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()