How can I find the number of arguments of a Python function? I need to know how many normal arguments it has and how many named arguments.

Example:

def someMethod(self, arg1, kwarg1=None):
    pass

This method has 2 arguments and 1 named argument.

The previously accepted answer has been deprecated as of Python 3.0. Instead of using inspect.getargspec you should now opt for the Signature class which superseded it.

Creating a Signature for the function is easy via the signature function:

from inspect import signature

def someMethod(self, arg1, kwarg1=None):
    pass

sig = signature(someMethod)

Now, you can either view its parameters quickly by string it:

str(sig)  # returns: '(self, arg1, kwarg1=None)'

or you can also get a mapping of attribute names to parameter objects via sig.parameters.

params = sig.parameters 
print(params['kwarg1']) # prints: kwarg1=20

Additionally, you can call len on sig.parameters to also see the number of arguments this function requires:

print(len(params))  # 3

Each entry in the params mapping is actually a Parameter object that has further attributes making your life easier. For example, grabbing a parameter and viewing its default value is now easily performed with:

kwarg1 = params['kwarg1']
kwarg1.default # returns: None

similarly for the rest of the objects contained in parameters.

As for Python 2.x users, while inspect.getargspec isn’t deprecated, the language will soon be :-). The Signature class isn’t available in the 2.x series and won’t be. So you still need to work with inspect.getargspec.

As for transitioning between Python 2 and 3, if you have code that relies on the interface of getargspec in Python 2 and switching to signature in 3 is too difficult, you do have the valuable option of using inspect.getfullargspec. It offers a similar interface to getargspec (a single callable argument) in order to grab the arguments of a function while also handling some additional cases that getargspec doesn’t:

from inspect import getfullargspec

def someMethod(self, arg1, kwarg1=None):
    pass

args = getfullargspec(someMethod)

As with getargspec, getfullargspec returns a NamedTuple which contains the arguments.

print(args)
FullArgSpec(args=['self', 'arg1', 'kwarg1'], varargs=None, varkw=None, defaults=(None,), kwonlyargs=[], kwonlydefaults=None, annotations={})

import inspect
inspect.getargspec(someMethod)

see the inspect module

someMethod.func_code.co_argcount

or, if the current function name is undetermined:

import sys

sys._getframe().func_code.co_argcount

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.

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

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.

inspect.getargspec() to meet your needs

from inspect import getargspec

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

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!

func.__code__.co_argcount gives you 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 appear before *args

Here is the 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

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.

I’m trying to call a function inside another function in python, but can’t find the right syntax. What I want to do is something like this:

def wrapper(func, args):
    func(args)

def func1(x):
    print(x)

def func2(x, y, z):
    return x+y+z

wrapper(func1, [x])
wrapper(func2, [x, y, z])

In this case first call will work, and second won’t. What I want to modify is the wrapper function and not the called functions.

To expand a little on the other answers:

In the line:

def wrapper(func, *args):

The * next to args means “take the rest of the parameters given and put them in a list called args“.

In the line:

    func(*args)

The * next to args here means “take this list called args and ‘unwrap’ it into the rest of the parameters.

So you can do the following:

def wrapper1(func, *args): # with star
    func(*args)

def wrapper2(func, args): # without star
    func(*args)

def func2(x, y, z):
    print x+y+z

wrapper1(func2, 1, 2, 3)
wrapper2(func2, [1, 2, 3])

In wrapper2, the list is passed explicitly, but in both wrappers args contains the list [1,2,3].

The simpliest way to wrap a function

    func(*args, **kwargs)

… is to manually write a wrapper that would call func() inside itself:

    def wrapper(*args, **kwargs):
        # do something before
        try:
            return func(*a, **kwargs)
        finally:
            # do something after

In Python function is an object, so you can pass it’s name as an argument of another function and return it. You can also write a wrapper generator for any function anyFunc():

    def wrapperGenerator(anyFunc, *args, **kwargs):
        def wrapper(*args, **kwargs):
            try:
                # do something before
                return anyFunc(*args, **kwargs)
            finally:
                #do something after
        return wrapper

Please also note that in Python when you don’t know or don’t want to name all the arguments of a function, you can refer to a tuple of arguments, which is denoted by its name, preceded by an asterisk in the parentheses after the function name:

    *args

For example you can define a function that would take any number of arguments:

    def testFunc(*args):
        print args    # prints the tuple of arguments

Python provides for even further manipulation on function arguments. You can allow a function to take keyword arguments. Within the function body the keyword arguments are held in a dictionary. In the parentheses after the function name this dictionary is denoted by two asterisks followed by the name of the dictionary:

    **kwargs

A similar example that prints the keyword arguments dictionary:

    def testFunc(**kwargs):
        print kwargs    # prints the dictionary of keyword arguments

You can use *args and **kwargs syntax for variable length arguments.

What do *args and **kwargs mean?

And from the official python tutorial

http://docs.python.org/dev/tutorial/controlflow.html#more-on-defining-functions

The literal answer to your question (to do exactly what you asked, changing only the wrapper, not the functions or the function calls) is simply to alter the line

func(args)

to read

func(*args)

This tells Python to take the list given (in this case, args) and pass its contents to the function as positional arguments.

