191

Sometimes it seems natural to have a default parameter which is an empty list. Yet Python produces unexpected behavior in these situations.

If for example, I have a function:

def my_func(working_list=[]):
    working_list.append("a")
    print(working_list)

The first time it is called, the default will work, but calls after that will update the existing list (with one "a" each call) and print the updated version.

So, what is the Pythonic way to get the behavior I desire (a fresh list on each call)?

Josh Correia
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John Mulder
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  • The same behavior happens for sets, although you need a slightly more complicated example for it to show up as a bug. – abeboparebop Nov 19 '15 at 12:28
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    As links die, let me explicitly point out that this is desired behaviour. Default variables are evaluated at function definition (which happens the first time it is called), and NOT each time the function is called. Consequently, if you mutate a mutable default argument any subsequent function call can only use the mutated object. – Moritz Jul 03 '18 at 16:50

9 Answers9

217
def my_func(working_list=None):
    if working_list is None: 
        working_list = []

    # alternative:
    # working_list = [] if working_list is None else working_list

    working_list.append("a")
    print(working_list)

The docs say you should use None as the default and explicitly test for it in the body of the function.

Nico Schlömer
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HenryR
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    Is it better to say: if working_list == None: or if working_list: ?? – John Mulder Dec 14 '08 at 11:31
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    This is the preferred way to do it in python, even if I don't like it since it's ugly. I'd say the best practice would be "if working_list is None". – e-satis Dec 14 '08 at 12:35
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    The preferred way in this example is to say: if working_list is None . The caller might have used an empty list-like object with a custom append. – tzot Dec 14 '08 at 12:45
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    Mohit Ranka: beware that not working_list is True if its length is 0. This leads to a inconsistent behaviour : if the functions receives a list with some element in it, its caller will have his list updated, and if the list is empty, it won't be touched. – vincent Dec 14 '08 at 13:21
  • This idiom helps in 2 cases: when you mutate an argument, and when you want the default to be determined at the moment your function is called. The latter is even more important. – Beni Cherniavsky-Paskin Sep 04 '11 at 21:40
  • Missing variable arguments are initialized as `()` and so in cases where no variable argument is given, `if args is None` will not work whereas `if not args` will. So what's the pythonic way in this case? Thanks. Here an example: https://pastebin.com/KnfN1iY1 – PatrickT Jun 07 '20 at 12:14
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    @PatrickT The right tool depends on the case — a varargs function is _very_ different from one that takes an (optional) list argument. Situations where you'd have to choose between them come up less often than you'd think. Varargs is great when the arg count changes, but is fixed when the code is WRITTEN. Like your example. If it's runtime-variable, or you wanted to call `f()` on a list, you'd have to call `f(*l)` which is gross. Worse, implementing `mate(['larch', 'finch', 'robin'], ['bumble', 'honey', 'queen'])` would SUCK w/ varargs. Much better if it's `def mate(birds=[], bees=[]):`. – FeRD Jun 11 '20 at 16:56
53

Other answers have already already provided the direct solutions as asked for, however, since this is a very common pitfall for new Python programmers, it's worth adding the explanation of why Python behaves this way, which is nicely summarized in The Hitchhikers Guide to Python under Mutable Default Arguments:

Python's default arguments are evaluated once when the function is defined, not each time the function is called (like it is in say, Ruby). This means that if you use a mutable default argument and mutate it, you will and have mutated that object for all future calls to the function as well.

Boris Verkhovskiy
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Zhenhua
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18

Not that it matters in this case, but you can use object identity to test for None:

if working_list is None: working_list = []

You could also take advantage of how the boolean operator or is defined in python:

working_list = working_list or []

Though this will behave unexpectedly if the caller gives you an empty list (which counts as false) as working_list and expects your function to modify the list he gave it.

bendin
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13

If the intent of the function is to modify the parameter passed as working_list, see HenryR's answer (=None, check for None inside).

