Python多处理映射函数错误

时间:2013-07-15 15:10:16

标签: python parallel-processing functional-programming multiprocessing functools

我有一个简单的多处理示例,我正在尝试创建。普通的map()函数版本可以工作,但是当改为Pool.map时,我收到一个奇怪的错误:

from multiprocessing import Pool
from functools import partial
x = [1,2,3]
y = 10
f = lambda x,y: x**2+y

# ordinary map works:
map(partial(f,y=y),x)
# [11, 14, 19]

# multiprocessing map does not
p = Pool(4)
p.map(partial(f, y=y), x)
Exception in thread Thread-2:
Traceback (most recent call last):
  File "/usr/lib/python2.7/threading.py", line 551, in __bootstrap_inner
    self.run()
  File "/usr/lib/python2.7/threading.py", line 504, in run
    self.__target(*self.__args, **self.__kwargs)
  File "/usr/lib/python2.7/multiprocessing/pool.py", line 319, in _handle_tasks
    put(task)
PicklingError: Can't pickle <type 'function'>: attribute lookup __builtin__.function failed

酸洗错误?究竟是什么?

1 个答案:

答案 0 :(得分:6)

Pool.map的参数必须是可选的。 Module-level functions are picklable,但partial(f, y=y)未在模块级别定义,因此无法选择。

有一个简单的解决方法:

def g(x, y=y):
    return f(x, y)

p.map(g, x)

使用functools.partial used to be unpickable创建的功能。 但是,使用Python2.7或更高版本,您还可以使用functools.partial定义g(在模块级别):

import multiprocessing as mp
import functools

def f(x, y):
    return x**2 + y

x = [1,2,3]
y = 10

g = functools.partial(f, y=y)

if __name__ == '__main__':
    p = mp.Pool()
    print(p.map(g, x))

收益[11, 14, 19]。但请注意,要获得此结果f必须使用def而不是lambda进行定义。我认为这是因为pickle relies on "fully qualified" name references查找函数对象值。