=(tests,))
for i in pool.imap_unordered(modify, args):
print("done %d" % i)
pool.close()
pool.join()
print(tests[:2])
if __name__ == '__main__':
main()
现在,我更进一步,将我原来的some_class 引入到游戏中,其中包含描述的字典属性some_dict。它不起作用:
import random
from multiprocessing import Pool, Manager
from pprint import pformat as pf
class some_class:
some_dict = {'some_key': None, 'some_other_key': None}
def some_routine(self):
self.some_dict.update({'some_key': 'some_value'})
def some_other_routine(self):
self.some_dict.update({'some_other_key': 77})
def __repr__(self):
return pf(self.some_dict)
def init(L):
global tests
tests = L
def modify(i_t_nn):
i, t, nn = i_t_nn
for method_name in nn:
getattr(t, method_name)()
tests[i] = t # copy back
return i
def main():
num_processes = num = 10 #note: num_processes and num may differ
manager = Manager()
tests = manager.list([some_class() for i in range(num)])
print(tests[:2])
args = ((i, t, ['some_routine', 'some_other_routine']) for i, t in enumerate(tests))
pool = Pool(processes=num_processes, initializer=init, initargs=(tests,))
for i in pool.imap_unordered(modify, args):
print("done %d" % i)
pool.close()
pool.join()
print(tests[:2])
if __name__ == '__main__':
main()
工作和不工作之间的差异真的很小,但我还是不明白:
diff --git a/test.py b/test.py
index b12eb56..0aa6def 100644
--- a/test.py
+++ b/test.py
@@ -1,15 +1,15 @@
import random
from multiprocessing import Pool, Manager
+from pprint import pformat as pf
-class Tester(object):
- def __init__(self, num=0.0, name='none'):
- self.num = num
- self.name = name
- def modify_me(self):
- self.num += random.normalvariate(mu=0, sigma=1)
- self.name = 'pla' + str(int(self.num * 100))
+class some_class:
+ some_dict = {'some_key': None, 'some_other_key': None}
+ def some_routine(self):
+ self.some_dict.update({'some_key': 'some_value'})
+ def some_other_routine(self):
+ self.some_dict.update({'some_other_key': 77})
def __repr__(self):
- return '%s(%r, %r)' % (self.__class__.__name__, self.num, self.name)
+ return pf(self.some_dict)
def init(L):
global tests
@@ -25,10 +25,10 @@ def modify(i_t_nn):
def main():
num_processes = num = 10 #note: num_processes and num may differ
manager = Manager()
- tests = manager.list([Tester(num=i) for i in range(num)])
+ tests = manager.list([some_class() for i in range(num)])
print(tests[:2])
- args = ((i, t, ['modify_me']) for i, t in enumerate(tests))
+ args = ((i, t, ['some_routine', 'some_other_routine']) for i, t in enumerate(tests))
这里发生了什么?
【问题讨论】:
标签: python python-3.x dictionary multiprocessing python-multiprocessing