Can a variable number of arguments be passed to a function?

Question

In a similar way to using varargs in C or C++:

fn(a, b)
fn(a, b, c, d, ...)

I refer the honorable lottness to podcast 53: itc.conversationsnetwork.org/shows/… — David Sykes
– David Sykes, Commented May 28, 2009 at 11:10
I gotta go with Mr Lott on this one. You can quickly get an authorative answer on this one in the Python docs, plus you'll get a feel for what else is there in the docs. It is to your benefit to get to know those docs if you plan on working in Python. — Brian Neal
– Brian Neal, Commented May 28, 2009 at 19:41
The episode #53 "let's go Rio" cannot be the one mentioned by @DavidSykes in 2009, since it was recorded in 2013. Strangely, there was another #53 also recorded in 2011, also after David's comment: open.spotify.com/episode/… — ouranos
– ouranos, Commented Mar 1, 2023 at 9:22
@ouranos Could it be this #53? stackoverflow.blog/podcast/page/46 — David Sykes
– David Sykes, Commented Mar 2, 2023 at 16:35

Nicolas Gervais · Accepted Answer · 2020-03-15 20:27:37Z

563

Yes. You can use *args as a non-keyword argument. You will then be able to pass any number of arguments.

def manyArgs(*arg):
  print "I was called with", len(arg), "arguments:", arg

>>> manyArgs(1)
I was called with 1 arguments: (1,)
>>> manyArgs(1, 2, 3)
I was called with 3 arguments: (1, 2, 3)

As you can see, Python will unpack the arguments as a single tuple with all the arguments.

For keyword arguments you need to accept those as a separate actual argument, as shown in Skurmedel's answer.

edited Mar 15, 2020 at 20:27

Nicolas Gervais

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answered May 28, 2009 at 7:51

unwind

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5 Comments

wilbbe01 Over a year ago

Also important...one may find a time when they have to pass an unknown number of arguments to a function. In a case like this call your "manyArgs" by creating a list called "args" and passing that to manyArgs like this "manyArgs(*args)"

Eric O. Lebigot Over a year ago

This is close, but this is unfortunately not general enough: manyArgs(x = 3) fails with TypeError. Skumedel's answer shows the solution to this. The key point is that the general signature of a function is f(*list_args, **keyword_args) (not f(*list_args)).

RoboAlex Over a year ago

The type of args is tuple. kwargs means keyword args. The type of kwargs is dictionary.

MasterControlProgram Over a year ago

Just for arg in args: for iterating through passed args

uuu777 Over a year ago

I have to pass variable number of integer parameters followed by optional string, can I call use xxx(*arg: int, log: str = "") ? The variety in the number of parameters is small 0, 1, 2 or 3, so I can use overloads but it seems like it is not not available in python.

Skurmedel · Accepted Answer · 2013-08-16 09:02:37Z

269

Adding to unwinds post:

You can send multiple key-value args too.

def myfunc(**kwargs):
    # kwargs is a dictionary.
    for k,v in kwargs.iteritems():
         print "%s = %s" % (k, v)

myfunc(abc=123, efh=456)
# abc = 123
# efh = 456

And you can mix the two:

def myfunc2(*args, **kwargs):
   for a in args:
       print a
   for k,v in kwargs.iteritems():
       print "%s = %s" % (k, v)

myfunc2(1, 2, 3, banan=123)
# 1
# 2
# 3
# banan = 123

They must be both declared and called in that order, that is the function signature needs to be *args, **kwargs, and called in that order.

edited Aug 16, 2013 at 9:02

answered May 28, 2009 at 7:59

Skurmedel

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6 Comments

Dannid Over a year ago

Not sure how you got myfunc(abc=123, def=456) to work, but in mine (2.7), 'def' can't be passed in this function without getting a SyntaxError. I assume this is because def has meaning in python. Try myfunc(abc=123,fgh=567) instead. (Otherwise, great answer and thanks for it!)

