Supply condition in function call

[Manuel Graune]

Peter Otten <__peter__ at web.de> writes:

> Cameron Simpson wrote:
>
>>   test1([0,1,2,3], [1,2,3,4], condition_test)
>> 
>> This passes the local variables inside test1() to "condition" as a single
>> parameter. Now, I grant that vars['i'] is a miracle of tediousness. So
>> consider this elaboration:
>> 
>>   from collections import namedtuple
>> 
>>   condition_test = lambda vars: vars.i + vars.j > 4
>> 
>>   def test1(a, b, condition):
>>     for i, j in zip(a,b):
>>       c = i + j
>>       vars = locals()
>>       varnames = list(vars.keys())
>
> instead or pass the list of arguments explicitly, optionally with some 
> inspect fallback:
>
> $ cat pass_condition_inspect.py
> import inspect
>
> def test3(a, b, condition, args=None):
>     if args is None:
>         args = inspect.getargspec(condition).args
>
>     for i, j in zip(a,b):
>         c = i + j
>         _locals = locals()
>         if condition(*[_locals[name] for name in args]):
>             print("Foo", i, j)
>
> def condition(c, i):
>     return i * i > c
>
> test3([1, 2, 3], [2, 3, 4], condition)
> print("---")
> # note reverse order of argument names
> test3([1, 2, 3], [2, 3, 4], condition, ["i", "c"]) 
> $ python3 pass_condition_inspect.py
> Foo 3 4
> ---
> Foo 1 2
> Foo 2 3
> Foo 3 4
>

I'm just comparing the different solutions. For most of them, the
logic and the costs/benefits are pretty clear to me. Can you
elaborate on your code? I'm not clear what you are achieving with
explicitly passing the arguments instead of simply passing some
variant of the complete locals()-dictionary.


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