我試圖找到一種方法來進(jìn)行for循環(huán)，如果for循環(huán)的迭代次數(shù)超過超時(shí)時(shí)間，則它會(huì)中斷并轉(zhuǎn)到下一次迭代。例子 ：timeout = 60for i in mylist: i += 1 if time > timeout: break
2 回答
當(dāng)年話下
TA貢獻(xiàn)1890條經(jīng)驗(yàn) 獲得超9個(gè)贊
我認(rèn)為您可以使用時(shí)間模塊，如下所示：
import time
#get the time at the start of the program
x = time.localtime(time.time())
start_time = time.strftime('%S', x)
#the loop
timeout = 5
for i in range(10000000):
i += 1
y = time.localtime(time.time())
now_time = time.strftime('%S', y)
run_time = int(now_time) - int(start_time)
print(run_time) #to see the run_time
if run_time > timeout:
break
一只名叫tom的貓
TA貢獻(xiàn)1906條經(jīng)驗(yàn) 獲得超3個(gè)贊
假設(shè)單次迭代不需要那么多，只需使用time模塊和 while 循環(huán)，如下所示：
mylist = [1,2,3]
import time
timeout = 60
time_start = time.time()
i = 0
while i < len(mylist) and time.time() - time_start < timeout:
# do your stuff
i += 1
if i == len(mylist):
# this code runs if the iteration has completed, pass does nothing
pass
else:
# and this code runs if there was a timeout
pass
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