在J.F.Sebastian的回答中了解了itertools之后，我决定更进一步，编写一个关注并行化的parmap包，在Python 2.7和Python 3.2（以及更高版本）中提供可以接受任意数量位置参数的map和starmap函数。

安装

pip install parmap

如何并行化：

import parmap
# If you want to do:
y = [myfunction(x, argument1, argument2) for x in mylist]
# In parallel:
y = parmap.map(myfunction, mylist, argument1, argument2)

# If you want to do:
z = [myfunction(x, y, argument1, argument2) for (x,y) in mylist]
# In parallel:
z = parmap.starmap(myfunction, mylist, argument1, argument2)

# If you want to do:
listx = [1, 2, 3, 4, 5, 6]
listy = [2, 3, 4, 5, 6, 7]
param = 3.14
param2 = 42
listz = []
for (x, y) in zip(listx, listy):
        listz.append(myfunction(x, y, param1, param2))
# In parallel:
listz = parmap.starmap(myfunction, zip(listx, listy), param1, param2)

我已经将parmap上传到PyPI和GitHub存储库。

例如，问题的答案如下：

import parmap

def harvester(case, text):
    X = case[0]
    text+ str(X)

if __name__ == "__main__":
    case = RAW_DATASET  # assuming this is an iterable
    parmap.map(harvester, case, "test", chunksize=1)

答案取决于版本和情况。最近版本的Python（从3.3开始）的最一般的答案首先由J.F.Sebastian在下面描述。1它使用Pool.starmap方法，接受一系列参数元组。然后，它会自动将每个元组中的参数解包，并将它们传递给给定的函数：

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，您需要编写一个助手函数来显式地解包参数。如果要与一起使用，还需要编写一个包装器，将Pool转换为上下文管理器。（感谢穆恩指出了这一点。）

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，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', ...

1.这大部分都是由他的答案激发的，而他的答案很可能应该被接受。但由于这本书一直停留在顶端，似乎最好为未来读者改进它。

将Python 3.3+与pool.starmap（）一起使用：

from multiprocessing.dummy import Pool as ThreadPool 

def write(i, x):
    print(i, "---", x)

a = ["1","2","3"]
b = ["4","5","6"] 

pool = ThreadPool(2)
pool.starmap(write, zip(a,b)) 
pool.close() 
pool.join()

结果：

1 --- 4
2 --- 5
3 --- 6

如果您喜欢，还可以zip（）更多参数：zip（a，b，c，d，e）

如果希望将常量值作为参数传递：

import itertools

zip(itertools.repeat(constant), a)

如果您的函数应该返回以下内容：

results = pool.starmap(write, zip(a,b))

这将提供一个包含返回值的列表。

在官方文档中，它只支持一个可迭代的参数。在这种情况下，我喜欢使用apply_async。如果是你，我会：

from multiprocessing import Process, Pool, Manager

text = "test"
def harvester(text, case, q = None):
 X = case[0]
 res = text+ str(X)
 if q:
  q.put(res)
 return res


def block_until(q, results_queue, until_counter=0):
 i = 0
 while i < until_counter:
  results_queue.put(q.get())
  i+=1

if __name__ == '__main__':
 pool = multiprocessing.Pool(processes=6)
 case = RAW_DATASET
 m = Manager()
 q = m.Queue()
 results_queue = m.Queue() # when it completes results will reside in this queue
 blocking_process = Process(block_until, (q, results_queue, len(case)))
 blocking_process.start()
 for c in case:
  try:
   res = pool.apply_async(harvester, (text, case, q = None))
   res.get(timeout=0.1)
  except:
   pass
 blocking_process.join()

我认为以下内容会更好：

def multi_run_wrapper(args):
   return add(*args)

def add(x,y):
    return x+y

if __name__ == "__main__":
    from multiprocessing import Pool
    pool = Pool(4)
    results = pool.map(multi_run_wrapper,[(1,2),(2,3),(3,4)])
    print results

输出

[3, 5, 7]

对于Python 2，可以使用此技巧

def fun(a, b):
    return a + b

pool = multiprocessing.Pool(processes=6)
b = 233
pool.map(lambda x:fun(x, b), range(1000))