Python: Prevent argument array being globally modified

Question

I have a problem with python here. If I pass an array through a recursive function that adds something to the array each time it is called, the array is modified in each instance

Code:

def test(n,myList):
    if n>0:
        myList.append("Test")
        print ( "BEFORE CALL Instance ", n, myList )
        test(n-1,myList)
        print ( "AFTER CALL Instance ", n, myList )
    else:
        return

Execution via test(5,[])

Results:

BEFORE CALL Instance  5 ['Test']
BEFORE CALL Instance  4 ['Test', 'Test']
BEFORE CALL Instance  3 ['Test', 'Test', 'Test']
BEFORE CALL Instance  2 ['Test', 'Test', 'Test', 'Test']
BEFORE CALL Instance  1 ['Test', 'Test', 'Test', 'Test', 'Test']
AFTER CALL Instance  1 ['Test', 'Test', 'Test', 'Test', 'Test']
AFTER CALL Instance  2 ['Test', 'Test', 'Test', 'Test', 'Test']
AFTER CALL Instance  3 ['Test', 'Test', 'Test', 'Test', 'Test']
AFTER CALL Instance  4 ['Test', 'Test', 'Test', 'Test', 'Test']
AFTER CALL Instance  5 ['Test', 'Test', 'Test', 'Test', 'Test']

Actual problem:

Each child function is modifying the array in all parent functions. How can I prevent this?

Copying the list into a new one and modifying it results in the same output as above

myListNew=myList
myListNew.append("Test")

jcdyer · Accepted Answer · 2010-01-26 18:50:08Z

5

Your attempted solution does not actually copy the array. It assigns a new name to the same array. If you want to create a new array, try this:

my_new_list = my_list[:]

That creates a list containing a slice of the old list, where the slice starts at the beginning, and ends at the end. In other words, a perfect copy of the list.

answered Jan 26, 2010 at 18:50

jcdyer

19.3k6 gold badges45 silver badges49 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

lamas Over a year ago

Is there some workaround on doing this on arrays (dictionaries) too? it says "unhashable type".

jcdyer Over a year ago

Check out the copy module in the library: another_d = copy.copy(d). Or try another_d = dict(d).

jcdyer Over a year ago

@sberry2A That works perfectly. My preference would be to do it inside test, to follow the general principle of not changing the data your caller provides. Then you don't have to worry about copying whenever you call test(). IOW, let them call test(n-1, myList), and then do newlist = mylist[:] right away inside, and do all work on newlist.

skool · Accepted Answer · 2010-12-30 01:41:48Z

1

You can use copy.deepcopy() to return a non-linked list. This also works for nested lists and dicts.

answered Dec 30, 2010 at 1:41

skool

111 bronze badge

Collectives™ on Stack Overflow

Python: Prevent argument array being globally modified

2 Answers 2

3 Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

3 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related