Is there a more efficient way of storing keyword arguments?

Question

So here is my code:

def formateUserData(FirstName = None, MiddleName = None, LastName = None, DOB = None, Gender = None): formatedUserData = {} dataFieldKeys = ['First Name', 'Middle Name', 'Last Name', 'DOB', 'Gender'] dataFieldValues = [FirstName, MiddleName, LastName, DOB, Gender] for key, value in zip(dataFieldKeys, dataFieldValues): if value: formatedUserData[key] = value return formatedUserData

As you can see, the keyword arguments has to be repeated 3 times (first in line 1, second in line 3 and third in line 4). Is there a way I could do the same thing without by just storing the keyword arguments once? Maybe by using some way to iterate through the keyword arguments if that possible? BTW, I'm looking to not use **kwargs.

Have a look at locals() or other answers as mentioned here: stackoverflow.com/questions/582056/… — ScootCork
– ScootCork, Commented Feb 2, 2021 at 21:37
You can pass an object insted of multiple args and then itarate on its attributes — idan ovadia
– idan ovadia, Commented Feb 2, 2021 at 21:38
The usual way here would be to level-up the abstraction, making a model class which has attributes for those fields. Then you just pass one instance, user, when you call this function. — wim
– wim, Commented Feb 2, 2021 at 21:48

Raphael · Accepted Answer · 2021-02-02 22:29:20Z

maybe this scratch inspires you to condsider dataclasses which I find very handy

from dataclasses import dataclass, fields from datetime import date from typing import Literal, Optional @dataclass class User: first_name: Optional[str] = None middle_name: Optional[str] = None last_name: Optional[str] = None dob: Optional[date] = None gender: Optional[Literal['male', 'female']] = None def not_none_dict(user: User) -> dict: not_none = {} for field in fields(user): value = getattr(user, field.name) if value: not_none[field.name] = value return not_none print( not_none_dict(User( first_name='Joe', last_name='Doe' )) ) >>> {'first_name': 'Joe', 'last_name': 'Doe'}

Glauco · Accepted Answer · 2021-02-03 09:18:55Z

For my point of view the shortest solution is to use locals():

def formateUserData2(FirstName = None, MiddleName = None, LastName = None, DOB = None, Gender = None): return {k:v for k,v in locals().items() if v}

But this solution don't consider the change of labels, so a binding is required

def formateUserData(FirstName = None, MiddleName = None, LastName = None, DOB = None, Gender = None): binding = {'First Name':FirstName, 'Middle Name':MiddleName, 'Last Name':LastName, 'DOB':DOB, 'Gender':Gender} return {k:v for k,v in binding.items() if v}

Use an explicit binding is a good practice, because it clarify what is your expectation.

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