optional arguments of functions in python

Question

I'm trying to write slicing code which gets a linked list, start,stop, step. My code is supposed to behave like same as using list[start:step:stop].

My problem starts when the user insert only 1 argument (let's says it was x) - then, x is supposed to get into stop and and not start. But, I have been told that optional arguments need to appear at the end of all the arguments.

Can someone tell me how can we insert only 1 input to the second argument while the first one is mandatory but the second one is not? By the way, i can't use built-in functions

Try to use Keyword arguements, where you can specify key=value type of arguements, and limit your positional/mandatory arguement inside your function/method — Siva Cn
– Siva Cn, Commented Dec 29, 2013 at 16:07
I can't use built-in functions.. that is the idea of the assigment. I will add this info to the description — user3130794
– user3130794, Commented Dec 29, 2013 at 16:07

Cameron White · Accepted Answer · 2013-12-29 16:42:39Z

3

You could write a LinkedList class that defines the __getitem__ function to get access to python's notation.

class LinkedList:

    # Implement the Linked ...

    def __getitem__(self, slice):

        start = slice.start
        stop = slice.stop
        step = slice.step

        # Implement the function

Then you can use LinkedList like you want

l = LinkedList()
l[1]
l[1:10]
l[1:10:2]

answered Dec 29, 2013 at 16:42

Cameron White

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Comments

henrebotha · Accepted Answer · 2013-12-29 17:01:28Z

1

Using optional (named) arguments:

def foo(start, stop=None, step=1):
    if stop == None:
        start, stop = 0, start
    #rest of the code goes here

Then foo(5) == foo(0,5,1), but foo(1,5) == foo(1,5,1). I think this works, anyway... :)

edited Dec 29, 2013 at 17:01

answered Dec 29, 2013 at 16:13

henrebotha

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1 Comment

kindall Over a year ago

I'd use None as sentinel, this will make things much clearer.

Christian Tapia · Accepted Answer · 2013-12-29 16:16:10Z

1

You could try:

def myF(*args):
    number_args = len(args)
    if number_args == 1:
        stop = ...
    elif number_args == 2:
        ...
    elif number_args == 3:
        ...
    else
        print "Error"

*args means that the arguments passed to the function myF will be stored in the variable args.

edited Dec 29, 2013 at 16:16

answered Dec 29, 2013 at 16:09

Christian Tapia

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7 Comments

Christian Tapia Over a year ago

The number of arguments is len(args). Read more about this on Internet. @volcano this is just an example.

Siva Cn Over a year ago

What If he passes 2 arguments and the very first one is supposed to be mandatory ???

volcano Over a year ago

@Crhistian, read about DRY on inetrnet. Even example should demonstrate efficient code

Christian Tapia Over a year ago

@volcano edited. @ Siva Cn I don't understand, maybe my bad English, can you give an example?

user3130794 Over a year ago

incredible.Didn't know I could do that. Thanks!

|

Siva Cn · Accepted Answer · 2013-12-29 16:12:41Z

0

def func(arg1=None, arg2=None, arg3=None):
    if not arg1:
        print 'arg1 is mandatory'
        return False
    ## do your stuff here

arg1 = 1
arg2 = 4
arg3 = 1

## works fine
func(arg1=arg1, arg2=arg2, arg3=arg3)

## Evaluated as False, Since the first argument is missing
func(arg2=arg2, arg3=arg3)

answered Dec 29, 2013 at 16:12

Siva Cn

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