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

Actually there is a "built-in", single line solution that doesn't require to introduce a new function. As I hoped to find such an answer here, I'm adding it:

>>> s = "123"
>>> i = int(s) if s.isdigit() else None
>>> print(i)
123

>>> s = "abc"
>>> i = int(s) if s.isdigit() else None
>>> print(i)
None

>>> s = ""
>>> i = int(s) if s.isdigit() else None
>>> print(i)
None

>>> s = "1a"
>>> i = int(s) if s.isdigit() else None
>>> print(i)
None

See also https://docs.python.org/3/library/stdtypes.html#str.isdigit

def intTryParse(value):
    try:
        return int(value), True
    except ValueError:
        return value, False

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