Mutable Default Arguments in Python

Introduction

Examples use python3 for python2 use func_defaults

We frequently use [] as default parameter.

>>> def foo(a=[]):
...   pass
... 
>>> foo.__defaults__
([],)
>>> 

But this is not a good practice as this lead to debugging rabbit hole

Understanding Why ?

The Default parameters are initialized during function definition time which occur at either module load time or during program execution. As we can see in the above example the foo function was defined and it’s default parameter shows the value of list which is empty ([],).

def foo(items, seq=[]):
  for item in items:
    if item >= 0:
      seq.append(item)
  return seq

Till now the only item in the __defaults__ is an empty list as can be seen

print(foo.__defaults__) # ([],)

After first Invocation

print(foo([2])) # [2]

print(foo.__defaults__) # ([2],)

From the output we can see that now the __defaults__ doesn’t shows that the default for seq which should be an empty list and this leads to error on subsequent invocations

print(foo([3])) # [2, 3]

print(foo([-1])) # [2, 3]

Why this happens because the list, being a mutable type, is still the same instance defined during function creation, but it is now populated from the interaction of the previous invocation.

General Solution

Shift the creation from definition time to runtime.

A Common idiom is to have the parameter default to None then check for that value in the body of the function

def foo(items,seq = None):
  seq = seq or []
  for item in items:
    if item >= 0:
      seq.append(item)
  return seq


print(foo([2])) #[2]
print(foo([3])) #[3]
print(foo([-1])) #[]

Point to Remember : Don’t use mutable types as default parameters