在Python多处理库中,是否有支持多个参数的pool.map变体?

import multiprocessing

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()

当前回答

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)

其他回答

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()

这是我用来将多个参数传递给pool.imap fork中使用的单参数函数的例程的示例:

from multiprocessing import Pool

# Wrapper of the function to map:
class makefun:
    def __init__(self, var2):
        self.var2 = var2
    def fun(self, i):
        var2 = self.var2
        return var1[i] + var2

# Couple of variables for the example:
var1 = [1, 2, 3, 5, 6, 7, 8]
var2 = [9, 10, 11, 12]

# Open the pool:
pool = Pool(processes=2)

# Wrapper loop
for j in range(len(var2)):
    # Obtain the function to map
    pool_fun = makefun(var2[j]).fun

    # Fork loop
    for i, value in enumerate(pool.imap(pool_fun, range(len(var1))), 0):
        print(var1[i], '+' ,var2[j], '=', value)

# Close the pool
pool.close()

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。

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。