我正在为多组可迭代项运行一个函数,在所有进程完成后立即返回所有结果的列表。
def fct(variable1, variable2):
# do an operation that does not necessarily take the same amount of
# time for different input variables and yields result1 and result2
return result1, result2
variables1 = [1,2,3,4]
variables2 = [7,8,9,0]
with ThreadPoolExecutor(max_workers = 8) as executor:
future = executor.map(fct,variables1,variables2)
print '[%s]' % ', '.join(map(str, future))
>>> [ (12,3) , (13,4) , (14,5) , (15,6) ]
如何在计算出中间结果后立即打印中间结果,例如variable1=1、variable2=7?
如果您想在完成时使用结果,而不保留原始可迭代的顺序,则可以将executor.submit
与concurrent.futures.as_completed
:一起使用
from concurrent.futures import ThreadPoolExecutor, as_completed
import time
import random
def fct(variable1, variable2):
time.sleep(random.randint(1,5))
return variable1+1, variable2+1
variables1 = [1,2,3,4]
variables2 = [7,8,9,0]
with ThreadPoolExecutor(max_workers = 8) as executor:
for out in as_completed([executor.submit(fct,*vars)
for vars in zip(variables1, variables2)]):
print(out.result())
输出(尽管由于random.randint
,在任何给定的运行中都可能有任何订单):
(4, 10)
(5, 1)
(2, 8)
(3, 9)
一旦Future
被标记为完成,as_completed
将从其输入列表中产生Future
,而不管它实际位于输入列表中的何处。这样,如果第二项在2秒后完成,但第一项需要15秒,那么您将在两秒后看到第二项的结果,而不需要等待15秒。根据您的具体用例,这可能是理想的行为,也可能不是理想的行为。
编辑:
请注意,您仍然可以通过这种方式以原始顺序获得输出。你只需要保存你给as_completed
:的列表
with ThreadPoolExecutor(max_workers = 8) as executor:
jobs = [executor.submit(fct, *vars)
for vars in zip(variables1, variables2)]
for out in as_completed(jobs):
print(out.result())
results = [r.result() for r in jobs]
print(results)
输出:
(5, 1)
(2, 8)
(3, 9)
(4, 10)
[(2, 8), (3, 9), (4, 10), (5, 1)]
map
已经这样做了,但join
需要使用整个可迭代项才能创建连接字符串。将其更改为for
循环将允许您增量打印:
for i in executor.map(fct, v1, v2):
print(str(i))
保持与join
代码相同的输出需要更多的工作,但无论如何都是可行的:
first = True
print("[ ", end="")
for i in executor.map(fct, v1, v2):
if first:
first = False
else:
print(" , ", end="")
print(str(i), end="")
print("]", end="")