Pythonic way to get the first passed kwarg without knowing its name?

Question

I have a function, which has a signature like this:

def func(**kwargs):

The user of that function will call the function with zero or one keyword arguments. If he passes one argument, the name will be foo_id, bar_id, baz_id etc., but I don't know the exact name he will use. The value of the passed argument will be some interger. I still want to take that argument's value and use it.

Currently I'm doing it like this, but I was wondering would there be a cleaner way to achieve this:

def func(**kwargs):
    if kwargs:
        target_id = list(kwargs.values())[0]
    else:
        target_id = None

    # use target_id here, no worries if it's None

I'm using Python 3.8, so backwards compatibility is not an issue.

What kind of API has un-predefined keyword arguments? Sounds like an oxymoron to me. — martineau, Dec 16 '19 at 17:19
@martineau Way to complex to fully explain here. But in short: this will be used in a base class for django graphene Mutations, which then will take one argument each. This argument is for consistency sake always the name of the model and `_id`. So no one actually calls this Python function, but the caller is the user of the graphql api. — ruohola, Dec 16 '19 at 17:26
Hmm, I understand you've left details out, but it still sounds like it could be an [XY problem](https://meta.stackexchange.com/questions/66377/what-is-the-xy-problem) with the real issue being with the the class hierarchy's design. — martineau, Dec 16 '19 at 17:40
@martineau This (like any other problem) could be tackled in a million different ways. But doing it with an abstract base class, which all schemas that need the functionality, derive to their own Mutations, gives me very little duplicate code and an extremely clean, consistent, and easy to use API. I'm quite pleased with this solution. — ruohola, Dec 16 '19 at 17:44

score 6 · Accepted Answer · answered Dec 16 '19 at 16:32

6

Here we are

def func(**kwargs):
    target_id = next(iter(kwargs.values()), None)

    print(target_id)


func(name_id='name')
func(value_id='value')
func(test_id='test')
func()

Outputs

python test.py
name
value
test
None

answered Dec 16 '19 at 16:32

Alexandr Shurigin

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Be sure to mention this works for Python 3.6 + – Jab Dec 16 '19 at 16:33
It should work in any Python version I believe, why not? – Alexandr Shurigin Dec 16 '19 at 16:34
2

Due to the way dicts are ordered in older versions. Or not therein – Jab Dec 16 '19 at 16:34
1

@Jab does it matter given the OP said there would be one or zero kwargs? – Mark Dec 16 '19 at 16:35
ah, you are talking about ordering. I thought about the syntax or etc. – Alexandr Shurigin Dec 16 '19 at 16:35
Ahh then no that's my mistake – Jab Dec 16 '19 at 16:35
Not sure about ordering case in the previous python versions :) Since we don't have ordering in the previous versions. Maybe it is not possible or possible only using inspection (or even traceback module). But looks like an interesting quesiton :) – Alexandr Shurigin Dec 16 '19 at 16:37
Although what if the value of the arg is `None` or falsey? Maybe consider using an `object` placeholder if so. – Jab Dec 16 '19 at 16:37
1

@Jab I specified my question to state that the value of the arg is always an integer. – ruohola Dec 16 '19 at 16:39
for integers, it works always, of course – Alexandr Shurigin Dec 16 '19 at 16:40

kaya3 · Answer 2 · 2019-12-16T16:43:16.563

2

Since the dictionary has only one item, and you don't need to keep it in the dictionary, the cleanest way is to use the dict.popitem method. This returns both the argument name and its value as a pair.

def func(**kwarg):
    if kwarg:
        name, value = kwarg.popitem()
        # ...
    else:
        # ...

Since the caller should supply at most one argument, I recommend explicitly raising an error if it is called with more arguments:

def func(**kwarg):
    if len(kwarg) > 1:
        raise TypeError(f'Expected at most 1 keyword arg, got {len(kwarg)}.')
    elif kwarg:
        name, value = kwarg.popitem()
        # ...
    else:
        # ...

edited Dec 16 '19 at 16:43

answered Dec 16 '19 at 16:39

kaya3

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`...will call the function with zero or one keyword arguments.` – Mark Dec 16 '19 at 16:41
1

@MarkMeyer Good catch, fixed. – kaya3 Dec 16 '19 at 16:43
very pythonic way :D – Alexandr Shurigin Dec 16 '19 at 16:44
1

I would think it would be more pythonic to put `popitem` in a `try` block and catch the `KeyError` if you need to. Of course that's a tiny detail. – Mark Dec 16 '19 at 16:44
3

not advocating this but `name, value = (kwarg or {None:None}).popitem()` – Chris_Rands Dec 16 '19 at 16:45
Regarding if/else vs. try/catch, my opinion is that *"explicit is better than implicit"*, so if/else is better here because the condition explicitly shows under what circumstances one will happen vs. the other. Try/except requires the reader to infer what the condition is from what exception could be thrown by the code in the try block, and if you write `except KeyError:` explicitly then the reader may still have to look up what error `popitem` throws on an empty dictionary to be sure it's correct. – kaya3 Dec 16 '19 at 16:50

Pythonic way to get the first passed kwarg without knowing its name?

2 Answers2