But if you didn't intend to mutate the argument, just use it as starting point for a list, you can simply copy it:

def myFunc(starting_list = []):
    starting_list = list(starting_list)
    starting_list.append("a")
    print starting_list

(or in this simple case just print starting_list + ["a"] but I guess that was just a toy example)

In general, mutating your arguments is bad style in Python. The only functions that are fully expected to mutate an object are methods of the object. It's even rarer to mutate an optional argument — is a side effect that happens only in some calls really the best interface?

  • If you do it from the C habit of "output arguments", that's completely unnecessary - you can always return multiple values as a tuple.

  • If you do this to efficiently build a long list of results without building intermediate lists, consider writing it as a generator and using result_list.extend(myFunc()) when you are calling it. This way your calling conventions remains very clean.

One pattern where mutating an optional arg is frequently done is a hidden "memo" arg in recursive functions:

def depth_first_walk_graph(graph, node, _visited=None):
    if _visited is None:
        _visited = set()  # create memo once in top-level call

    if node in _visited:
        return
    _visited.add(node)
    for neighbour in graph[node]:
        depth_first_walk_graph(graph, neighbour, _visited)
Beni Cherniavsky-Paskin
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4

I might be off-topic, but remember that if you just want to pass a variable number of arguments, the pythonic way is to pass a tuple *args or a dictionary **kargs. These are optional and are better than the syntax myFunc([1, 2, 3]).

If you want to pass a tuple:

def myFunc(arg1, *args):
  print args
  w = []
  w += args
  print w
>>>myFunc(1, 2, 3, 4, 5, 6, 7)
(2, 3, 4, 5, 6, 7)
[2, 3, 4, 5, 6, 7]

If you want to pass a dictionary:

def myFunc(arg1, **kargs):
   print kargs
>>>myFunc(1, option1=2, option2=3)
{'option2' : 2, 'option1' : 3}
Mapad
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1

Quote from https://docs.python.org/3/reference/compound_stmts.html#function-definitions

Default parameter values are evaluated from left to right when the function definition is executed. This means that the expression is evaluated once, when the function is defined, and that the same “pre-computed” value is used for each call. This is especially important to understand when a default parameter is a mutable object, such as a list or a dictionary: if the function modifies the object (e.g. by appending an item to a list), the default value is in effect modified. This is generally not what was intended. A way around this is to use None as the default, and explicitly test for it in the body of the function, e.g.:

def whats_on_the_telly(penguin=None):
    if penguin is None:
        penguin = []
    penguin.append("property of the zoo")
    return penguin
Norman
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0

There have already been good and correct answers provided. I just wanted to give another syntax to write what you want to do which I find more beautiful when you for instance want to create a class with default empty lists:

class Node(object):
    def __init__(self, _id, val, parents=None, children=None):
        self.id = _id
        self.val = val
        self.parents = parents if parents is not None else []
        self.children = children if children is not None else []

This snippet makes use of the if else operator syntax. I like it especially because it's a neat little one-liner without colons, etc. involved and it nearly reads like a normal English sentence. :)

In your case you could write

def myFunc(working_list=None):
    working_list = [] if working_list is None else working_list
    working_list.append("a")
    print working_list
drssdinblck
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0

Perhaps the simplest thing of all is to just create a copy of the list or tuple within the script. This avoids the need for checking. For example,

    def my_funct(params, lst = []):
        liste = lst.copy()
         . . 
24b4Jeff
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-4

I took the UCSC extension class Python for programmer

Which is true of: def Fn(data = []):

a) is a good idea so that your data lists start empty with every call.

b) is a good idea so that all calls to the function that do not provide any arguments on the call will get the empty list as data.

c) is a reasonable idea as long as your data is a list of strings.

d) is a bad idea because the default [] will accumulate data and the default [] will change with subsequent calls.

Answer:

d) is a bad idea because the default [] will accumulate data and the default [] will change with subsequent calls.

Community
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Peter Chen
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