How can I create a Python function dynamically which takes a specified number of arguments?

Question

I create a function dynamically this way:

def create_function(value):
    def _function():
        print value
return _function

f1 = create_func(1)
f1()

which works fine and prints '1'.

but my problem is slightly different, say there is a variable called no_of_arguments which contains the number of arguments the function that is being returned (_function() ) takes.

def create_function():
    no_of_arguments = int(raw_input()) #provided by user
    def _function(a,b,c,....):

'this function has to accept a certain number of arguments, specified in the variable no_of_arguments'

        #do something here
return _function

f1 = create_func()
f1(a,b,c......)

I don't understand what you really want, but there are *args and **kwargs keywords for functions with arbitrary parameters in Python. May be they will be helpfull. — Denis Nikanorov
– Denis Nikanorov, Commented Jul 2, 2013 at 7:30

Ashwini Chaudhary · Accepted Answer · 2013-07-02 07:29:47Z

1

Use * in the function arguments to make it accept any number of positional arguments.

def func(*args):
    if len(args) == 1:
       print args[0]
    else:
       print args
...        
>>> func(1)
1
>>> func(1,2)
(1, 2)
>>> func(1,2,3,4)
(1, 2, 3, 4)

answered Jul 2, 2013 at 7:29

Ashwini Chaudhary

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Comments

Ignacio Vazquez-Abrams · Accepted Answer · 2013-07-02 07:29:20Z

0

A function can be defined as taking any (minimum) number of arguments by preceding one with a *, which will then result in the name being bound to a tuple containing the appropriate arguments.

def foo(a, b, *c):
  print a, b, c

foo(1, 2, 3, 4, 5)

You will need to limit/check the number of values passed this way yourself though.

answered Jul 2, 2013 at 7:29

Ignacio Vazquez-Abrams

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Comments

TerryA · Accepted Answer · 2013-07-02 07:29:56Z

0

You could use *args:

def create_function():
    no_of_arguments = int(raw_input()) #provided by user
    def _function(*args):
        if len(args) == no_of_arguments:
            dostuff()
        else:
            print "{} arguments were not given!".format(no_of_arguments)
    return _function

Running it as an example:

>>> f1 = create_function()
4 # The input
>>> f1('hi','hello','hai','cabbage')
>>> f1('hey')
4 arguments were not given!

answered Jul 2, 2013 at 7:29

TerryA

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iklubov · Accepted Answer · 2013-07-02 07:36:35Z

0

As far as i understood, you need to pass different number of arguments to the function You can pass different number of arguments using * like that:

def create_function():
    no_of_arguments = (argList) #tuple of arguments
    def _function(*argList):

answered Jul 2, 2013 at 7:36

iklubov

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