问题描述
在尝试将字符串解析为整数时,我必须编写以下函数才能正常失败.我想 Python 有内置的东西可以做到这一点,但我找不到它.如果没有,是否有一种更 Pythonic 的方式,不需要单独的函数?
I had to write the following function to fail gracefully when trying to parse a string to an integer. I would imagine Python has something built in to do this, but I can't find it. If not, is there a more Pythonic way of doing this that doesn't require a separate function?
def try_parse_int(s, base=10, val=None): try: return int(s, base) except ValueError: return val
我最终使用的解决方案是修改了@sharjeel 的答案.以下内容在功能上相同,但我认为更具可读性.
The solution I ended up using was a modification of @sharjeel's answer. The following is functionally identical, but, I think, more readable.
def ignore_exception(exception=Exception, default_val=None): """Returns a decorator that ignores an exception raised by the function it decorates. Using it as a decorator: @ignore_exception(ValueError) def my_function(): pass Using it as a function wrapper: int_try_parse = ignore_exception(ValueError)(int) """ def decorator(function): def wrapper(*args, **kwargs): try: return function(*args, **kwargs) except exception: return default_val return wrapper return decorator
推荐答案
这是一个非常常规的场景,所以我编写了一个ignore_exception"装饰器,它适用于各种抛出异常而不是优雅地失败的函数:
This is a pretty regular scenario so I've written an "ignore_exception" decorator that works for all kinds of functions which throw exceptions instead of failing gracefully:
def ignore_exception(IgnoreException=Exception,DefaultVal=None): """ Decorator for ignoring exception from a function e.g. @ignore_exception(DivideByZero) e.g.2. ignore_exception(DivideByZero)(Divide)(2/0) """ def dec(function): def _dec(*args, **kwargs): try: return function(*args, **kwargs) except IgnoreException: return DefaultVal return _dec return dec
在您的情况下的用法:
sint = ignore_exception(ValueError)(int) print sint("Hello World") # prints none print sint("1340") # prints 1340