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}
========

将所有参数存储为元组数组。

该示例表示，通常调用函数为：

def mainImage(fragCoord: vec2, iResolution: vec3, iTime: float) -> vec3:

而是传递一个元组并解压缩参数：

def mainImage(package_iter) -> vec3:
    fragCoord = package_iter[0]
    iResolution = package_iter[1]
    iTime = package_iter[2]

预先使用循环构建元组：

package_iter = []
iResolution = vec3(nx, ny, 1)
for j in range((ny-1), -1, -1):
    for i in range(0, nx, 1):
        fragCoord: vec2 = vec2(i, j)
        time_elapsed_seconds = 10
        package_iter.append((fragCoord, iResolution, time_elapsed_seconds))

然后通过传递元组数组来执行所有using map：

array_rgb_values = []

with concurrent.futures.ProcessPoolExecutor() as executor:
    for val in executor.map(mainImage, package_iter):
        fragColor = val
        ir = clip(int(255* fragColor.r), 0, 255)
        ig = clip(int(255* fragColor.g), 0, 255)
        ib = clip(int(255* fragColor.b), 0, 255)

        array_rgb_values.append((ir, ig, ib))

我知道Python有*和**用于开箱，但我还没有尝试过。

使用高级库并发期货也比使用低级多处理库更好。

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]
>>>

另一种方法是将列表列表传递给单参数例程：

import os
from multiprocessing import Pool

def task(args):
    print "PID =", os.getpid(), ", arg1 =", args[0], ", arg2 =", args[1]

pool = Pool()

pool.map(task, [
        [1,2],
        [3,4],
        [5,6],
        [7,8]
    ])

然后可以用自己喜欢的方法构造一个参数列表。

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]

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。