Python 回调函数如何在多处理 map_async 中工作？
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How does the callback function work in multiprocessing map_async?
提问by user2727768
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.
调试我的代码花了我整整一个晚上，我终于找到了这个棘手的问题。请看下面的代码。
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()
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]]?
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])并且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取A=[0,0,1]而不是[[0],[0,1]]?
采纳答案by falsetru
Callback is called once with the result ([[0], [0, 1]]) if you use map_async.
[[0], [0, 1]]如果您使用 map_async，则使用结果 ( )调用一次回调。
>>> 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]]
Use apply_asyncif you want callback to be called for each time.
apply_async如果您希望每次都调用回调，请使用。
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()
