How not to stop the execution of other function in python in case of Exception/Error

Question

I have a script in python which works as shown below. Each function performs a completely different task and not related to each other. My problem is if function2() is having an issue during the execution process then function3(), function4(), function5() will not execute. I know you will say to handle this by catching the exception (try..except) but then i have to catch every exception which is not i am looking for. In a nutshell how do i code where my other functions are not impacted if any of the function is having issue. Ideally it should exclude that problematic function and let the other function to execute.

def function1():
    some code

def function2():
    some code

def function3():
    some code

def function4():
    some code

def function5():
    some code

if __name__ == '__main__':
    function1()
    function2()
    function3()
    function4()
    function5()

Answer 1

As of Python 3.4, a new context manager as contextlib.suppress is added which as per the doc:

Return a context manager that suppresses any of the specified exceptions if they occur in the body of a with statement and then resumes execution with the first statement following the end of the with statement.

In order to suppress all the exceptions, you may use it as:

from contextlib import suppress

if __name__ == '__main__':

    func_list = [function1, function2, function3, function4, function5]

    for my_func in func_list:
        with suppress(Exception):  # `Exception` to suppress all the exceptions
            my_func()  # Any exception raised by `my_func()` will be suppressed