This trick works on both “sides” of the function call, so a function defined like this:

def func2(*args):
    return sum(args)

would be able to accept as many positional arguments as you throw at it, and place them all into a list called args.

I hope this helps to clarify things a little. Note that this is all possible with dicts/keyword arguments as well, using ** instead of *.

You need to use arguments unpacking..

def wrapper(func, *args):
    func(*args)

def func1(x):
    print(x)

def func2(x, y, z):
    print x+y+z

wrapper(func1, 1)
wrapper(func2, 1, 2, 3)

A small addition to previous answers, since I couldn’t find a solution for a problem, which is not worth opening a new question, but led me here.

Here is a small code snippet, which combines lists, zip() and *args, to provide a wrapper that can deal with an unknown amount of functions with an unknown amount of arguments.

def f1(var1, var2, var3):
    print(var1+var2+var3)

def f2(var1, var2):
    print(var1*var2)

def f3():
    print('f3, empty')

def wrapper(a,b, func_list, arg_list):
    print(a)
    for f,var in zip(func_list,arg_list):
        f(*var)
    print(b)

f_list = [f1, f2, f3]
a_list = [[1,2,3], [4,5], []]

wrapper('begin', 'end', f_list, a_list)

Keep in mind, that zip() does not provide a safety check for lists of unequal length, see zip iterators asserting for equal length in python.

Is it possible to pass functions with arguments to another function in Python?

Say for something like:

def perform(function):
    return function()

But the functions to be passed will have arguments like:

action1()
action2(p)
action3(p,r)

Do you mean this?

def perform( fun, *args ):
    fun( *args )

def action1( args ):
    something

def action2( args ):
    something

perform( action1 )
perform( action2, p )
perform( action3, p, r )

This is what lambda is for:

def Perform(f):
    f()

Perform(lambda: Action1())
Perform(lambda: Action2(p))
Perform(lambda: Action3(p, r))

You can use the partial function from functools like so.

from functools import partial

def perform(f):
    f()

perform(Action1)
perform(partial(Action2, p))
perform(partial(Action3, p, r))

Also works with keywords

perform(partial(Action4, param1=p))

Use functools.partial, not lambdas! And ofc Perform is a useless function, you can pass around functions directly.

for func in [Action1, partial(Action2, p), partial(Action3, p, r)]:
  func()

(months later) a tiny real example where lambda is useful, partial not:
say you want various 1-dimensional cross-sections through a 2-dimensional function, like slices through a row of hills.
quadf( x, f ) takes a 1-d f and calls it for various x.
To call it for vertical cuts at y = -1 0 1 and horizontal cuts at x = -1 0 1,

fx1 = quadf( x, lambda x: f( x, 1 ))
fx0 = quadf( x, lambda x: f( x, 0 ))
fx_1 = quadf( x, lambda x: f( x, -1 ))
fxy = parabola( y, fx_1, fx0, fx1 )

f_1y = quadf( y, lambda y: f( -1, y ))
f0y = quadf( y, lambda y: f( 0, y ))
f1y = quadf( y, lambda y: f( 1, y ))
fyx = parabola( x, f_1y, f0y, f1y )

As far as I know, partial can’t do this —

quadf( y, partial( f, x=1 ))
TypeError: f() got multiple values for keyword argument 'x'

(How to add tags numpy, partial, lambda to this ?)

This is called partial functions and there are at least 3 ways to do this. My favorite way is using lambda because it avoids dependency on extra package and is the least verbose. Assume you have a function add(x, y) and you want to pass add(3, y) to some other function as parameter such that the other function decides the value for y.

Use lambda

# generic function takes op and its argument
def runOp(op, val):
    return op(val)

# declare full function
def add(x, y):
    return x+y

# run example
def main():
    f = lambda y: add(3, y)
    result = runOp(f, 1) # is 4

Create Your Own Wrapper

Here you need to create a function that returns the partial function. This is obviously lot more verbose.

# generic function takes op and its argument
def runOp(op, val):
    return op(val)

# declare full function
def add(x, y):
    return x+y

# declare partial function
def addPartial(x):
    def _wrapper(y):
        return add(x, y)
    return _wrapper

# run example
def main():
    f = addPartial(3)
    result = runOp(f, 1) # is 4

Use partial from functools

This is almost identical to lambda shown above. Then why do we need this? There are few reasons. In short, partial might be bit faster in some cases (see its implementation) and that you can use it for early binding vs lambda’s late binding.

from functools import partial

# generic function takes op and its argument
def runOp(op, val):
    return op(val)

# declare full function
def add(x, y):
    return x+y

# run example
def main():
    f = partial(add, 3)
    result = runOp(f, 1) # is 4

Here is a way to do it with a closure:

    def generate_add_mult_func(func):
        def function_generator(x):
            return reduce(func,range(1,x))
        return function_generator

    def add(x,y):
        return x+y

    def mult(x,y):
        return x*y

    adding=generate_add_mult_func(add)
    multiplying=generate_add_mult_func(mult)

    print adding(10)
    print multiplying(10)