Python creating objects in a loop append items to same list instead of a new list

Question

I have a code that creates an object inside a loop. The object has a list that contains different type of object.

class A:
    a = ''
    def __init__(self):
        pass

class B:
    def __init__(self):
        pass
    list_of_A = list()

for i in range(3):
    _b = B()
    for j in range(2):
        _a = A()
        _b.list_of_A.append(_a)
    print(len(_b.list_of_A))

The output is: 2 4 6

What I expected was: 2 2 2

I tried deleting _b at the end of the inner loop. But didn't work. How should I make sure the loop creates a new object of B.

Move list_of_A = list() into __init__ and prepend self., i.e. def __init__(self): self.list_of_A = list(). BTW, list() can be []. — iz_
– iz_, Commented Dec 1, 2019 at 5:30

Austin · Accepted Answer · 2019-12-01 05:31:04Z

1

One way is to make list_of_A an instance variable, so it creates new list everytime you create an object:

class B:
    def __init__(self):
        self.list_of_A = list()   # or = []

When you have the following arrangement:

class B:
    def __init__(self):
        pass
    list_of_A = list()

list_of_A is shared by all objects of that class (because this is a class variable), and so you are appending to the same list object.

answered Dec 1, 2019 at 5:31

Austin

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Comments

Ajay Dabas · Accepted Answer · 2019-12-01 05:34:17Z

1

Do this:

class B:
    def __init__(self):
        self.list_of_A = list()

Reason: Class Variable(or Static Variable) vs Instance Variable

In your code, list_of_A is a class variable(or static variable) so it is same for all the objects(or instances) of class B. In the code suggested above, list_of_A is an instance variable so it is different and unique for each object(or instance) of class B.

answered Dec 1, 2019 at 5:34

Ajay Dabas

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