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
你可以使用Python重试包。 重试
它是用Python编写的,目的是简化向几乎任何东西添加重试行为的任务。
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
在Python装饰器库中也有类似的东西。
请记住,它不测试异常,而是测试返回值。它会重新尝试,直到被修饰的函数返回True。
稍微修改一下版本就可以了。
我喜欢laurent-laporte的回答。下面是我的版本,它包装在一个类与静态方法和一些例子。我实现了重试计数作为另一种重试方式。还增加了kwargs。
from typing import List
import time
class Retry:
@staticmethod
def onerror_retry(exception, callback, retries: int = 0, timeout: float = 0, timedelta: float = 0,
errors: List = None, **kwargs):
"""
@param exception: The exception to trigger retry handling with.
@param callback: The function that will potentially fail with an exception
@param retries: Optional total number of retries, regardless of timing if this threshold is met, the call will
raise the exception.
@param timeout: Optional total amount of time to do retries after which the call will raise an exception
@param timedelta: Optional amount of time to sleep in between calls
@param errors: A list to receive all the exceptions that were caught.
@param kwargs: An optional key value parameters to pass to the function to retry.
"""
for retry in Retry.__onerror_retry(exception, callback, retries, timeout, timedelta, errors, **kwargs):
if retry: retry(**kwargs) # retry will be None when all retries fail.
@staticmethod
def __onerror_retry(exception, callback, retries: int = 0, timeout: float = 0, timedelta: float = 0,
errors: List = None, **kwargs):
end_time = time.time() + timeout
continues = 0
while True:
try:
yield callback(**kwargs)
break
except exception as ex:
print(ex)
if errors:
errors.append(ex)
continues += 1
if 0 < retries < continues:
print('ran out of retries')
raise
if timeout > 0 and time.time() > end_time:
print('ran out of time')
raise
elif timedelta > 0:
time.sleep(timedelta)
err = 0
#
# sample dumb fail function
def fail_many_times(**kwargs):
global err
err += 1
max_errors = kwargs.pop('max_errors', '') or 1
if err < max_errors:
raise ValueError("I made boo boo.")
print("Successfully did something.")
#
# Example calls
try:
#
# retries with a parameter that overrides retries... just because
Retry.onerror_retry(ValueError, fail_many_times, retries=5, max_errors=3)
err = 0
#
# retries that run out of time, with 1 second sleep between retries.
Retry.onerror_retry(ValueError, fail_many_times, timeout=5, timedelta=1, max_errors=30)
except Exception as err:
print(err)
以下是我关于如何解决这个问题的想法:
j = 19
def calc(y):
global j
try:
j = j + 8 - y
x = int(y/j) # this will eventually raise DIV/0 when j=0
print("i = ", str(y), " j = ", str(j), " x = ", str(x))
except:
j = j + 1 # when the exception happens, increment "j" and retry
calc(y)
for i in range(50):
calc(i)
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