Creating a function that takes a list or integer as argument (Python)

Question

I'm trying to create a function that gives the average of either a list of numbers or just integers as variables. So far I have:

def average(*args):
    if type(args) is list:
        for x in args:
            print sum(x) / float(len(x))
    else:
        for x in args:
            args = list(args)
            print sum(x) / float(len(x))

When I input a list, like

average([1, 3, 5, 2])

it works great. But when I enter in

average(1, 3, 5, 2)

it gives "TypeError: 'int' object is not iterable". I've checked other questions but none of the solutions seem to work. I've tried to check if it's a list with type() and isinstance() but whenever I get one of them to work, the other throws out an error.

Why are you attempting to sum(x) and len(x) passing a simple int? Furthermore, try to add print(type(args)) and you will discover that is a tuple not a list — gmas80
– gmas80, Commented Oct 26, 2014 at 23:24

Padraic Cunningham · Accepted Answer · 2014-10-26 23:31:16Z

6

args is a tuple so check if args[0] is a list the sum the contents of args[0], if just ints are passed in just sum args:

def average(*args):
    if isinstance(args[0],list):
        print(sum(args[0]) / float(len(args[0])))
    else:
        print (sum(args) / float(len(args)))


In [2]: average(1, 3, 5, 2)
2.75

In [3]: average([1, 3, 5, 2])
2.75

If you want to accept tuples,use collections.Iterable:

from collections import Iterable
def average(*args):
    if isinstance(args[0],Iterable):
        print(sum(args[0]) / float(len(args[0])))
    else:
        print (sum(args) / float(len(args)))

In [5]: average([1, 3, 5, 2])
2.75

In [6]: average(1, 3, 5, 2)
2.75

In [7]: average((1, 3, 5, 2))
2.75

edited Oct 26, 2014 at 23:31

answered Oct 26, 2014 at 23:26

Padraic Cunningham

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Comments

ch3ka · Accepted Answer · 2014-10-26 23:16:10Z

1

the second print sum(x) / float(len(x)) calls len() on x, which is an integer.

I think you mean something like:

  else:
        print sum(args) / float(len(args))

answered Oct 26, 2014 at 23:16

ch3ka

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Comments

Weafs.py · Accepted Answer · 2014-10-26 23:30:20Z

1

Demo on repl.it

def average(*args):
    if type(args) is tuple:
        r = 0
        for x in args:
            for y in x:
                r += y
        print r / float(len(x))
    else:
        print sum(args) / float(len(args))

average([2, 3, 8, 1, 9])

edited Oct 26, 2014 at 23:30

answered Oct 26, 2014 at 23:16

Weafs.py

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3 Comments

sfowler12 Over a year ago

When I do it this way, it accepts the "average()" argument but not the "average([])" list argument.

gmas80 Over a year ago

Try to modify to read: if type(args) is tuple: rather than if type(args) is list: :)

sfowler12 Over a year ago

Lol now it works for "average([])" and not "average()". This has been the bane of my past four hours.

Pabitra Pati · Accepted Answer · 2014-10-27 09:34:26Z

The if portion in your code i.e. if type(args) is list: is never going to be executed. Because, whenever non-keyword arguments are passed to a function using *args, the value received as args is always a tuple. So type(args) is list is always going to be False. What needs to be done is iterate through the arguments passed through *args and check if the type of argument, if it's list then you have passed a list i.e. average([1,3,5,7]) else it will be integer when integers are passed i.e. average(1,3,5,7). So following piece of code will work :-

def average(*args):
    for x in args:
        if type(x) == list:
            # List is passed
            return sum(x) / float(len(x))
    # Assuming inputs passed restricted to list and numbers
    # Numbers are passed as argument
    return sum(args) / float(len(args))

Here this will provide you the output for both types of input :-

>>> average([1,3,5,7])
4.0
>>> average(1,3,5,7)
4.0

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