Blu*_*oon 4 python python-decorators
我正在尝试编写一个装饰器以重复N次错误函数,并且之间的睡眠时间不断增加。到目前为止,这是我的尝试:
def exponential_backoff(seconds=10, attempts=10):
def our_decorator(func):
def function_wrapper(*args, **kwargs):
for s in range(0, seconds*attempts, attempts):
sleep(s)
try:
return func(*args, **kwargs)
except Exception as e:
print(e)
return function_wrapper
return our_decorator
@exponential_backoff
def test():
for a in range(100):
if a - random.randint(0,1) == 0:
print('success count: {}'.format(a))
pass
else:
print('error count {}'.format(a))
'a' + 1
test()
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我不断收到错误:
TypeError: our_decorator() missing 1 required positional argument: 'func'
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Sra*_*raw 10
了解什么是装饰器:
@exponential_backoff
def test():
pass
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等于:
def test():
pass
test = exponential_backoff(test)
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在这种情况下,test是def our_decorator(func):。这就是为什么你TypeError在调用test().
更进一步:
@exponential_backoff()
def test():
pass
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等于:
def test():
pass
test = exponential_backoff()(test)
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在这种情况下,现在test就是您所需要的。
此外,functools.wraps帮助您将原始函数的所有属性复制到修饰函数。例如函数名或文档字符串:
from functools import wraps
def exponential_backoff(func):
# @wraps(func)
def function_wrapper(*args, **kwargs):
pass
return function_wrapper
@exponential_backoff
def test():
pass
print(test) # <function exponential_backoff.<locals>.function_wrapper at 0x7fcc343a4268>
# uncomment `@wraps(func)` line:
print(test) # <function test at 0x7fcc343a4400>
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您应该使用:
@exponential_backoff()
def test():
...
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整体装饰器的设计并非使参数为可选,因此()在使用时必须提供。
如果要使装饰器允许参数列表成为可选的示例,请参见:
您可能还考虑使用wrapt包使装饰器更轻松,更坚固。
您可以选择@Graham Dumpleton 提供的解决方案,也可以像这样修改您的装饰器:
from functools import wraps, partial
def exponential_backoff(func=None, seconds=10, attempts=10):
if func is None:
return partial(exponential_backoff, seconds=seconds, attempts=attempts)
@wraps(func)
def function_wrapper(*args, **kwargs):
for s in range(0, seconds*attempts, attempts):
sleep(s)
try:
return func(*args, **kwargs)
except Exception as e:
print(e)
return function_wrapper
@exponential_backoff
def test():
for a in range(100):
if a - random.randint(0,1) == 0:
print('success count: {}'.format(a))
pass
else:
print('error count {}'.format(a))
'a' + 1
test()
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编辑 我的答案并不完全正确,请参阅@GrahamDumpleton的答案,它展示了如何使我的解决方案尝试可行(即此链接)。立即修复,谢谢@GrahamDumpleton!
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