Create a function for getting use input of arguments for another function

Question

I need to take arguments from a function (f1) and supply mapped arguments to those arguments to another function (f2).

For this I created the argument dictionary against f1 and f2 arguments as under:

def f1(variable,*args):
    arg_dict = {
                'a' : [args1,args2,args3,args4]
                'b' : [args5,args6]
                'c' : [args7,'{:.2f}'.format]
               }
    f2(*args)
    #---do this---
    #---print this---
    return #---this---

But I am unable to think of an efficient and simple way to supply those arguments to f2 inside the f1 function.

Say I do this:

f1(var1,'a','c')

Then the f2 should be run as:

f2(args1,args2,args3,args4,args7,'{:.2f}'.format)

I also thought of asking the f1 arguments as a single value or list under a variable:

def f1(var, param):
    #
    #

So something like this:

f1(var1, ['a','c'])

It should assign ['a','c'] to param under f1. How to proceed next?

Green Cloak Guy · Accepted Answer · 2021-04-02 12:53:12Z

1

If I'm understanding your problem correctly, then all you need is a way to filter arg_dict by the sets of arguments you actually want to pass, and flatten that set. You can do that by creating a new list and .extend()ing it in a for loop:

def f1(variable,*args):
    arg_dict = {
                'a' : [args1,args2,args3,args4]
                'b' : [args5,args6]
                'c' : [args7,'{:.2f}'.format]
               }
    pass_args = []
    for arg in args:
        if arg in arg_dict:
            pass_args.extend(arg_dict[arg])
    f2(*pass_args)
    #---do this---
    #---print this---
    return #---this---

answered Apr 2, 2021 at 12:53

Green Cloak Guy

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1 Comment

Veki Over a year ago

Yes. Thank you. I didn't know adding that * in f2 will do the trick. Thanks again.

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