回调函数在多处理 map_async 中如何工作?

问题描述

调试代码花了我一晚上的时间，终于发现了这个棘手的问题.请看下面的代码.

It cost me a whole night to debug my code, and I finally found this tricky problem. Please take a look at the code below.

from multiprocessing import Pool

def myfunc(x):
    return [i for i in range(x)]

pool=Pool()

A=[]
r = pool.map_async(myfunc, (1,2), callback=A.extend)
r.wait()

我以为我会得到 A=[0,0,1]，但输出是 A=[[0],[0,1]].这对我来说没有意义，因为如果我有 A=[]、A.extend([0]) 和 A.extend([0,1]) 会给我A=[0,0,1].回调可能以不同的方式工作.所以我的问题是如何获得 A=[0,0,1] 而不是 [[0],[0,1]]?

I thought I would get A=[0,0,1], but the output is A=[[0],[0,1]]. This does not make sense to me because if I have A=[], A.extend([0]) and A.extend([0,1]) will give me A=[0,0,1]. Probably the callback works in a different way. So my question is how to get A=[0,0,1] instead of [[0],[0,1]]?

推荐答案

如果使用 map_async，则调用一次回调并返回结果 ([[0], [0, 1]]).

Callback is called once with the result ([[0], [0, 1]]) if you use map_async.

>>> from multiprocessing import Pool
>>> def myfunc(x):
...     return [i for i in range(x)]
... 
>>> A = []
>>> def mycallback(x):
...     print('mycallback is called with {}'.format(x))
...     A.extend(x)
... 
>>> pool=Pool()
>>> r = pool.map_async(myfunc, (1,2), callback=mycallback)
>>> r.wait()
mycallback is called with [[0], [0, 1]]
>>> print(A)
[[0], [0, 1]]

使用 apply_async如果您希望每次都调用回调.

Use apply_async if you want callback to be called for each time.

pool=Pool()
results = []
for x in (1,2):
    r = pool.apply_async(myfunc, (x,), callback=mycallback)
    results.append(r)
for r in results:
    r.wait()

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