Function which return dictionary of arguments in python

Question

Function func(*args, **kwargs) should return dictionary, and all elements of that dictionary should be numeric or string variables. If argument of function is dictionary, then function should return dictionary with all elements of argument dictionary, and other arguments. For example:

arg1 = { 'x': 'X', 'y': 'Y' }  
arg2 = 2
arg3 = { 'p': arg1, 'pi': 3.14 }
func(arg2, arg3, arg4=4)

should return this:

{ 'x': 'X', 'y': 'Y', 'arg2': 2, 'pi': 3.14, 'arg4': 4 }

How to do that? Recursion is not desirable.

A function func(*args) won't be able to determine and return the name arg2! — deceze
– deceze ♦, Commented Jun 9, 2017 at 7:52
@Tbaki it's not missing, arg3 is dictionary so we have to go deeper and find out if there are numeric or string elements in arg3. — nimi
– nimi, Commented Jun 9, 2017 at 8:02
@deceze there is locals() function where names of arguments could be determined, am I wrong? — nimi
– nimi, Commented Jun 9, 2017 at 8:03
No, locals won't help you. Only a stack trace would, but please for the love of all that is holy don't go there. — deceze
– deceze ♦, Commented Jun 9, 2017 at 8:04

Chłop Z Lasu · Accepted Answer · 2017-06-09 08:01:14Z

1

When you pass arg2 to the function, the function does not know that it was named arg2 when you passed it in, it just know its value is 2. You would have to get stack trace to know who called that function, scrape source code to get the name and then it would be simple.

But what if I call it like this:

func(2)

what output do you expect then?

answered Jun 9, 2017 at 8:01

Chłop Z Lasu

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4 Comments

deceze Over a year ago

"and then it would be simple" – For special definitions of "simple"… ಠ_ಠ

nimi Over a year ago

I am new in python, so don't blame me. What if as *args could be given only dictionaries?

Chłop Z Lasu Over a year ago

read about update method of class dict: tutorialspoint.com/python/dictionary_update.htm

nimi Over a year ago

That isn't what I've been focused on. Maurice gave simple solution.

Maurice Meyer · Accepted Answer · 2017-06-09 08:05:20Z

0

You could do something like this - but that solution needs argument numbers ascending (arg1, arg2, ..., argN).

def func(*args, **kwargs):
    res = {}
    for i, a in enumerate(args):
        if isinstance(a, dict):
            res.update(a)
        else:
            res['arg%s' % (i+1)] = a
    res.update(kwargs)
    return res


arg1 = { 'x': 'X', 'y': 'Y' }  
arg2 = 2
arg3 = { 'p': arg1, 'pi': 3.14 }

myDict = func(arg1, arg2, arg3, arg4=4)
print(myDict)

returns:

{'x': 'X', 'y': 'Y', 'arg2': 2, 'p': {'x': 'X', 'y': 'Y'}, 'pi': 3.14, 'arg4': 4}

answered Jun 9, 2017 at 8:05

Maurice Meyer

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2 Comments

nimi Over a year ago

That's pretty close. One more thing i would like to be there is the element with key 'p' unpacked. Variables with same names should be in in myDict only if their values are the same, if that's not the case then neither of them should be there.

Maurice Meyer Over a year ago

This should be done before you pass the args to the function ... I mean, when you create the arg-variables.

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