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)

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]

这可能是另一种选择。技巧在于包装器函数，它返回传递给pool.map的另一个函数。下面的代码读取一个输入数组，对于其中的每个（唯一）元素，返回该元素在数组中出现的次数（即计数）。例如，如果输入是

np.eye(3) = [ [1. 0. 0.]
              [0. 1. 0.]
              [0. 0. 1.]]

然后零出现6次，一出现3次

import numpy as np
from multiprocessing.dummy import Pool as ThreadPool
from multiprocessing import cpu_count


def extract_counts(label_array):
    labels = np.unique(label_array)
    out = extract_counts_helper([label_array], labels)
    return out

def extract_counts_helper(args, labels):
    n = max(1, cpu_count() - 1)
    pool = ThreadPool(n)
    results = {}
    pool.map(wrapper(args, results), labels)
    pool.close()
    pool.join()
    return results

def wrapper(argsin, results):
    def inner_fun(label):
        label_array = argsin[0]
        counts = get_label_counts(label_array, label)
        results[label] = counts
    return inner_fun

def get_label_counts(label_array, label):
    return sum(label_array.flatten() == label)

if __name__ == "__main__":
    img = np.ones([2,2])
    out = extract_counts(img)
    print('input array: \n', img)
    print('label counts: ', out)
    print("========")
           
    img = np.eye(3)
    out = extract_counts(img)
    print('input array: \n', img)
    print('label counts: ', out)
    print("========")
    
    img = np.random.randint(5, size=(3, 3))
    out = extract_counts(img)
    print('input array: \n', img)
    print('label counts: ', out)
    print("========")

你应该得到：

input array: 
 [[1. 1.]
 [1. 1.]]
label counts:  {1.0: 4}
========
input array: 
 [[1. 0. 0.]
 [0. 1. 0.]
 [0. 0. 1.]]
label counts:  {0.0: 6, 1.0: 3}
========
input array: 
 [[4 4 0]
 [2 4 3]
 [2 3 1]]
label counts:  {0: 1, 1: 1, 2: 2, 3: 2, 4: 3}
========

从Python 3.4.4中，您可以使用multiprocessing.get_context（）获取上下文对象，以使用多个启动方法：

import multiprocessing as mp

def foo(q, h, w):
    q.put(h + ' ' + w)
    print(h + ' ' + w)

if __name__ == '__main__':
    ctx = mp.get_context('spawn')
    q = ctx.Queue()
    p = ctx.Process(target=foo, args=(q,'hello', 'world'))
    p.start()
    print(q.get())
    p.join()

或者你只是简单地替换

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

具有：

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

pool.map是否有支持多个参数的变体？

Python 3.3包含pool.starmap（）方法：

#!/usr/bin/env python3
from functools import partial
from itertools import repeat
from multiprocessing import Pool, freeze_support

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

def main():
    a_args = [1,2,3]
    second_arg = 1
    with Pool() as pool:
        L = pool.starmap(func, [(1, 1), (2, 1), (3, 1)])
        M = pool.starmap(func, zip(a_args, repeat(second_arg)))
        N = pool.map(partial(func, b=second_arg), a_args)
        assert L == M == N

if __name__=="__main__":
    freeze_support()
    main()

对于旧版本：

#!/usr/bin/env python2
import itertools
from multiprocessing import Pool, freeze_support

def func(a, b):
    print a, b

def func_star(a_b):
    """Convert `f([1,2])` to `f(1,2)` call."""
    return func(*a_b)

def main():
    pool = Pool()
    a_args = [1,2,3]
    second_arg = 1
    pool.map(func_star, itertools.izip(a_args, itertools.repeat(second_arg)))

if __name__=="__main__":
    freeze_support()
    main()

输出

1 1
2 1
3 1

注意这里是如何使用itertools.izip（）和itertools.crepeat（）的。

由于@unsubu提到的错误，您不能在Python 2.6上使用functools.partial（）或类似功能，因此应该显式定义简单包装函数func_tar（）。另请参阅uptimebox建议的解决方法。

有一个叫做pathos的多处理分支（注意：使用GitHub上的版本），它不需要starmap——map函数镜像Python map的API，因此map可以接受多个参数。

使用pathos，您通常也可以在解释器中执行多处理，而不是陷入__main__块。Pathos将在经过一些轻微的更新后发布——主要是转换为Python3.x。

  Python 2.7.5 (default, Sep 30 2013, 20:15:49)
  [GCC 4.2.1 (Apple Inc. build 5566)] on darwin
  Type "help", "copyright", "credits" or "license" for more information.
  >>> def func(a,b):
  ...     print a,b
  ...
  >>>
  >>> from pathos.multiprocessing import ProcessingPool
  >>> pool = ProcessingPool(nodes=4)
  >>> pool.map(func, [1,2,3], [1,1,1])
  1 1
  2 1
  3 1
  [None, None, None]
  >>>
  >>> # also can pickle stuff like lambdas
  >>> result = pool.map(lambda x: x**2, range(10))
  >>> result
  [0, 1, 4, 9, 16, 25, 36, 49, 64, 81]
  >>>
  >>> # also does asynchronous map
  >>> result = pool.amap(pow, [1,2,3], [4,5,6])
  >>> result.get()
  [1, 32, 729]
  >>>
  >>> # or can return a map iterator
  >>> result = pool.imap(pow, [1,2,3], [4,5,6])
  >>> result
  <processing.pool.IMapIterator object at 0x110c2ffd0>
  >>> list(result)
  [1, 32, 729]

pathos有几种方法可以让你得到星图的精确行为。

>>> def add(*x):
...   return sum(x)
...
>>> x = [[1,2,3],[4,5,6]]
>>> import pathos
>>> import numpy as np
>>> # use ProcessPool's map and transposing the inputs
>>> pp = pathos.pools.ProcessPool()
>>> pp.map(add, *np.array(x).T)
[6, 15]
>>> # use ProcessPool's map and a lambda to apply the star
>>> pp.map(lambda x: add(*x), x)
[6, 15]
>>> # use a _ProcessPool, which has starmap
>>> _pp = pathos.pools._ProcessPool()
>>> _pp.starmap(add, x)
[6, 15]
>>>