Check if a function raises NotImplementedError before calling it in Python

Question

I have the following simplified scheme:

class NetworkAnalyzer(object):
    def __init__(self):
       print('is _score_funct implemented?')

    @staticmethod
    def _score_funct(network):
        raise NotImplementedError

class LS(NetworkAnalyzer):
    @staticmethod
    def _score_funct(network):
        return network

and I am looking for what I should use instead of print('is _score_funct implemented?') in order to figure out if a subclass has already implemented _score_funct(network) or not.

Note: If there is a more pythonic/conventional way of structuring the code, I would also appreciate its mention. The reason I defined it this way is, some NetworkAnalyzer subclasses have _score_funct in their definition, and the ones that dont have it will have different initialization of variables although they will have the same structure

abstract methods? stackoverflow.com/questions/4382945/abstract-methods-in-python — Jean-François Fabre
– Jean-François Fabre ♦, Commented Apr 18, 2018 at 15:09
Well you could take a look at the AST and look if there is only a single raise statement and the only constant is NotImplementedError. — Wombatz
– Wombatz, Commented Apr 18, 2018 at 15:13
I would suggest you figure out the minimal behavior for each subclass and turn that into an abstract base class. See the abc docs . Research Liskov Substitution Principle . — Ben
– Ben, Commented Apr 18, 2018 at 15:16
Why not just not implement the function in the base class and in __init__ use getattr() and callable() to check if it exists and can be called? — Attie
– Attie, Commented Apr 18, 2018 at 15:16
I havent used abstract methods as I had little understanding of metaclasses. I will read the documentation to have it more conventional. By then, I have tried @Jean-FrançoisFabre's answer and it works — ozgeneral
– ozgeneral, Commented Apr 18, 2018 at 15:18

Duncan · Accepted Answer · 2018-04-18 15:28:22Z

4

Use an abstract base class and you won't be able to instantiate the class unless it implements all of the abstract methods:

import abc

class NetworkAnalyzerInterface(abc.ABC):
    @staticmethod
    @abc.abstractmethod
    def _score_funct(network):
        pass

class NetworkAnalyzer(NetworkAnalyzerInterface):
    def __init__(self):
        pass

class LS(NetworkAnalyzer):
    @staticmethod
    def _score_funct(network):
        return network

class Bad(NetworkAnalyzer):
    pass

ls = LS()   # Ok
b = Bad()   # raises TypeError: Can't instantiate abstract class Bad with abstract methods _score_funct

answered Apr 18, 2018 at 15:28

Duncan

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2 Comments

ozgeneral Over a year ago

Do I need the NetworkAnalyzerInterface in this example by any conventional reason or can I just bypass it by NetworkAnalyzer(abc.ABC)

Duncan Over a year ago

You can probably just make NetworkAnalyzer itself the abstract class in this case. I would just prefer to keep interface and implementation code separate although in the example I ended up with there's no implementation code.

Jean-François Fabre · Accepted Answer · 2018-04-18 15:25:52Z

I'm not a metaclass/class specialist but here's a method that works in your simple case (not sure it works as-is in a complex/nested class namespace):

To check if the method was overridden, you could try a getattr on the function name, then check the qualified name (class part is enough using string partitionning):

class NetworkAnalyzer(object):
    def __init__(self):
        funcname = "_score_funct"
        d = getattr(self,funcname)
        print(d.__qualname__.partition(".")[0] == self.__class__.__name__)

if _score_funct is defined in LS, d.__qualname__ is LS._score_funct, else it's NetworkAnalyzer._score_funct.

That works if the method is implemented at LS class level. Else you could replace by:

d.__qualname__.partition(".")[0] != "NetworkAnalyzer"

Of course if the method is overridden with some code which raises an NotImplementedError, that won't work... This method doesn't inspect methods code (which is hazardous anyway)

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