我有一个循环,开头为for I,范围为(0,100)。正常情况下,它可以正常运行,但有时由于网络条件而出现故障。目前,我已经设置它,以便在失败时,它将继续在except子句中(继续到I的下一个数字)。
我是否可以将相同的数字重新分配给i,并再次运行失败的循环迭代?
我有一个循环,开头为for I,范围为(0,100)。正常情况下,它可以正常运行,但有时由于网络条件而出现故障。目前,我已经设置它,以便在失败时,它将继续在except子句中(继续到I的下一个数字)。
我是否可以将相同的数字重新分配给i,并再次运行失败的循环迭代?
当前回答
Decorator是一个很好的方法。
from functools import wraps
import time
class retry:
def __init__(self, success=lambda r:True, times=3, delay=1, raiseexception=True, echo=True):
self.success = success
self.times = times
self.raiseexception = raiseexception
self.echo = echo
self.delay = delay
def retry(fun, *args, success=lambda r:True, times=3, delay=1, raiseexception=True, echo=True, **kwargs):
ex = Exception(f"{fun} failed.")
r = None
for i in range(times):
if i > 0:
time.sleep(delay*2**(i-1))
try:
r = fun(*args, **kwargs)
s = success(r)
except Exception as e:
s = False
ex = e
# raise e
if not s:
continue
return r
else:
if echo:
print(f"{fun} failed.", "args:", args, kwargs, "\nresult: %s"%r)
if raiseexception:
raise ex
def __call__(self, fun):
@wraps(fun)
def wraper(*args, retry=0, **kwargs):
retry = retry if retry>0 else self.times
return self.__class__.retry(fun, *args,
success=self.success,
times=retry,
delay=self.delay,
raiseexception = self.raiseexception,
echo = self.echo,
**kwargs)
return wraper
一些用法示例:
@retry(success=lambda x:x>3, times=4, delay=0.1)
def rf1(x=[]):
x.append(1)
print(x)
return len(x)
> rf1()
[1]
[1, 1]
[1, 1, 1]
[1, 1, 1, 1]
4
@retry(success=lambda x:x>3, times=4, delay=0.1)
def rf2(l=[], v=1):
l.append(v)
print(l)
assert len(l)>4
return len(l)
> rf2(v=2, retry=10) #overwite times=4
[2]
[2, 2]
[2, 2, 2]
[2, 2, 2, 2]
[2, 2, 2, 2, 2]
5
> retry.retry(lambda a,b:a+b, 1, 2, times=2)
3
> retry.retry(lambda a,b:a+b, 1, "2", times=2)
TypeError: unsupported operand type(s) for +: 'int' and 'str'
其他回答
在for循环中执行while True,将try代码放入其中,只有当代码成功时才退出while循环。
for i in range(0,100):
while True:
try:
# do stuff
except SomeSpecificException:
continue
break
使用while和计数器:
count = 1
while count <= 3: # try 3 times
try:
# do_the_logic()
break
except SomeSpecificException as e:
# If trying 3rd time and still error??
# Just throw the error- we don't have anything to hide :)
if count == 3:
raise
count += 1
最清晰的方法是显式地设置i。例如:
i = 0
while i < 100:
i += 1
try:
# do stuff
except MyException:
continue
我最近用我的python解决了这个问题,我很高兴与stackoverflow的访问者分享,如果需要请给予反馈。
print("\nmonthly salary per day and year converter".title())
print('==' * 25)
def income_counter(day, salary, month):
global result2, result, is_ready, result3
result = salary / month
result2 = result * day
result3 = salary * 12
is_ready = True
return result, result2, result3, is_ready
i = 0
for i in range(5):
try:
month = int(input("\ntotal days of the current month: "))
salary = int(input("total salary per month: "))
day = int(input("Total Days to calculate> "))
income_counter(day=day, salary=salary, month=month)
if is_ready:
print(f'Your Salary per one day is: {round(result)}')
print(f'your income in {day} days will be: {round(result2)}')
print(f'your total income in one year will be: {round(result3)}')
break
else:
continue
except ZeroDivisionError:
is_ready = False
i += 1
print("a month does'nt have 0 days, please try again")
print(f'total chances left: {5 - i}')
except ValueError:
is_ready = False
i += 1
print("Invalid value, please type a number")
print(f'total chances left: {5 - i}')
只有当try子句成功时才增加循环变量