How to get method parameter names?

Question

Given the Python function:

def a_method(arg1, arg2):     pass 

How can I extract the number and names of the arguments. I.e., given that I have a reference to func, I want the func.[something] to return ("arg1", "arg2").

The usage scenario for this is that I have a decorator, and I wish to use the method arguments in the same order that they appear for the actual function as a key. I.e., how would the decorator look that printed "a,b" when I call a_method("a", "b")?

Answer 1

Take a look at the inspect module - this will do the inspection of the various code object properties for you.

>>> inspect.getfullargspec(a_method) (['arg1', 'arg2'], None, None, None) 

The other results are the name of the *args and **kwargs variables, and the defaults provided. ie.

>>> def foo(a, b, c=4, *arglist, **keywords): pass >>> inspect.getfullargspec(foo) (['a', 'b', 'c'], 'arglist', 'keywords', (4,)) 

Note that some callables may not be introspectable in certain implementations of Python. For Example, in CPython, some built-in functions defined in C provide no metadata about their arguments. As a result, you will get a ValueError if you use inspect.getfullargspec() on a built-in function.

Since Python 3.3, you can use inspect.signature() to see the call signature of a callable object:

>>> inspect.signature(foo) <Signature (a, b, c=4, *arglist, **keywords)> 

Answer 2

In CPython, the number of arguments is

a_method.func_code.co_argcount 

and their names are in the beginning of

a_method.func_code.co_varnames 

These are implementation details of CPython, so this probably does not work in other implementations of Python, such as IronPython and Jython.

One portable way to admit "pass-through" arguments is to define your function with the signature func(*args, **kwargs). This is used a lot in e.g. matplotlib, where the outer API layer passes lots of keyword arguments to the lower-level API.