python multiprocessing pool.map用于多个参数

Python多处理库中,是否存在pool.map的变体,它支持多个参数?

text = "test"

def harvester(text, case):

X = case[0]

text+ str(X)

if __name__ == '__main__':

pool = multiprocessing.Pool(processes=6)

case = RAW_DATASET

pool.map(harvester(text,case),case, 1)

pool.close()

pool.join()

回答:

答案取决于版本和情况。JF Sebastian首先描述了最近版本的Python(从3.3开始)的最一般的答案。1它使用Pool.starmap方法,该方法接受一个参数元组序列。然后,它会自动将每个元组的参数解包,并将其传递给给定的函数:

import multiprocessing

from itertools import product

def merge_names(a, b):

return '{} & {}'.format(a, b)

if __name__ == '__main__':

names = ['Brown', 'Wilson', 'Bartlett', 'Rivera', 'Molloy', 'Opie']

with multiprocessing.Pool(processes=3) as pool:

results = pool.starmap(merge_names, product(names, repeat=2))

print(results)

# Output: ['Brown & Brown', 'Brown & Wilson', 'Brown & Bartlett', ...

对于早期版本的Python,您需要编写一个辅助函数来显式解压缩参数。如果要使用with,则还需要编写一个包装器以变为Pool上下文管理器。(感谢muon指出这一点。)

import multiprocessing

from itertools import product

from contextlib import contextmanager

def merge_names(a, b):

return '{} & {}'.format(a, b)

def merge_names_unpack(args):

return merge_names(*args)

@contextmanager

def poolcontext(*args, **kwargs):

pool = multiprocessing.Pool(*args, **kwargs)

yield pool

pool.terminate()

if __name__ == '__main__':

names = ['Brown', 'Wilson', 'Bartlett', 'Rivera', 'Molloy', 'Opie']

with poolcontext(processes=3) as pool:

results = pool.map(merge_names_unpack, product(names, repeat=2))

print(results)

# Output: ['Brown & Brown', 'Brown & Wilson', 'Brown & Bartlett', ...

在更简单的情况下,使用固定的第二个参数,您也可以使用partial,但仅在Python 2.7+中使用。

import multiprocessing

from functools import partial

from contextlib import contextmanager

@contextmanager

def poolcontext(*args, **kwargs):

pool = multiprocessing.Pool(*args, **kwargs)

yield pool

pool.terminate()

def merge_names(a, b):

return '{} & {}'.format(a, b)

if __name__ == '__main__':

names = ['Brown', 'Wilson', 'Bartlett', 'Rivera', 'Molloy', 'Opie']

with poolcontext(processes=3) as pool:

results = pool.map(partial(merge_names, b='Sons'), names)

print(results)

# Output: ['Brown & Sons', 'Wilson & Sons', 'Bartlett & Sons', ...

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