問題描述

調試代碼花了我一晚上的時間，終于發現了這個棘手的問題.請看下面的代碼.

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()

我以為我會得到 A=[0,0,1]，但輸出是 A=[[0],[0,1]].這對我來說沒有意義，因為如果我有 A=[]、A.extend([0]) 和 A.extend([0,1]) 會給我A=[0,0,1].回調可能以不同的方式工作.所以我的問題是如何獲得 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]]?

推薦答案

如果使用 map_async，則調用一次回調并返回結果 ([[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如果您希望每次都調用回調.

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()

這篇關于回調函數在多處理 map_async 中如何工作?的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！