Is there a built-in or more Pythonic way to try to parse a string to an integer

Question

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 

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 

Answer 1

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 

Usage in your case:

sint = ignore_exception(ValueError)(int) print sint("Hello World") # prints none print sint("1340") # prints 1340