Comparing with type(my_function)

Question

I have a function that should operate according to the type of argument passed into it, simply illustrated:

def operate_according_to_type(argument_passed): 
    if type(argument_passed) == str:
        do string stuff
    elif type(argument_passed) == dict:
        do dict stuff
    elif type(argument_passed) == function:
        argument_passed()

def my_function(): pass

operate_according_to_type("Hello world")
operate_according_to_type({"foo": "bar"})
operate_according_to_type(my_function)

Now while type("Hello world"), type({"foo": "bar"}) and type(my_function) will return respectively <class 'str'>, <class 'dict'> and <class 'function'>, I can't seem to compare to function just as I can to str, the word is not even "reserved".

How should I proceed? Should I proceed at all or is this just plain hazardous?

also related: https://stackoverflow.com/questions/54009407/python-find-the-type-of-a-function/54009756#54009756 — Paritosh Singh, Mar 10 '20 at 11:58

Riccardo Bucco · Accepted Answer · 2020-03-10T11:28:59.553

1

You can check if the object is callable using the callable built-in function:

...
elif callable(argument_passed):
    argument_passed()

More details can be found here.

edited Mar 10 '20 at 11:28

answered Mar 10 '20 at 11:23

Riccardo Bucco

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Thanks! Sorry for the duplicate. Is a case like the example considered bad practice overall or is it fine? – Paulo Modulo Mar 10 '20 at 14:23
@PierreMDL checking the type of a parameter is never a bad practice :) – Riccardo Bucco Mar 10 '20 at 14:34

Comparing with type(my_function)

1 Answers1