How to get method parameter names?

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)>

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.

The Python 3 version is:

def _get_args_dict(fn, args, kwargs):    args_names = fn.__code__.co_varnames[:fn.__code__.co_argcount]    return {**dict(zip(args_names, args)), **kwargs}

The method returns a dictionary containing both args and kwargs.