我知道这是一个老问题。但是,我想添加我已经准备好的解决方案。最好的方法是编写一个retry
装饰器,当异常发生时会重试。此外,您还可以设置自定义指数延迟。文档字符串解释了如何使用装饰器。干得好:
import logging
import time
from functools import partial, wraps
def retry(func=None, exception=Exception, n_tries=5, delay=5, backoff=1, logger=False):
"""Retry decorator with exponential backoff.
Parameters
----------
func : typing.Callable, optional
Callable on which the decorator is applied, by default None
exception : Exception or tuple of Exceptions, optional
Exception(s) that invoke retry, by default Exception
n_tries : int, optional
Number of tries before giving up, by default 5
delay : int, optional
Initial delay between retries in seconds, by default 5
backoff : int, optional
Backoff multiplier e.g. value of 2 will double the delay, by default 1
logger : bool, optional
Option to log or print, by default False
Returns
-------
typing.Callable
Decorated callable that calls itself when exception(s) occur.
Examples
--------
>>> import random
>>> @retry(exception=Exception, n_tries=4)
... def test_random(text):
... x = random.random()
... if x < 0.5:
... raise Exception("Fail")
... else:
... print("Success: ", text)
>>> test_random("It works!")
"""
if func is None:
return partial(
retry,
exception=exception,
n_tries=n_tries,
delay=delay,
backoff=backoff,
logger=logger,
)
@wraps(func)
def wrapper(*args, **kwargs):
ntries, ndelay = n_tries, delay
while ntries > 1:
try:
return func(*args, **kwargs)
except exception as e:
msg = f"{str(e)}, Retrying in {ndelay} seconds..."
if logger:
logging.warning(msg)
else:
print(msg)
time.sleep(ndelay)
ntries -= 1
ndelay *= backoff
return func(*args, **kwargs)
return wrapper