imap无序值访问

Question

我有一个更复杂的情况:

依赖性:

#formula
def cube_rand(x):
c = x**3
time.sleep(random.random())
return (x,c)
#dataframe to be used
dict_input = {
'col1': ['a','b','c'],
'col2': [1,2,3]
}
df_test = pd.DataFrame(dict_input)

imap实例:

from multiprocessing import Pool
start = time.time()
p=Pool(8)
results = p.imap_unordered(cube_rand, (index for index,row in df_test.iterrows()))
print(results)
done = time.time()
print("processing time: ", done-start)

我想为定义的数据框架中的每一行无序地打印成对("index"，"index^3")。但是，我收到以下输出:

<multiprocessing.pool.IMapUnorderedIterator object at 0x7fb5af0cbd50>
processing time:  0.15177679061889648

有关于如何解决这个问题的猜测吗?

感谢

Answer 1

您需要遍历imap_unordered调用的结果:

for x in p.imap_unordered(cube_rand, (index for index,row in df_test.iterrows())):
print(x)

输出:

(0, 0)
(2, 8)
(1, 1)
processing time:  1.1946678161621094