Skurmedel Over a year ago

@Dannid: No idea either haha... doesn't work on 2.6 or 3.2 either. I'll rename it.

Nikhil Prabhu Over a year ago

When you say 'called in that order', do you mean passed in that order? Or something else?

Skurmedel Over a year ago

@NikhilPrabhu: Yeah. Keyword arguments must come last when you call it. They always come last if you have non-keyword arguments in the call too.

MasterControlProgram Over a year ago

For Python 3: Just exchange print a with print(a), and kwargs.iteritems(): with kwargs.items().

|

wizzwizz4 · Accepted Answer · 2017-12-10 21:02:55Z

28

If I may, Skurmedel's code is for python 2; to adapt it to python 3, change iteritems to items and add parenthesis to print. That could prevent beginners like me to bump into: AttributeError: 'dict' object has no attribute 'iteritems' and search elsewhere (e.g. Error “ 'dict' object has no attribute 'iteritems' ” when trying to use NetworkX's write_shp()) why this is happening.

def myfunc(**kwargs):
for k,v in kwargs.items():
   print("%s = %s" % (k, v))

myfunc(abc=123, efh=456)
# abc = 123
# efh = 456

and:

def myfunc2(*args, **kwargs):
   for a in args:
       print(a)
   for k,v in kwargs.items():
       print("%s = %s" % (k, v))

myfunc2(1, 2, 3, banan=123)
# 1
# 2
# 3
# banan = 123

edited Dec 10, 2017 at 21:02

wizzwizz4

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answered Feb 23, 2017 at 5:29

calocedrus

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Comments

Vladtn · Accepted Answer · 2012-02-26 01:02:05Z

13

Adding to the other excellent posts.

Sometimes you don't want to specify the number of arguments and want to use keys for them (the compiler will complain if one argument passed in a dictionary is not used in the method).

def manyArgs1(args):
  print args.a, args.b #note args.c is not used here

def manyArgs2(args):
  print args.c #note args.b and .c are not used here

class Args: pass

args = Args()
args.a = 1
args.b = 2
args.c = 3

manyArgs1(args) #outputs 1 2
manyArgs2(args) #outputs 3

Then you can do things like

myfuns = [manyArgs1, manyArgs2]
for fun in myfuns:
  fun(args)

edited Feb 26, 2012 at 1:02

answered Feb 26, 2012 at 0:56

Vladtn

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Comments

Rob · Accepted Answer · 2013-05-28 07:31:33Z

1

def f(dic):
    if 'a' in dic:
        print dic['a'],
        pass
    else: print 'None',

    if 'b' in dic:
        print dic['b'],
        pass
    else: print 'None',

    if 'c' in dic:
        print dic['c'],
        pass
    else: print 'None',
    print
    pass
f({})
f({'a':20,
   'c':30})
f({'a':20,
   'c':30,
   'b':'red'})
____________

the above code will output

None None None
20 None 30
20 red 30

This is as good as passing variable arguments by means of a dictionary

edited May 28, 2013 at 7:31

Rob

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answered May 28, 2013 at 7:00

B N Venkataraman

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1 Comment

metaperture Over a year ago

This is terrible code. Far better would be: f = lambda **dic: ' '.join(dic.get(key, 'None') for key in 'abc')

transversality condition · Accepted Answer · 2020-05-19 13:05:30Z

1

Another way to go about it, besides the nice answers already mentioned, depends upon the fact that you can pass optional named arguments by position. For example,

def f(x,y=None):
    print(x)
    if y is not None:
        print(y)

Yields

In [11]: f(1,2)
1
2

In [12]: f(1)
1

answered May 19, 2020 at 13:05

transversality condition

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1 Comment

Stephen Ellwood Over a year ago

Since the question implies a variable number of items this answer is not really applicable. More generally you can mix a fixed number items passed by position and a variable number of arguments pass by value. e.g. def test(a b, **kwargs).

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