Python argument values

Question

I am following the Python tutorial.

def f(a, L=[]):
 L.append(a)
 return L

When I call f

print(f(1))
print(f(2))
print(f(3, L=[]))
print(f(4))

I get

[1]
[1, 2]
[3]
[1, 2, 4]

Why it returns [1, 2, 4] after [3] instead of [3, 4]?

Answer 1

The problem is that you declare L=[] in the arguments. This creates the List that is assigned to L only once. For the third case where you have

print(f(3, L=[]))

you pass an empty list to the function for the parameter L. This does not overwrite L but is only valid for that specific function call. For the next call, it takes the default argument list again, which contains at that point in time [1,2].

It is (as far as I know) not possible to reassign the default parameter after initialization.

This question may also be of interest to you as it talks about mutable default arguments

Answer 2

If you want to get the same result you need to initialize a list and pass it to the function.

def f(a, L):
 L.append(a)
 return L
 
L = []
print(f(1, L))
print(f(2, L))
L = []
print(f(3, L))
print(f(4, L))

The output is

[1] 
[1, 2] 
[3] 
[3, 4] 

Answer 3

You're getting a different value because of L=[] default value and it is mutable. The tutorial link you provided there clearly stated that

• Default value is evaluated only once
• the default is a mutable object such as a list, dictionary, or instances of most classes

So, When you pass another L=[] in your f() function then L=[] is initiating for a new one and return a single result. But after calling f(4) it's getting to default function and return update result

 def f(a, L=[]):
 L.append(a)
 return L
 
 print(f(1))
 print(f(2))
 print(f(3, L=[]))
 print(f(4))

• Gui: Thanks, but I still do not understand why it returns [3] and then [1, 2, 4]. Could you be more didactic please?
• mhhabib: @Gui well assume you have to make sum the value of a if it is 5 else print the value that is not 5. Now compare with f(a, L=[]) where L=[] default. If the condition meets then it will append to L=[] list or print the value that comes from the user end.

Answer 4

Looks like that you are creating a new list object in memory by resetting a new value object of L at the third time it runs, so it stores the the a variable value in the new list object instead of appending to the old one that has been in memory already, but in step four and since L=[] has hasn't been changed this time it still points to the same old list object in memory, so step 4 a variable value gets stored there as well as step one and two.

here is a simple example: if we modified the code a little bit to debug what's happening here, you will see that the L variable get's assigned to a new list object in memory at step 3.

def f(a, L=[]):
 L.append(a)
 print(f'Your function list id: {id(L)}, now gets assigned a value of: {a}')
 return L
print(f(1))
print(f(2))
print(f(3, L=[]))
print(f(4))