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How to keep Dict[str, List] default method parameter as local to the method?


zljubisic at gmail.com wrote:

> Hi,
> 
> consider this example:
> 
> from typing import Dict, List
> 
> 
> class chk_params:
>     def execute(self, params: Dict[str, List] = None):
>         if params is None:
>             params = {}
> 
>         for k, v in params.items():
>             params[k] = [val + 10 for val in v]
> 
>         return params.values()
> 
> params = {
>     'A' : [1],
>     'B': [2],
>     'C': [3],
>     'D': [4],
>     'E': [5]
> }
> print(params)
> params_obj = chk_params()
> 
> print(params_obj.execute(params = params))
> print(params)
> 
> Output is:
> {'A': [1], 'B': [2], 'C': [3], 'D': [4], 'E': [5]}
> dict_values([[11], [12], [13], [14], [15]])
> {'A': [11], 'B': [12], 'C': [13], 'D': [14], 'E': [15]}
> 
> I expected that last print statement will show original parameters A=1,
> B=2... but this is not the case.
> 
> How to construct the method parameters, with params parameter as type of
> Dict[str, List], but at the same time keep params as local dictionary to
> the chk_params.execute() method?

The type annotations are only a distraction here; let's give classical 
Python a chance with

- no type annotations
- no class
- dangerous default, for the thrill of it;)

def execute(params={}):
    return [[val + 10 for val in v] for v in params.values()]

Clean and simple.  If you insist on getting dict_values():

def execute(params={}):
    return {k: [val + 10 for val in v] for k, v in params.items()}.values()

The relevant takeaway is: if you don't want to modify a value, don't  modify 
it. Do not wait for a const modifier to save you.