How can I find the number of arguments of a Python function?

import inspect
inspect.getargspec(someMethod)

see the inspect module

inspect.getargspec()

Get the names and default values of a function’s arguments. A tuple of four things is returned: (args, varargs, varkw, defaults). args is a list of the argument names (it may contain nested lists). varargs and varkw are the names of the * and ** arguments or None. defaults is a tuple of default argument values or None if there are no default arguments; if this tuple has n elements, they correspond to the last n elements listed in args.

Changed in version 2.6: Returns a named tuple ArgSpec(args, varargs, keywords, defaults).

See can-you-list-the-keyword-arguments-a-python-function-receives.

someMethod.func_code.co_argcount

or, if the current function name is undetermined:

import sys

sys._getframe().func_code.co_argcount

Adding to the above, I’ve also seen that the most of the times help() function really helps

For eg, it gives all the details about the arguments it takes.

help(<method>)

gives the below

method(self, **kwargs) method of apiclient.discovery.Resource instance
Retrieves a report which is a collection of properties / statistics for a specific customer.

Args:
  date: string, Represents the date in yyyy-mm-dd format for which the data is to be fetched. (required)
  pageToken: string, Token to specify next page.
  parameters: string, Represents the application name, parameter name pairs to fetch in csv as app_name1:param_name1, app_name2:param_name2.

Returns:
  An object of the form:

    { # JSON template for a collection of usage reports.
    "nextPageToken": "A String", # Token for retrieving the next page
    "kind": "admin#reports#usageReports", # Th

As other answers suggest, getargspec works well as long as the thing being queried is actually a function. It does not work for built-in functions such as open, len, etc, and will throw an exception in such cases:

TypeError: <built-in function open> is not a Python function

The below function (inspired by this answer) demonstrates a workaround. It returns the number of args expected by f:

from inspect import isfunction, getargspec
def num_args(f):
  if isfunction(f):
    return len(getargspec(f).args)
  else:
    spec = f.__doc__.split('n')[0]
    args = spec[spec.find('(')+1:spec.find(')')]
    return args.count(',')+1 if args else 0

The idea is to parse the function spec out of the __doc__ string. Obviously this relies on the format of said string so is hardly robust!

inspect.getargspec() to meet your needs

from inspect import getargspec

def func(a, b):
    pass
print len(getargspec(func).args)

Good news for folks who want to do this in a portable way between Python 2 and Python 3.6+: use inspect.getfullargspec() method. It works in both Python 2.x and 3.6+

As Jim Fasarakis Hilliard and others have pointed out, it used to be like this:
1. In Python 2.x: use inspect.getargspec()
2. In Python 3.x: use signature, as getargspec() and getfullargspec() were deprecated.

However, starting Python 3.6 (by popular demand?), things have changed towards better:

From the Python 3 documentation page:

inspect.getfullargspec(func)

Changed in version 3.6: This method was previously documented as deprecated in favour of signature() in Python 3.5, but that decision has been reversed in order to restore a clearly supported standard interface for single-source Python 2/3 code migrating away from the legacy getargspec() API.

In:

import inspect 

class X:
    def xyz(self, a, b, c): 
        return 

print(len(inspect.getfullargspec(X.xyz).args))

Out:

4

Note: If xyz wasn’t inside class X and had no "self" and just "a, b, c", then it would have printed 3.

For python below 3.5, you may want to replace inspect.getfullargspec by inspect.getargspec in the code above.

func.__code__.co_argcount gives you the number of any arguments BEFORE *args

func.__kwdefaults__ gives you a dict of the keyword arguments AFTER *args

func.__code__.co_kwonlyargcount is equal to len(func.__kwdefaults__)

func.__defaults__ gives you the values of optional arguments that appears before *args

Here is a simple illustration:

>>> def a(b, c, d, e, f=1, g=3, h=None, *i, j=2, k=3, **L):
    pass

>>> a.__code__.co_argcount
7
>>> a.__defaults__
(1, 3, None)
>>> len(a.__defaults__)
3
>>> 
>>> 
>>> a.__kwdefaults__
{'j': 2, 'k': 3}
>>> len(a.__kwdefaults__)
2
>>> a.__code__.co_kwonlyargcount
2

The accepted answer by Dimitris Fasarakis Hilliard suggests getting parameters in the string format but I think one can make a mistake when parsing this string and thus I created rather a list of the parameters directly using the inspect module

import inspect
def my_function(a,b,c):
    #some code
    pass

result=list(inspect.signature(my_function).parameters.keys())
print(result)
['a','b','c']

Assuming you may be dealing with class based methods or simply functions, you could do something like the following.

This will automatically subtract one input if the input is a class method (and therefore includes self).

import types
def get_arg_count(fn):
    extra_method_input_count=1 if isinstance(fn, types.MethodType) else 0
    return fn.__code__.co_argcount-extra_method_input_count

Then you can apply as you need to functions or methods:

def fn1(a, b, c):
    return None

class cl1:
    def fn2(self, a, b, c):
        return None

print(get_arg_count(fn1)) #=> 3
print(get_arg_count(cl1().fn2)) #=> 3

You get the number of arguments by (replace "function" by the name of your function):

function.__code__.co_argcount ## 2

And the names for the arguments by:

function.__code__.co_varnames ## ('a', 'b')

This is a solution to getting the number of mandatory arguments of a function (*)

Many of the solutions proposed here do not work for this purpose if some more uncommon parameter specifications are used (positional-only parameters with defaults, keyword-only parameters without defaults, etc.)

from typing import Callable, Any
import inspect

def get_mandatory_argcount(f: Callable[..., Any]) -> int:
    """Get the number of mandatory arguments of a function."""
    sig = inspect.signature(f)
    
    def parameter_is_mandatory(p: inspect.Parameter) -> bool:
        return p.default is inspect.Parameter.empty and p.kind not in (
            inspect.Parameter.VAR_POSITIONAL,
            inspect.Parameter.VAR_KEYWORD,
        )
    
    return sum(parameter_is_mandatory(p) for p in sig.parameters.values())

# mandatory keyword-only
def f1(b=2, *args, c, d=1, **kwds): pass
print(get_mandatory_argcount(f1))

# positional only with default
def f2(a=1, /, b=3, *args, **kwargs): pass
print(get_mandatory_argcount(f2))

(*) I would have liked to put this as an answer to Programmatically determining amount of parameters a function requires – Python instead, but for some reason this question is marked as duplicate to this one despite it asking specifically about the number of required arguments whereas this question only asks about the general number of arguments.