python catch exception and continue try block

Question

Can I return to executing try-block after exception occurs? (The goal is to write less) For Example:

try:
    do_smth1()
except:
    pass

try:
    do_smth2(         


        

Answer 1

special_func to avoid try-except repetition:

def special_func(test_case_dict):
    final_dict = {}
    exception_dict = {}

    def try_except_avoider(test_case_dict):

        try:
            for k,v in test_case_dict.items():
                final_dict[k]=eval(v) #If no exception evaluate the function and add it to final_dict

        except Exception as e:
            exception_dict[k]=e #extract exception
            test_case_dict.pop(k)
            try_except_avoider(test_case_dict) #recursive function to handle remaining functions

        finally:  #cleanup
            final_dict.update(exception_dict)
            return final_dict #combine exception dict and  final dict

    return try_except_avoider(test_case_dict) 

Run code:

def add(a,b):
    return (a+b)
def sub(a,b):
    return (a-b)
def mul(a,b):
    return (a*b)

case = {"AddFunc":"add(8,8)","SubFunc":"sub(p,5)","MulFunc":"mul(9,6)"}
solution = special_func(case)

Output looks like:

{'AddFunc': 16, 'MulFunc': 54, 'SubFunc': NameError("name 'p' is not defined")}

To convert to variables:

locals().update(solution)

Variables would look like:

AddFunc = 16, MulFunc = 54, SubFunc = NameError("name 'p' is not defined")