How do I stop Python modifying multiple variables?

Question

When I try to modify a value in a list, it is also modifying a separate value in another list. I don't believe this would be a bug, but I would like to know how to treat these variables independently.

I have tried this for single variables instead of lists, and this problem does not occur. I could set the lists separately, but that seems unnecessary for when there are a large number of lists.

numbers = [1,2,3,4,5]
list_A = numbers
list_B = numbers

print("list A:",list_A)
print("list B:",list_B)
list_A[2] = 10
print("list A:",list_A)
print("list B:",list_B)

I would expect an output of:
list_A:[1, 2, 3, 4, 5]
list_B:[1, 2, 3, 4, 5]
list_A: [1, 2, 10, 4, 5]
list_B: [1, 2, 3, 4, 5]

but instead get this:
list_A:[1, 2, 3, 4, 5]
list_B: [1, 2, 3, 4, 5]
list_A: [1, 2, 10, 4, 5]
list_B: [1, 2, 10, 4, 5]

where both lists have been modified

stackoverflow.com/questions/2612802/how-to-clone-or-copy-a-list — Jainil Patel
– Jainil Patel, Commented Aug 4, 2019 at 12:26

Jainil Patel · Accepted Answer · 2019-08-04 12:21:50Z

3

numbers = [1,2,3,4,5]
list_A = numbers.copy()
list_B = numbers.copy()

print("list A:",list_A)
print("list B:",list_B)
list_A[2] = 10
print("list A:",list_A)
print("list B:",list_B)

this is python so you have to do this.

for further information see How to clone or copy a list?

answered Aug 4, 2019 at 12:21

Jainil Patel

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1 Comment

Jainil Patel Over a year ago

stackoverflow.com/questions/2612802/how-to-clone-or-copy-a-list

U13-Forward · Accepted Answer · 2019-08-04 12:25:22Z

2

As you do here:

list_A = numbers
list_B = numbers

You're copying the whole object, however it will change all the lists, so you would need:

list_A = numbers.copy()
list_B = numbers.copy()

Or:

list_A = numbers[:]
list_B = numbers[:]

Or:

list_A = list(numbers)
list_B = list(numbers)

answered Aug 4, 2019 at 12:25

U13-Forward

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1 Comment

Ben Morrison Over a year ago

I must have misclicked earlier, sorry. Thanks for the help :)

clubby789 · Accepted Answer · 2019-08-04 12:24:24Z

1

Creating a list with newList = oldList means the two lists are linked, as they're both references to the same list. Use newList = list(oldList) to create a new list.

answered Aug 4, 2019 at 12:24

clubby789

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Comments

Maroder · Accepted Answer · 2019-08-04 12:26:37Z

1

The reason you are seeing this is because Python is assigning the reference and not the value for structures by default. To achieve what you want you need to use a way of copying like shown in the reply above.

You will see this if you try to print the variables

answered Aug 4, 2019 at 12:26

Maroder

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Alireza HI · Accepted Answer · 2019-08-04 12:37:03Z

0

You are accessing the array using the reference. You have multiple options to do that. The simplest one would be make the array again by iterating:

numbers = [1,2,3,4,5]
list_A = [i for i in numbers]
list_B = [i for i in numbers]

With this you are creating array with different reference.

Also you can read this document: https://docs.python.org/2/library/copy.html

You can try these too:

# shallow copy of array, means that it handles the simple copy but not the nested ones(if you have a `list` in `list`)
list_A = copy.copy(numbers)
list_B = copy.copy(numbers)

# deep copy of array, so handles nested ones too
list_A = copy.deepcopy(numbers)
list_B = copy.deepcopy(numbers)

In your situation both of them work

edited Aug 4, 2019 at 12:37

answered Aug 4, 2019 at 12:29

Alireza HI

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