如何在多处理池中获得传递参数以返回值?

Question

例如:

from multiprocessing import Pool
from functools import partial
numbers = []
for i in range(100):
numbers.append(i)
def add_one(number, new_numbers):
new = number + 1
new_numbers.append(new)
new_numbers = []
process_pool = Pool(2)
add_one_helper = partial(add_one, new_numbers=new_numbers)
process_pool.map(add_one_helper, numbers)
print(new_numbers)

我假设new_numbers将打印:1,2,3,4…，但在print()中为空。如何在调用后获得new_numbers来填充?

Answer 1

首先是观察。你有:

numbers = []
for i in range(100):
numbers.append(i)

这是使用一个循环来创建一个值为0,1,2…99. 但是，您可以同样简单且更有效地指定:

numbers = list(range(100))

但是实际上传递给Pool.map方法的iterable参数并不需要是一个列表(但是，如果它不支持__len__方法，它将自动转换为列表)。因此，您可以简单地将range(100)作为可迭代参数传递给map调用，而不是创建numbers列表。现在开始讨论手头的问题:

Pool.map方法返回一个列表的所有调用worker函数的返回值(在本例中add_one)。所以你只需要让add_one返回它想要附加到新列表的值:

from multiprocessing import Pool
def add_one(number):
return number + 1
process_pool = Pool(2)
new_numbers = process_pool.map(add_one, range(100))
print(new_numbers)

打印:

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100]