Manipulating default argument value in python

Question

Consider the code below. It involves application of a very interesting python concept of using default argument during recursion[reversing a list] A = [1,2,3,4]

def recur(A,Q=[]): 
    if len(A)==1: 
        Q.append(A[0])
        return 
    recur(A[1:],Q)
    Q.append(A[0])
    return Q
print(recur(A)) # 4 3 2 1

Issue is, when I call this function again, Q will not be empty. Is there a way I can set Q=[] when this recursive call finishes ?

NOTE: my intent is not to reverse a list, it's just for demonstration. Thanks

Does this answer your question? "Least Astonishment" and the Mutable Default Argument — ti7
– ti7, Commented Jul 21, 2022 at 5:56
Yes it does. I remember I learned about this fact from this link only, but couldn't find it again. thanks a lot! — the_he_man
– the_he_man, Commented Jul 21, 2022 at 14:07

petezurich · Accepted Answer · 2022-07-21 06:02:41Z

2

When you first run the script, you're initializing and making Q immutable, regardless of calling the function or future input. You'll need to recreate Q as a new list every time within the function.

Try this:

def recur(A,Q=None):
    if Q == None:
        Q = []
    if len(A)==1: 
        Q.append(A[0])
        return 
    recur(A[1:],Q)
    Q.append(A[0])
    return Q

print(recur([4, 3, 2, 1])) # 1 2 3 4
print(recur([1, 2, 3, 4])) # 4 3 2 1

edited Jul 21, 2022 at 6:02

petezurich

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answered Jul 21, 2022 at 5:47

HWW

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Comments

VPfB · Accepted Answer · 2022-07-21 05:59:06Z

0

Or completely without a default arg:

def _recur(A, Q):
    ...

def recur(A):
    return _recur(A, [])

answered Jul 21, 2022 at 5:59

VPfB

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