April 13, 2019
7:59 a.m.
On Fri, Apr 12, 2019 at 11:10 AM Viktor Roytman <viktor.roytman@gmail.com> wrote:
The standard approach I have encountered in this scenario is to pass in the keyword arguments explicitly like so
func( a=kwargs_dict["a"], b=kwargs_dict["b"], c=kwargs_dict["c"], )
func(**{k:v for k, v in d.items() if k in ('a','b','c')) Or you can `def dict_filter(d, yes)` to the the above. -- Juancarlo *Añez*