我有以下情況：res = []def longfunc(arg): # function runs arg number of steps # each step can take 500 ms to 2 seconds to complete # longfunc keeps adding result of each step into the array res def getResult(arg,timeout): # should call longfunc() # if longfunc() has not provided result by timeout milliseconds then return None # if there is partial result in res by timeout milliseconds then return res # if longfunc() ends before timeout milliseconds then return complete result of longfunc i.e. res arrayresult = getResult(2, 500)我正在考慮使用multiprocessing.Process()放入longfunc()一個單獨的進程，然后啟動另一個線程休眠幾timeout毫秒。我無法弄清楚如何在主線程中從它們兩個中獲得結(jié)果并決定哪個先出現(xiàn)。對這種方法或其他方法的任何建議表示贊賞。
2 回答
函數(shù)式編程
TA貢獻1807條經(jīng)驗 獲得超9個贊
您可以使用time.perf_counter，您的代碼將看到：
import time
ProcessTime = time.perf_counter #this returns nearly 0 when first call it if python version <= 3.6
ProcessTime()
def longfunc(arg, timeout):
start = ProcessTime()
while True
# Do anything
delta = start + timeout - ProcessTime()
if delta > 0:
sleep(1)
else:
return #Error or False
您可以更改 While 對于每個任務(wù)的 for 循環(huán)，檢查超時
一只萌萌小番薯
TA貢獻1795條經(jīng)驗 獲得超7個贊
如果你正在應(yīng)用 multiprocessing 那么你必須簡單地應(yīng)用p.join(timeout=5)where p in a process 這是一個簡單的例子
import time
from itertools import count
from multiprocessing import Process
def inc_forever():
print('Starting function inc_forever()...')
while True:
time.sleep(1)
print(next(counter))
def return_zero():
print('Starting function return_zero()...')
return 0
if __name__ == '__main__':
# counter is an infinite iterator
counter = count(0)
p1 = Process(target=inc_forever, name='Process_inc_forever')
p2 = Process(target=return_zero, name='Process_return_zero')
p1.start()
p2.start()
p1.join(timeout=5)
p2.join(timeout=5)
p1.terminate()
p2.terminate()
if p1.exitcode is None:
print(f'Oops, {p1} timeouts!')
if p2.exitcode == 0:
print(f'{p2} is luck and finishes in 5 seconds!')
我想這可能對你有幫助
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