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

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

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