Call a function with argument list in python

Question

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.

Answer 1

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

http://stackoverflow.com/questions/287085/what-does-args-and-kwargs-mean

And from the official python tutorial

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

Answer 2

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)

Answer 3

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].

Answer 4

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 *.

Answer 5

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 wraper 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 funther 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 arter 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