Updating locals() that have a predefined value
python
Solution
It's mentioned in the `locals` documentation:
Note: The contents of this dictionary should not be modified; changes may not affect the values of local and free variables used by the interpreter.
AFAIK, there's no reliable way to define/change local variables.
Problem
Consider the following example: ``` def main(): a = 'predefined' variables = {'a':'dynamic'} locals().update(variables) print a if __name__ == '__main__': main() ``` When running the script, I would expect to see: ``` dynamic ``` but I see ``` predefined ``` Why? How can I get the dynamic value instead? Update: The reason why I ask: I have a program that takes many input arguments, with lengthy variable names. I was hoping to simply "unpack" whatever the `argparse` parser receives in a single call to `locals().update(...)` ``` def main(): a = 'predefined' parser = argparse.ArgumentParser(description='My program') parser.add_argument('-a', type=int, default=a, required=False); # Hoping to avoid typing lines like the following for every parameter: # a = parser.parse_args().a input_variables = vars(parser.parse_args()) locals().update(input_variables) # Process stuff using the parameter names directly, e.g. print a ```