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 h

Answer 1

In python 3, below is to make *args and **kwargs into a dict (use OrderedDict for python < 3.6 to maintain dict orders):

from functools import wraps

def display_param(func):
    @wraps(func)
    def wrapper(*args, **kwargs):

        param = inspect.signature(func).parameters
        all_param = {
            k: args[n] if n < len(args) else v.default
            for n, (k, v) in enumerate(param.items()) if k != 'kwargs'
        }
        all_param .update(kwargs)
        print(all_param)

        return func(**all_param)
    return wrapper

Answer 2

To update a little bit Brian's answer, there is now a nice backport of inspect.signature that you can use in older python versions: funcsigs. So my personal preference would go for

try:  # python 3.3+
    from inspect import signature
except ImportError:
    from funcsigs import signature

def aMethod(arg1, arg2):
    pass

sig = signature(aMethod)
print(sig)

For fun, if you're interested in playing with Signature objects and even creating functions with random signatures dynamically you can have a look at my makefun project.

Answer 3

What about dir() and vars() now?

Seems doing exactly what is being asked super simply…

Must be called from within the function scope.

But be wary that it will return all local variables so be sure to do it at the very beginning of the function if needed.

Also note that, as pointed out in the comments, this doesn't allow it to be done from outside the scope. So not exactly OP's scenario but still matches the question title. Hence my answer.