回调函数如何在多处理map_async中工作? [英] How does the callback function work in multiprocessing map_async?
问题描述
我花了一整夜的时间调试代码,终于找到了这个棘手的问题.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]
,但是输出是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取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,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]]
如果要回调到,请使用 apply_async
每次都会被调用.
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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